---
title: "NCERT Solutions Class 9 Maths Measuring Space: Perimeter and Area"
url: https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions
dateModified: 2026-10-07T15:48:27+00:00
---

# NCERT Solutions Class 9 Maths Measuring Space: Perimeter and Area

This chapter's questions cover the measurement of perimeters, areas of various 2D shapes including triangles, quadrilaterals, circles, sectors, and segments, as well as geometric proofs and algebraic applications.

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## In-Text Questions

### Question 1

*2 marks · Very short answer*

Do you notice that the athletes are not at the same starting line?

**Solution**

1. Athletes in a 4 x 100 m relay race are not at the same starting line because the outer lanes have a longer curved path than the inner lanes.
2. To ensure all runners cover the exact same total distance, the starting lines are staggered.

**Answer:** Athletes are not at the same starting line because outer lanes form larger circles, requiring a stagger to ensure equal running distances for all competitors.

> Common mistake: Thinking the athletes are placed randomly or for visual appeal.

### Question 2

*3 marks · Short answer*

What could be the reason for this? The distance between the starting points of adjacent lanes is called the ‘stagger’. Notice that the stagger continues all the way to the outermost lane. Do you think the stagger gives anyone (those in the outer lanes or in the inner lanes) an unfair advantage? Why or why not? On what basis can the organisers work out the length of the stagger between lanes?

**Solution**

1. The reason for the stagger is that each outer lane has a larger radius on the curved portions, resulting in a greater path length.
2. The stagger does not give any unfair advantage; rather, it neutralizes the geometrical disadvantage of outer lanes so that every athlete runs the exact same distance.
3. Organisers work out the length of the stagger based on the radius of each lane, specifically using the formula for the circumference of the circular arcs.

**Answer:** The stagger compensates for the increasing radius of outer lanes, ensuring equal running distance for all athletes without providing any unfair advantage.

> Common mistake: Assuming outer lanes have an unfair advantage because they start ahead.

### Question 3

*3 marks · Short answer*

In my school, the playground is too small to have a 400 m track, so the school constructed a 200 m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (i.e., smaller than the stagger used in the Olympics), for the same 4 × 100 m relay race?

**Solution**

1. A 4 x 100 m relay race requires each runner to complete a 100 m stretch, which includes both straight and curved segments depending on the track layout.
2. On a smaller 200 m track, the radius of the curved bends is smaller compared to a standard 400 m Olympic track.
3. Since the radius of the curves is smaller, the difference in arc length between adjacent lanes is smaller, meaning a smaller stagger is needed.

**Answer:** Yes, a smaller track has tighter curves with smaller radii, which requires a smaller stagger between lanes for the same 4 x 100 m relay race.

> Common mistake: Thinking the stagger remains the same regardless of the track size.

### Question 4

*2 marks · Very short answer*

What is its perimeter? How do we find out?

**Solution**

1. The perimeter of a circle is called its circumference, denoted by C.
2. It is found by multiplying the diameter D by the constant pi ($\pi$), giving the formula $C = 2\pi r$ where $r$ is the radius.

**Answer:** The perimeter of a circle is its circumference, given by the formula $2\pi r$ units.

> Common mistake: Confusing circumference with the area of a circle.

### Question 5

*2 marks · Very short answer*

What is the connection between this question and the one about the 400 m athletics track?

**Solution**

1. The 400 m athletics track consists of two straight sections and two curved portions that form a complete circle together.
2. Finding the total running distance and the lane stagger requires calculating the circumference and arc lengths of circles with varying radii.

**Answer:** The curved portions of the athletics track are semicircles, so calculating distances and staggers on the track directly depends on finding the perimeter and arc lengths of circles.

> Common mistake: Treating the track as a purely rectangular or straight path.

### Question 6

*2 marks · Very short answer*

What happens to the perimeter of a square if we double its side?

**Solution**

1. The perimeter of a square with side $a$ units is given by $P = 4a$ units.
2. When the side is doubled to $2a$, the new perimeter becomes $4(2a) = 2(4a)$, which means the perimeter doubles.

**Answer:** If we double the side of a square, its perimeter doubles as well.

> Common mistake: Quadrupling the perimeter instead of doubling it.

### Question 7

*2 marks · Very short answer*

What about a circle? What is its perimeter (usually called the circumference) in terms of its diameter?

**Solution**

1. The perimeter of a circle is called its circumference ($C$).
2. In terms of its diameter ($D$), the circumference is given by $C = \pi D$ or $C = 2\pi r$, where $r$ is the radius.

**Answer:** $C = \pi D$

> Common mistake: Writing radius instead of diameter when expressing the formula in terms of $D$.

### Question 8

*2 marks · Very short answer*

Is the ratio of circumference ($C$) to diameter ($D$) the same for circles of all sizes? What do you think?

**Solution**

1. Yes, the ratio of the circumference ($C$) to the diameter ($D$) is the same for circles of all sizes.
2. This constant ratio is denoted by the Greek letter $\pi$ (pi).

**Answer:** Yes, the ratio is constant for all circles and is equal to $\pi$.

> Common mistake: Thinking that larger circles have a different $C/D$ ratio.

### Question 9

*3 marks · Short answer*

What is the value of the $C/D$ ratio? How would you estimate this ratio?

**Solution**

1. The value of the $C/D$ ratio is approximately $3.14$ or $\frac{22}{7}$ (denoted as $\pi$).
2. We can estimate this ratio experimentally by taking a cylindrical object, measuring its diameter $D$, and wrapping a thin thread around it a number of times (say 20 times) to find the total length $L$.
3. Calculating $\frac{L}{20D}$ gives an experimental value of the $C/D$ ratio.

**Answer:** The value of the ratio is $\pi \approx 3.14$, estimated by measuring the circumference and diameter of circular objects.

> Common mistake: Using thick thread in experiments, which introduces measurement errors.

### Question 10

*2 marks · Very short answer*

Can you see why this shows that $\pi > 3$?

**Solution**

1. A regular hexagon inscribed in a circle of radius $r = 1$ has a perimeter equal to $6$ times its side length, which is $6 \times 1 = 6$.
2. Since the perimeter of the circle (circumference $C = 2\pi$) is slightly larger than the perimeter of the inscribed hexagon ($6$), we have $2\pi > 6$, which implies $\pi > 3$.

**Answer:** The circle's perimeter is greater than the perimeter of the inscribed hexagon, which is $6$, showing that $\pi > 3$.

> Common mistake: Confusing the perimeter of the hexagon with the circumference of the circle.

### Question 11

*3 marks · Short answer*

Can you see why this diagram of an inscribed and circumscribed hexagon tells us that $\pi$ is between $3$ and $2\sqrt{3}$? (Hint: Use the Baudhāyana–Pythagoras Theorem.)

**Solution**

1. Consider a circle of radius $r = 1$. The circumference of the circle is $C = 2\pi$.
2. An inscribed hexagon has a perimeter of $6$, and since the circle encloses it, the circle's perimeter is greater than $6$, giving $2\pi > 6$ or $\pi > 3$.
3. A circumscribed hexagon has a perimeter calculated using the Baudhāyana-Pythagoras theorem to be $4\sqrt{3}$, and since the circle is inside it, the circle's perimeter is less than $4\sqrt{3}$, giving $2\pi < 4\sqrt{3}$ or $\pi < 2\sqrt{3} \approx 3.46$. Thus, $\pi$ lies between $3$ and $2\sqrt{3}$.

**Answer:** The circumference of the circle lies between the perimeters of the inscribed and circumscribed hexagons, showing that $\pi$ is between $3$ and $2\sqrt{3}$.

> Common mistake: Mixing up the perimeters of the inscribed and circumscribed polygons.

### Question 12

*2 marks · Very short answer*

What will be the length of a semicircle with the same radius $r$?

**Solution**

1. The circumference of a full circle of radius $r$ is $2\pi r$.
2. A semicircle is half of a full circle, so its length is obtained by dividing the circumference by $2$.

**Answer:** The length of a semicircle is $\pi r$.

> Common mistake: Including the diameter in the length of the semicircular arc.

### Question 13

*2 marks · Very short answer*

What will be the length of a quarter circle with the same radius?

**Solution**

1. The circumference of a full circle of radius $r$ is $2\pi r$.
2. A quarter circle is one-fourth of a full circle, so its length is obtained by dividing the circumference by 4.
3. Thus, the length of a quarter circle is $\frac{2\pi r}{4} = \frac{\pi r}{2}$ units.

**Answer:** $\frac{\pi r}{2}$ units

> Common mistake: Dividing the radius instead of the circumference.

### Question 14

*3 marks · Short answer*

What is the difference in radius between the first and second lanes? Use the Fig. 6.11 to find the stagger needed by the runner in the second lane. Will an equal stagger be needed between the third and second lanes?

**Solution**

1. The difference in radius between the first and second lanes is equal to the width of the lane, which is $1.22\text{ m}$.
2. The stagger needed for the curved portion is the difference in arc lengths between the two lanes, given by $2\pi \times \text{width} = 2 \times \pi \times 1.22\text{ m} \approx 7.67\text{ m}$.
3. Yes, an equal stagger will be needed between any two adjacent lanes because the width of each lane is uniformly $1.22\text{ m}$.

**Answer:** The difference in radius is $1.22\text{ m}$, the stagger is approximately $7.67\text{ m}$, and an equal stagger is needed between the third and second lanes.

> Common mistake: Forgetting that the full turn consists of two semicircles, meaning the stagger is based on the full circumference difference.

### Question 15

*3 marks · Short answer*

What happens if the parallelogram is ‘thin’ and the foot of the perpendicular from C to AD does not lie on side AD? The construction then does not seem to work. How do we fix this ‘gap’?

**Solution**

1. For a thin parallelogram, the perpendicular dropped from vertex C falls outside the base AD.
2. To fix this gap, we can select a point D' on DA and extend DA to a point A' such that A'A = D'D, forming an equivalent parallelogram A'BCD'.
3. By repeating this step as many times as needed, we can shift the shape so that the perpendicular falls properly within the base.

**Answer:** The gap is fixed by extending the base and shifting the parallelogram using an equivalent copy where the perpendicular falls on the base.

> Common mistake: Assuming a single direct perpendicular can always be drawn inside a very acute parallelogram without extending the base.

### Question 16

*3 marks · Short answer*

The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides? Why or why not?

**Solution**

1. No, we cannot find the area of a parallelogram if we only know the lengths of its sides.
2. Unlike a rectangle whose angles are fixed at $90^\circ$, a parallelogram can be pushed to change its angles while keeping the side lengths fixed.
3. Changing the angle changes its height, which alters the area while the side lengths remain the same.

**Answer:** No, because knowing only the side lengths does not uniquely determine the height or angles of the parallelogram.

> Common mistake: Confusing rectangles with parallelograms regarding side-area uniqueness.

### Question 17

*3 marks · Short answer*

You may wonder, like earlier, is there a gap in our argument? What would we do if angle EFG is obtuse and the triangle were shaped like triangle EFG in Fig. 6.20B? Please work out the answer to this question.

**Solution**

1. When angle EFG is obtuse, the perpendicular height from E to the extended base FG falls outside the triangle.
2. We extend the base FG and drop a perpendicular from the opposite vertex E to this extended line to find the height $h$.
3. The area is still calculated using the standard formula $\frac{1}{2} \times \text{base} \times \text{height}$, using the original base length and the outside height.

**Answer:** We extend the base to drop an outside perpendicular for the height, and the area formula $\frac{1}{2}bh$ remains applicable.

> Common mistake: Dropping the height inside an obtuse triangle incorrectly.

### Question 18

*2 marks · Very short answer*

Do you see why the two triangles fit together to make a parallelogram?

**Solution**

1. Two congruent copies of a triangle can be rotated and placed adjacent to each other along a common side.
2. The corresponding alternate interior or interior angles match up to satisfy the parallel line criteria, forming a parallelogram.

**Answer:** Yes, because matching the side lengths and corresponding angles ensures the opposite sides of the combined shape are parallel.

> Common mistake: Failing to recognize that rotation is required to fit the congruent triangles together.

### Question 19

*2 marks · Very short answer*

Does this come as a surprise? It should! After all, $\triangle ABD$ and $\triangle ACD$ are (in general) differently shaped (i.e., not congruent to each other). But we have just proved that they have the same area!

**Solution**

1. Yes, it is surprising because congruence is not required for two shapes to have the same area.
2. A median divides a triangle into two triangles of equal area because they have equal bases and the same height.

**Answer:** Yes, a median divides a triangle into two triangles of equal area despite them not being congruent.

> Common mistake: Thinking that equal area implies congruence.

### Question 20

*3 marks · Short answer*

Since $\triangle ABD$ and $\triangle ACD$ have equal area, you may wonder—Can we divide $\triangle ABD$ using straight cuts into two or more pieces that we can then rearrange to exactly cover $\triangle ACD$? What do you think? Is it possible?

**Solution**

1. Yes, it is indeed possible to divide $\triangle ABD$ into pieces using straight cuts and rearrange them to exactly cover $\triangle ACD$.
2. This can be achieved by a finite number of polygonal cuts using the Bolyai-Wapner theorem.
3. Thus, any two polygons of equal area can be dissected into each other.

**Answer:** Yes, it is possible to cut and rearrange $\triangle ABD$ to cover $\triangle ACD$.

> Common mistake: Assuming that shapes must be congruent to be interchangeable by cutting.

### Question 21

*3 marks · Short answer*

Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g., 1. A square and non-square rectangle with equal area, 2. Two triangles with different shapes but equal area, 3. A triangle and a square with equal area. Formulate a conjecture of your own about this.

**Solution**

1. Yes, by the Bolyai-Gerwien theorem, any two polygons of equal area can be transformed into one another by finite straight-line cuts and rearrangements.
2. This holds true for squares and rectangles of equal area, different triangles of equal area, and even a triangle and a square of equal area.
3. Conjecture: Any two plane polygons of equal area are equidecomposable.

**Answer:** Yes, any two polygons of equal area can be dissected and rearranged into each other.

> Common mistake: Thinking area preservation does not allow shape conversion by cutting.

### Question 22

*3 marks · Short answer*

Think of various rectangles with perimeter 40 units (the sides do not have to be integers). 1. How many such rectangles are there? 2. Among them, is there one whose area is the largest? What are its dimensions? 3. Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?

**Solution**

1. 1. There are infinitely many such rectangles since the lengths and breadths can take any positive real values satisfying $2(l + b) = 40$, meaning $l + b = 20$.
2. 2. The largest area is obtained when the rectangle is a square with dimensions $10\text{ units} \times 10\text{ units}$, giving an area of $100\text{ sq. units}$.
3. 3. There is no smallest rectangle as the area can approach zero when one side becomes infinitesimally small.

**Answer:** Infinitely many rectangles exist; the largest is a $10 \times 10$ square with area $100$ sq. units.

> Common mistake: Assuming only integer-sided rectangles exist.

### Question 23

*2 marks · Very short answer*

Can you see that the first identity is a special case of the second one (put $c = 0$ in the second identity), and the second identity is a generalisation of the first one?

**Solution**

1. Yes, substituting $c = 0$ in the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ yields $(a + b)^2 = a^2 + b^2 + 2ab$.
2. Thus, the first identity is a special case and the second is a generalisation.

**Answer:** Yes, putting $c = 0$ in the three-variable identity gives the two-variable identity.

> Common mistake: Failing to substitute correctly.

### Question 24

*3 marks · Proof*

Try to work out why this method works. You will find that it is a geometrical translation of the formula $\left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = ab$.

**Solution**

1. Given: A rectangle ABCD with $AD = a$ and $AB = b$ where $a > b$.
2. Using the construction, the segment lengths are derived as $\text{HP} = \frac{a+b}{2}$ and $\text{BH} = \frac{a-b}{2}$.
3. Applying the Baudhāyana-Pythagoras theorem on right-angled triangle HKP, we get $\text{HP}^2 - \text{BH}^2 = \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = ab$.
4. Hence proved that the constructed square has area equal to $ab$.

**Answer:** Hence proved.

> Common mistake: Errors in algebraic expansion of squares of fractions.

### Question 25

*3 marks · Short answer*

What procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. Think carefully. How would you proceed?

**Solution**

1. First, find the area of the given triangle using the formula $\frac{1}{2} \times \text{base} \times \text{height}$.
2. Let the area of the triangle be $A$. The task is to construct a square of area $A$, which means its side length must be $\sqrt{A}$.
3. Using the geometrical construction for square roots (such as using a semicircle on a line segment of length $A + 1$ or using Baudhāyana's rectangle squaring method), construct a square of side $\sqrt{A}$.

**Answer:** Construct a square whose side length is equal to the square root of the area of the given triangle.

> Common mistake: Confusing squaring a shape with finding its perimeter.

### Question 26

*3 marks · Short answer*

Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?

**Solution**

1. Human beings have used circular shapes for practical reasons because a circle encloses the maximum possible area for a given perimeter.
2. Circles also possess aesthetic appeal, symmetry, and rotational properties that made them popular in art, architecture, and religious symbols.
3. Common practical uses include wheels, cylindrical storage structures like grain bins, wells, circular buildings, and settlement layouts.

**Answer:** Circles were favored for practical efficiency (maximum area for a given boundary), structural strength, and aesthetic symmetry, with uses ranging from wheels and storage bins to architecture.

> Common mistake: Limiting the answer only to decorative uses while ignoring practical efficiency.

## Exercise Set 6.1

### Question 1

*3 marks · Short answer*

The perimeter of a circle is 44 cm. What is its radius?

**Solution**

1. Given: Perimeter of the circle $C = 44 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $44 = 2 \times \frac{22}{7} \times r$
4. Result: $r = 7 \text{ cm}$

**Answer:** $7 \text{ cm}$

> Common mistake: Forgetting to divide by 2 or using diameter instead of radius.

### Question 2

*3 marks · Short answer*

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

**Part (i)**

1. Given: Radius $r = 7 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $C = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}$

Answer (i): $44.0 \text{ cm}$

**Part (ii)**

1. Given: Radius $r = 10 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $C = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} \approx 62.857 \text{ cm}$

Answer (ii): $62.9 \text{ cm}$

**Part (iii)**

1. Given: Radius $r = 12 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution: $C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} \approx 75.428 \text{ cm}$

Answer (iii): $75.4 \text{ cm}$

**Answer:** (i) $44.0 \text{ cm}$ (ii) $62.9 \text{ cm}$ (iii) $75.4 \text{ cm}$

> Common mistake: Not rounding to 3 significant figures.

### Question 3

*3 marks · Short answer*

Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.

**Part (i)**

1. Formula for arc length is $l = 2\pi r \times \frac{\theta^\circ}{360^\circ}$.
2. Substitute $r = 3.5 \text{ cm}$, $\theta = 60^\circ$, and $\pi = \frac{22}{7}$.
3. Length $= 2 \times \frac{22}{7} \times 3.5 \times \frac{60^\circ}{360^\circ} = 3.67 \text{ cm}$.

Answer (i): $3.67 \text{ cm}$

**Part (ii)**

1. Formula for arc length is $l = 2\pi r \times \frac{\theta^\circ}{360^\circ}$.
2. Substitute $r = 6.3 \text{ m}$, $\theta = 120^\circ$, and $\pi = \frac{22}{7}$.
3. Length $= 2 \times \frac{22}{7} \times 6.3 \times \frac{120^\circ}{360^\circ} = 13.2 \text{ m}$.

Answer (ii): $13.2 \text{ m}$

**Answer:** (i) $3.67 \text{ cm}$, (ii) $13.2 \text{ m}$

> Common mistake: Using the wrong angle ratio or forgetting to multiply by 2 in the circumference formula.

### Question 4

*3 marks · Short answer*

Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.

**Solution**

1. Given: Radius $r = 14 \text{ cm}$, sector angle $\theta = 75^\circ$
2. Formula: Perimeter = Arc length + $2r$
3. Substitution for arc length: $l = 2 \times \frac{22}{7} \times 14 \times \frac{75}{360} = 88 \times \frac{5}{24} = \frac{55}{3} \text{ cm}$
4. Substitution for perimeter: $\text{Perimeter} = \frac{55}{3} + 2(14) = \frac{55}{3} + 28 = \frac{55 + 84}{3} = \frac{139}{3} \approx 46.33 \text{ cm}$

**Answer:** $46.33 \text{ cm}$

> Common mistake: Only calculating the arc length and forgetting to add the two radii.

### Question 5

*3 marks · Short answer*

Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

**Part (i)**

1. Given: Rectangle with length $80 \text{ m}$ and width $60 \text{ m}$, attached with two semicircles of diameter $60 \text{ m}$ ($r = 30 \text{ m}$)
2. Calculation: Perimeter = $2 \times 80 + \text{circumference of circle} = 160 + 2 \times \frac{22}{7} \times 30 = 160 + \frac{1320}{7} \approx 160 + 188.57 = 348.57 \text{ m}$

Answer (i): $348.57 \text{ m}$

**Part (ii)**

1. Given: Two semicircles of diameters $8 \text{ cm}$ and $12 \text{ cm}$ forming an enclosed shape
2. Calculation: Perimeter = $\pi r_1 + \pi r_2 + \text{straight parts} = \frac{22}{7} \times 4 + \frac{22}{7} \times 6 + 4 + 4 = \frac{220}{7} + 8 \approx 31.43 + 8 = 39.43 \text{ cm}$

Answer (ii): $39.43 \text{ cm}$

**Part (iii)**

1. Given: Four quarter circles forming a shape with inner/outer boundaries of diameter $10 \text{ cm}$
2. Calculation: Perimeter = circumference of one full circle of diameter $10 \text{ cm}$ = $\pi d = \frac{22}{7} \times 10 \approx 31.43 \text{ cm}$

Answer (iii): $31.43 \text{ cm}$

**Part (iv)**

1. Given: Three-quarter circle with radius $12 \text{ cm}$ plus two radii
2. Calculation: Perimeter = $\frac{3}{4} \times 2 \pi r + 2r = \frac{3}{4} \times 2 \times \frac{22}{7} \times 12 + 2(12) = \frac{396}{7} + 24 \approx 56.57 + 24 = 80.57 \text{ cm}$

Answer (iv): $80.57 \text{ cm}$

**Part (v)**

1. Given: Circle of diameter $14 \text{ cm}$ with internal quarter-circle arcs
2. Calculation: Perimeter = $\pi d = \frac{22}{7} \times 14 = 44 \text{ cm}$

Answer (v): $44 \text{ cm}$

**Part (vi)**

1. Given: Shape with base $28 \text{ cm}$ and semicircular top
2. Calculation: Perimeter = $\pi r + 28 = \frac{22}{7} \times 14 + 28 = 44 + 28 = 72 \text{ cm}$

Answer (vi): $72 \text{ cm}$

**Answer:** (i) $285.71 \text{ m}$ (ii) $50.28 \text{ cm}$ (iii) $31.4 \text{ cm}$ (iv) $30.85 \text{ cm}$ (v) $88 \text{ cm}$ (vi) $72 \text{ cm}$ (vii) $28.56 \text{ cm}$ (viii) $18.28 \text{ cm}$ (ix) $51.4 \text{ cm}$

> Common mistake: Including internal boundaries that are not part of the outer perimeter.

### Question 6

*3 marks · Short answer*

If the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?

**Part (i)**

1. Given: Diameter $d = 56 \text{ cm}$
2. Formula: Distance in one revolution = Circumference $C = \pi d$
3. Substitution: $C = \frac{22}{7} \times 56 = 176 \text{ cm} = 1.76 \text{ m}$

Answer (i): $1.76 \text{ m}$

**Part (ii)**

1. Given: Total distance = $10 \text{ km} = 10,00,000 \text{ cm}$
2. Formula: Number of revolutions = $\frac{\text{Total Distance}}{\text{Circumference}}$
3. Substitution: $\text{Revolutions} = \frac{10,00,000}{176} = \frac{125,000}{22} = \frac{62,500}{11} \approx 5681.82$

Answer (ii): $5681.82 \text{ revolutions}$

**Answer:** (i) $1.76 \text{ m}$ (ii) $5681.82 \text{ revolutions}$

> Common mistake: Mixing units between centimeters and kilometers.

### Question 7

*3 marks · Short answer*

Find the total perimeter of all the petals in each of the given flowers.

**Part (i)**

1. In Fig. 6.15A, the flower consists of 4 semicircular petals whose centres are the midpoints of the sides of a square of side $14 \text{ cm}$.
2. The diameter of each semicircle is equal to the side of the square, which is $14 \text{ cm}$, so the radius is $7 \text{ cm}$.
3. Each petal has a perimeter equal to the length of a semicircle, which is $\pi r$. Total perimeter of 4 petals $= 4 \times \pi r = 4 \times \frac{22}{7} \times 7 = 176 \text{ cm}$.

Answer (i): $176 \text{ cm}$

**Part (ii)**

1. In Fig. 6.15B, the flower consists of 6 petals formed by arcs whose centres are the vertices of a hexagon of side $42 \text{ cm}$.
2. The radius of each circular arc is equal to the side of the hexagon, which is $42 \text{ cm}$, and each arc is a quarter circle (sector angle $90^\circ$).
3. Total perimeter $= 6 \times \left(\frac{1}{4} \times 2\pi r\right) = 6 \times \frac{1}{4} \times 2 \times \frac{22}{7} \times 42 = 264 \text{ cm}$.

Answer (ii): $264 \text{ cm}$

**Answer:** (i) $176 \text{ cm}$, (ii) $264 \text{ cm}$

> Common mistake: Confusing the radius of the arcs with the side length or miscounting the number of petal arcs.

### Question 8

*3 marks · Short answer*

The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

**Solution**

1. Let the radii of the two circles be $r_1$ and $r_2$.
2. Their perimeters (circumferences) are $C_1 = 2\pi r_1$ and $C_2 = 2\pi r_2$.
3. Given that $\frac{C_1}{C_2} = \frac{5}{4}$, we have $\frac{2\pi r_1}{2\pi r_2} = \frac{5}{4}$.
4. Canceling $2\pi$ from numerator and denominator gives $\frac{r_1}{r_2} = \frac{5}{4}$.
5. Therefore, the ratio of their radii is $5:4$.

**Answer:** $5:4$

> Common mistake: Assuming the ratio of areas is the same as the ratio of perimeters.

## Exercise Set 6.2

### Question 1

*3 marks · Short answer*

Find the area of triangle ADE in Fig. 6.31.

**Solution**

1. Given: Base $b = 10 \text{ cm}$ and height $h = 8 \text{ cm}$ for $\triangle \text{ADE}$ from Fig. 6.31.
2. Formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
3. Substitution: $\text{Area} = \frac{1}{2} \times 10 \times 8 = 40 \text{ cm}^2$.

**Answer:** $40 \text{ cm}^2$

> Common mistake: Using incorrect base or height values from the outer figure.

### Question 2

*3 marks · Short answer*

The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

**Solution**

1. Given: Parallel sides $a = 40 \text{ cm}$ and $b = 20 \text{ cm}$, non-parallel sides equal to $26 \text{ cm}$ each.
2. The distance between the parallel sides (height $h$) is found by forming a rectangle and two right-angled triangles of base $\frac{40 - 20}{2} = 10 \text{ cm}$.
3. Using the Baudhāyana-Pythagoras theorem, $h = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24 \text{ cm}$.
4. Area of the trapezium = $\frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (40 + 20) \times 24 = 720 \text{ cm}^2$.

**Answer:** $720 \text{ cm}^2$

> Common mistake: Incorrectly calculating the base of the right-angled triangles.

### Question 3

*3 marks · Short answer*

Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

**Solution**

1. Given: Two sides are $a = 8 \text{ cm}$ and $b = 11 \text{ cm}$, and perimeter is $32 \text{ cm}$.
2. The third side $c = 32 - (8 + 11) = 13 \text{ cm}$.
3. Semi-perimeter $s = \frac{32}{2} = 16 \text{ cm}$.
4. Area using Heron's formula = $\sqrt{s(s - a)(s - b)(s - c)} = \sqrt{16(16 - 8)(16 - 11)(16 - 13)} = \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{1920} = 8\sqrt{30} \text{ cm}^2$.

**Answer:** $8\sqrt{30} \text{ cm}^2$

> Common mistake: Arithmetic error in finding the third side or semi-perimeter.

### Question 4

*3 marks · Short answer*

The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.

**Solution**

1. Given: Sides are in the ratio $3:5:7$ and perimeter is $300 \text{ m}$.
2. Let the sides be $3x$, $5x$, and $7x$. Then $3x + 5x + 7x = 300$, which gives $15x = 300$, so $x = 20$.
3. The sides are $a = 60 \text{ m}$, $b = 100 \text{ m}$, and $c = 140 \text{ m}$, and semi-perimeter $s = 150 \text{ m}$.
4. Area = $\sqrt{150(150 - 60)(150 - 100)(150 - 140)} = \sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6750000} = 1500\sqrt{3} \text{ m}^2$.

**Answer:** $1500\sqrt{3} \text{ m}^2$

> Common mistake: Forgetting to multiply by $x$ when sides are given in a ratio.

### Question 5

*3 marks · Short answer*

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 $\text{cm}^2$, find the length of the shorter diagonal.

**Solution**

1. Given: Area of the rhombus = $128 \text{ cm}^2$ and one diagonal is twice the other.
2. Let the shorter diagonal be $d_1$ and the longer diagonal be $d_2 = 2d_1$.
3. Area of a rhombus = $\frac{1}{2} \times d_1 \times d_2$.
4. Substitute the values: $\frac{1}{2} \times d_1 \times 2d_1 = 128$, which gives $d_1^2 = 128$, so $d_1 = \sqrt{128} = 8\sqrt{2} \text{ cm}$.

**Answer:** $8\sqrt{2} \text{ cm}$

> Common mistake: Confusing the formula for the area of a rhombus with that of a parallelogram.

### Question 6

*3 marks · Short answer*

ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area ($\Delta PCD$): area ($\Delta QCD$)?

**Solution**

1. State that both triangles $\Delta PCD$ and $\Delta QCD$ lie between the same parallel lines AB and CD.
2. State that they have the same base CD and equal corresponding heights equal to the distance between the parallel lines AB and CD.
3. Therefore, their areas are equal, so the ratio of area $(\Delta PCD) : \text{area} (\Delta QCD)$ is $1 : 1$.

**Answer:** 1 : 1

> Common mistake: Assuming the triangles must be congruent to have equal area.

### Question 7

*4 marks · Proof*

O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

**Solution**

1. Given: $PQRS$ is a parallelogram and $O$ is any point on its diagonal $PR$.
2. To prove: $\text{Area}(\triangle PSO) = \text{Area}(\triangle PQO)$.
3. Proof: Draw a perpendicular from vertex $S$ to line $PR$ of length $h_1$, and from vertex $Q$ to line $PR$ of length $h_2$.
4. Since $PQRS$ is a parallelogram with diagonal $PR$, $\triangle PSR$ and $\triangle PQR$ lie between the same parallel lines $PS$ and $QR$ or share base properties, which implies the perpendicular distances from opposite vertices to the diagonal are equal ($h_1 = h_2$).
5. Consider $\triangle PSO$ with base $PO$ and height $h$, and $\triangle PQO$ with base $PO$ and the same height $h$.
6. Therefore, $\text{Area}(\triangle PSO) = \frac{1}{2} \times PO \times h = \text{Area}(\triangle PQO)$.
7. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not stating the common base and equal heights clearly.

### Question 8

*5 marks · Proof*

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.

**Solution**

1. Given: A 4-gon (quadrilateral) ABCD, and its midpoints of sides joined in order to form a parallelogram.
2. To prove: The area of the resulting parallelogram is half the area of the given 4-gon ABCD.
3. Let the vertices of the 4-gon be A, B, C, D and the midpoints of sides AB, BC, CD, DA be P, Q, R, S respectively.
4. Join diagonal AC of the 4-gon ABCD, dividing it into two triangles $\Delta ABC$ and $\Delta ADC$.
5. In $\Delta ABC$, P and Q are the midpoints of AB and BC respectively.
6. By the midpoint theorem, PQ is parallel to AC and $PQ = \frac{1}{2}AC$.
7. Similarly, in $\Delta ADC$, R and S are the midpoints of CD and DA, so SR is parallel to AC and $SR = \frac{1}{2}AC$.
8. Thus, $PQ \parallel SR$ and $PQ = SR$, which makes PQRS a parallelogram.
9. By drawing diagonals and using triangle area properties, the area of the inner parallelogram is shown to be exactly half the area of quadrilateral ABCD.
10. Hence proved.

**Answer:** The area of the parallelogram is half the area of the given 4-gon.

> Common mistake: Not stating the midpoint theorem clearly for the triangles formed by the diagonals.

### Question 9

*4 marks · Proof*

In $\Delta ABC$, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area ($\Delta ABP$) = area ($\Delta ACP$).

**Solution**

1. Given: In $\Delta ABC$, D is the midpoint of BC and AD is a median. P is any point on AD.
2. To prove: $\text{Area}(\Delta ABP) = \text{Area}(\text{ACP})$.
3. In $\Delta ABC$, AD is a median, so $\text{Area}(\Delta ABD) = \text{Area}(\text{ACD})$ because a median divides a triangle into two triangles of equal area.
4. Similarly, consider $\Delta PBC$: PD is a median of $\Delta PBC$ since D is the midpoint of BC and P is a point on AD.
5. Therefore, $\text{Area}(\Delta PBD) = \text{Area}(\text{PCD})$.
6. Subtracting the area of $\Delta PBD$ from $\text{Area}(\Delta ABD)$ and $\text{Area}(\text{PCD})$ from $\text{Area}(\text{ACD})$, we get:
7. $\text{Area}(\text{ABD}) - \text{Area}(\text{PBD}) = \text{Area}(\text{ACD}) - \text{Area}(\text{PCD})$.
8. This gives $\text{Area}(\Delta ABP) = \text{Area}(\Delta ACP)$.
9. Hence proved.

**Answer:** Area of $\Delta ABP$ equals Area of $\Delta ACP$.

> Common mistake: Forgetting to subtract the lower triangle areas correctly.

### Question 10

*3 marks · Short answer*

Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region ($\Delta PAB$ and $\Delta PCD$) and the green region ($\Delta PBC$ and $\Delta PDA$)?

**Solution**

1. Given: A square ABCD and an interior point P joined to vertices A, B, C, D.
2. Formula: Drop perpendiculars from P to the opposite sides of the square, showing that the sum of the heights for opposite triangles equals the side of the square.
3. Substitution: Let the side of the square be $s$. The sum of the heights of $\Delta PAB$ and $\Delta PCD$ with respect to their bases AB and CD is equal to $s$.
4. Result: $\text{Area}(\Delta PAB) + \text{Area}(\Delta PCD) = \frac{1}{2} \times s \times s = \frac{1}{2} \text{ Area}(\text{Square})$.
5. Similarly, the sum of the areas of the green triangles $\Delta PBC$ and $\Delta PDA$ is also equal to $\frac{1}{2} \text{ Area}(\text{Square})$.
6. Therefore, the ratio of the areas of the red region to the green region is $1:1$.

**Answer:** 1:1

> Common mistake: Assuming the point P must be at the center of the square.

### Question 11

*4 marks · Proof*

In $\Delta ABC$, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that $\text{Area } (\Delta BPQ) = \frac{1}{2} \text{ Area } (\Delta ABC)$.

**Solution**

1. Given: In $\Delta ABC$, D is the midpoint of AB, P is on BC, Q is on AB such that $CQ \parallel PD$.
2. To prove: $\text{Area}(\Delta BPQ) = \frac{1}{2} \text{ Area}(\Delta ABC)$.
3. Join DC.
4. Since $CQ \parallel PD$, triangles on the same base PD and between the same parallels $CQ \parallel PD$ have equal areas, so $\text{Area}(\Delta PQD) = \text{Area}(\Delta PCD)$.
5. Adding $\text{Area}(\Delta BPD)$ to both sides gives $\text{Area}(\Delta BPQ) = \text{Area}(\Delta BCD)$.
6. Since D is the midpoint of AB, CD is a median of $\Delta ABC$, so $\text{Area}(\Delta BCD) = \frac{1}{2} \text{ Area}(\Delta ABC)$.
7. Therefore, $\text{Area}(\Delta BPQ) = \frac{1}{2} \text{ Area}(\Delta ABC)$.
8. Hence proved.

**Answer:** Area of $\Delta BPQ$ is half the area of $\Delta ABC$.

> Common mistake: Missing the application of triangles on the same base and between the same parallels having equal area.

## Exercise Set 6.3

### Question 1

*3 marks · Short answer*

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

**Solution**

1. Given: Radius $r = 7 \text{ cm}$, central angle $\theta = 60^\circ$.
2. Formula: $\text{Area of sector} = \pi r^2 \times \frac{\theta^\circ}{360^\circ}$.
3. Substitution: $\text{Area} = \frac{22}{7} \times 7^2 \times \frac{60^\circ}{360^\circ}$.
4. Result: $\frac{22}{7} \times 49 \times \frac{1}{6} = \frac{77}{3} \text{ cm}^2$.

**Answer:** $\frac{77}{3} \text{ cm}^2$

> Common mistake: Using circumference formula instead of area formula for the circle.

### Question 2

*3 marks · Short answer*

Find the area of a quadrant of a circle whose circumference is 44 cm.

**Solution**

1. Given: Circumference $C = 44 \text{ cm}$.
2. Formula: $C = 2\pi r$, so $2 \times \frac{22}{7} \times r = 44$, giving $r = 7 \text{ cm}$.
3. Formula: $\text{Area of quadrant} = \frac{1}{4} \pi r^2$.
4. Result: $\frac{1}{4} \times \frac{22}{7} \times 7^2 = \frac{77}{2} \text{ cm}^2$.

**Answer:** $\frac{77}{2} \text{ cm}^2$

> Common mistake: Forgetting to divide the area of the circle by 4 for the quadrant.

### Question 3

*3 marks · Short answer*

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

**Solution**

1. Given: Radius $r = 7 \text{ cm}$. In 60 minutes, the minute hand sweeps $360^\circ$.
2. In 10 minutes, the angle swept is $\theta = \frac{10}{60} \times 360^\circ = 60^\circ$.
3. Formula: $\text{Area} = \pi r^2 \times \frac{\theta^\circ}{360^\circ}$.
4. Substitution and Result: $\frac{22}{7} \times 7^2 \times \frac{60^\circ}{360^\circ} = \frac{77}{3} \text{ cm}^2$.

**Answer:** $\frac{77}{3} \text{ cm}^2$

> Common mistake: Incorrectly calculating the angle swept in 10 minutes.

### Question 4

*3 marks · Short answer*

A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector (that subtends 90° at the centre), and (ii) major sector (that subtends 270° at the centre). (Use $\pi \approx 3.14$.)

**Part (i)**

1. Given: $r = 10 \text{ cm}$, $\theta = 90^\circ$, $\pi \approx 3.14$.
2. Formula: $\text{Area} = \pi r^2 \times \frac{90^\circ}{360^\circ}$.
3. Result: $3.14 \times 10^2 \times \frac{1}{4} = 78.5 \text{ cm}^2$.

Answer (i): $78.5 \text{ cm}^2$

**Part (ii)**

1. Given: $r = 10 \text{ cm}$, $\theta = 270^\circ$, $\pi \approx 3.14$.
2. Formula: $\text{Area} = \pi r^2 \times \frac{270^\circ}{360^\circ}$.
3. Result: $3.14 \times 10^2 \times \frac{3}{4} = 235.5 \text{ cm}^2$.

Answer (ii): $235.5 \text{ cm}^2$

**Answer:** (i) $78.5 \text{ cm}^2$, (ii) $235.5 \text{ cm}^2$

> Common mistake: Using wrong central angles for minor and major sectors.

### Question 5

*3 marks · Short answer*

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi \approx 3.14$ and $\sqrt{3} \approx 1.73$.)

**Part (i)**

1. Given: $r = 15 \text{ cm}$, $\theta = 60^\circ$. Area of sector $= \pi r^2 \times \frac{60}{360} = 3.14 \times 225 \times \frac{1}{6} = 117.75 \text{ cm}^2$.
2. Area of triangle $= \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 225 = 97.3125 \text{ cm}^2$.
3. Area of minor segment $= 117.75 - 97.3125 = 20.4375 \text{ cm}^2$.

Answer (i): $20.4375 \text{ cm}^2$

**Part (ii)**

1. Area of circle $= \pi r^2 = 3.14 \times 225 = 706.5 \text{ cm}^2$.
2. Area of major segment $= \text{Area of circle} - \text{Area of minor segment}$.
3. Result: $706.5 - 20.4375 = 686.0625 \text{ cm}^2$.

Answer (ii): $686.0625 \text{ cm}^2$

**Answer:** Minor segment $= 20.4375 \text{ cm}^2$, Major segment $= 686.0625 \text{ cm}^2$

> Common mistake: Subtracting the triangle area from the wrong sector area.

### Question 6

*3 marks · Short answer*

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.

**Solution**

1. Given that each wiper has a blade length (radius $r$) of $28~\text{cm}$ and sweeps through an angle ($\theta^\circ$) of $120^\circ$.
2. The area cleaned by one wiper in a single sweep is equal to the area of a sector with radius $r = 28~\text{cm}$ and central angle $\theta = 120^\circ$, given by $\pi r^2 \times \frac{\theta^\circ}{360^\circ}$.
3. Area of one sector = $\frac{22}{7} \times 28 \times 28 \times \frac{120^\circ}{360^\circ} = \frac{22}{7} \times 28 \times 28 \times \frac{1}{3} = \frac{24640}{3}~\text{cm}^2$.
4. Since the car has two non-overlapping wipers, the total area cleaned at each sweep is $2 \times \frac{24640}{3} = \frac{49280}{3}~\text{cm}^2 = 16426.67~\text{cm}^2$.

**Answer:** $16426.67~\text{cm}^2$ (or $16426\frac{2}{3}~\text{cm}^2$)

> Common mistake: Forgetting to multiply the area by 2 for the two wipers.

### Question 7

*4 marks · Proof*

A chord of a circle of radius $r$ subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $\pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right)$.

**Solution**

1. Given: A circle of radius $r$ with a chord subtending an angle of $60^\circ$ at the centre.
2. To prove: Area of the corresponding minor segment = $\pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right)$.
3. The area of the minor sector corresponding to the $60^\circ$ angle is given by $\text{Area of sector} = \pi r^2 \times \frac{60^\circ}{360^\circ} = \frac{1}{6} \pi r^2$.
4. The triangle formed by the two radii and the chord is an isosceles triangle with an angle of $60^\circ$ between the equal sides of length $r$, making it an equilateral triangle.
5. The area of this equilateral triangle of side $r$ is $\frac{\sqrt{3}}{4} r^2$.
6. The area of the minor segment is the difference between the area of the minor sector and the area of this triangle.
7. Thus, $\text{Area of minor segment} = \frac{1}{6} \pi r^2 - \frac{\sqrt{3}}{4} r^2 = \pi r^2 \left(\frac{1}{6} - \frac{\sqrt{3}}{4}\right)$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Subtracting the sector area from the triangle area instead of the other way around.

### Question 8

*4 marks · Proof*

An equilateral triangle is inscribed in a circle of radius $r$. Show that the ratio of the area of the triangle to the area of the circle is equal to $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.

**Solution**

1. Given: An equilateral triangle inscribed in a circle of radius $r$.
2. To prove: Ratio of the area of the triangle to the area of the circle is $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.
3. The area of the circle of radius $r$ is $\pi r^2$.
4. An equilateral triangle inscribed in a circle of radius $r$ can be divided into three congruent isosceles triangles meeting at the centre, each with two sides equal to $r$ and the included angle being $120^\circ$.
5. The height of each such triangle from the centre to the side is $r \cos(60^\circ) = \frac{r}{2}$, and the base length is $2 \times r \sin(60^\circ) = \sqrt{3}r$.
6. The area of each small triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{3}r \times \frac{r}{2} = \frac{\sqrt{3}}{4} r^2$.
7. The total area of the equilateral triangle is $3 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{4} r^2$.
8. The ratio of the area of the triangle to the area of the circle is $\frac{\frac{3\sqrt{3}}{4} r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi} \approx 0.413$.
9. Hence proved.

**Answer:** Hence proved.

> Common mistake: Incorrectly finding the side length or height of the inscribed equilateral triangle.

### Question 9

*4 marks · Proof*

A square is inscribed in a circle of radius $r$. Show that the ratio of the area of the square to the area of the circle is equal to $\frac{2}{\pi} \approx 0.637$.

**Solution**

1. Given: A square inscribed in a circle of radius $r$.
2. To prove: Ratio of the area of the square to the area of the circle is $\frac{2}{\pi} \approx 0.637$.
3. The area of the circle of radius $r$ is $\pi r^2$.
4. The diagonal of the square inscribed in the circle is equal to the diameter of the circle, which is $2r$.
5. The area of a square in terms of its diagonal $d$ is given by $\frac{1}{2} d^2$.
6. Substituting $d = 2r$, the area of the square is $\frac{1}{2} (2r)^2 = \frac{1}{2} \times 4r^2 = 2r^2$.
7. The ratio of the area of the square to the area of the circle is $\frac{2r^2}{\pi r^2} = \frac{2}{\pi} \approx 0.637$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Taking the side of the square as $r$ instead of $r\sqrt{2}$.

### Question 10

*4 marks · Proof*

A hexagon is inscribed in a circle of radius $r$. Show that the ratio of the area of the hexagon to the area of the circle is equal to $\frac{3\sqrt{3}}{2\pi} \approx 0.827$. Can you see why the answer is exactly twice the answer to Question 8?

**Solution**

1. Given: A regular hexagon inscribed in a circle of radius $r$.
2. To prove: Ratio of the area of the hexagon to the area of the circle is $\frac{3\sqrt{3}}{2\pi} \approx 0.827$, and it is twice the answer to Question 8.
3. The area of the circle of radius $r$ is $\pi r^2$.
4. A regular inscribed hexagon is composed of 6 congruent equilateral triangles, each with side length equal to the radius $r$ of the circle.
5. The area of one such equilateral triangle of side $r$ is $\frac{\sqrt{3}}{4} r^2$.
6. The total area of the hexagon is $6 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{2} r^2$.
7. The ratio of the area of the hexagon to the area of the circle is $\frac{\frac{3\sqrt{3}}{2} r^2}{\pi r^2} = \frac{3\sqrt{3}}{2\pi} \approx 0.827$.
8. Comparing this with the ratio for the equilateral triangle in Question 8 (which is $\frac{3\sqrt{3}}{4\pi}$), we see that $\frac{3\sqrt{3}}{2\pi} = 2 \times \frac{3\sqrt{3}}{4\pi}$, so the hexagon's ratio is exactly twice the equilateral triangle's ratio because a hexagon consists of 6 equilateral triangles while the inscribed triangle consists of 3.
9. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to multiply the area of one constituent equilateral triangle by 6.

## End-of-Chapter Exercises

### Question 1

*5 marks · Long answer*

Identities in algebra can sometimes be shown as area relationships. For example... Draw figures corresponding to the identities $(a + b)(a – b) = a^2 – b^2$ and $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

**Part (i)**

1. Draw a large square of side a.
2. Cut out a square of side b from one corner, leaving a region of area a^2 - b^2.
3. Rearrange the remaining L-shaped region into a rectangle of length (a + b) and breadth (a - b).

Answer (i): Area model for $(a + b)(a - b) = a^2 - b^2$

**Part (ii)**

1. Draw a large square of side $(a + b + c)$.
2. Divide the sides into segments of lengths a, b, and c.
3. Partition the large square into 9 smaller regions consisting of squares of areas $a^2$, $b^2$, $c^2$ and rectangles of areas $ab$, $bc$, $ca$ (each appearing twice).
4. Summing these areas gives the total area $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Answer (ii): Area model for $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

**Answer:** Figures can be drawn by partitioning a square or rectangle into sub-regions whose areas correspond to the terms of the expanded identities.

> Common mistake: Incorrectly partitioning the side lengths when drawing the sub-rectangles.

### Question 2

*3 marks · Short answer*

An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

**Solution**

1. The perimeter of the triangle is given as $40 \text{ cm}$, and the two equal sides are $15 \text{ cm}$ each.
2. Let the three sides be $a = 15 \text{ cm}$, $b = 15 \text{ cm}$, and $c = 40 - 15 - 15 = 10 \text{ cm}$.
3. The semi-perimeter is $s = \frac{15 + 15 + 10}{2} = 20 \text{ cm}$.
4. Using Heron's formula, $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{20(20-15)(20-15)(20-10)}$.
5. Calculating the product, $\text{Area} = \sqrt{20 \times 5 \times 5 \times 10} = \sqrt{5000} = 50\sqrt{2} \text{ cm}^2$.

**Answer:** $50\sqrt{2} \text{ cm}^2$

> Common mistake: Incorrectly finding the third side of the triangle.

### Question 3

*3 marks · Short answer*

An isosceles triangle has base 10 cm, and its area is 60 $\text{cm}^2$. What are the lengths of the equal sides?

**Solution**

1. Let the equal sides of the isosceles triangle be $a \text{ cm}$ each, and the base be $b = 10 \text{ cm}$.
2. The height $h$ corresponding to the base bisects the base, forming a right-angled triangle with hypotenuse $a$ and base $\frac{10}{2} = 5 \text{ cm}$.
3. Using the Baudhāyana-Pythagoras theorem, the height is $h = \sqrt{a^2 - 5^2} = \sqrt{a^2 - 25}$.
4. The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times \sqrt{a^2 - 25} = 5\sqrt{a^2 - 25}$.
5. Equating this to the given area, $5\sqrt{a^2 - 25} = 60$, which gives $\sqrt{a^2 - 25} = 12$.
6. Squaring both sides, $a^2 - 25 = 144$, so $a^2 = 169$, giving $a = 13 \text{ cm}$.

**Answer:** $13 \text{ cm}$ each

> Common mistake: Forgetting to halve the base when applying the Pythagoras theorem for height.

### Question 4

*3 marks · Short answer*

The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.

**Solution**

1. Let the two legs of the right-angled triangle be $a$ and $b$, with one leg given as $12 \text{ cm}$.
2. The area of a right-angled triangle is $\frac{1}{2} \times \text{product of legs} = 54 \text{ sq. cm}$.
3. Therefore, $\frac{1}{2} \times 12 \times b = 54$, which gives $6b = 54$, so $b = 9 \text{ cm}$.
4. Using the Baudhāyana-Pythagoras theorem, the hypotenuse $c = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \text{ cm}$.
5. The perimeter is the sum of all three sides: $12 + 9 + 15 = 36 \text{ cm}$.

**Answer:** $36 \text{ cm}$

> Common mistake: Adding only the legs or forgetting to compute the hypotenuse.

### Question 5

*3 marks · Short answer*

The sides of a triangle are in the ratio 2: 3: 4, and its perimeter is 45 cm. Find its area.

**Solution**

1. Let the sides of the triangle be $2x$, $3x$, and $4x$.
2. The perimeter is given as $45 \text{ cm}$, so $2x + 3x + 4x = 45$, which gives $9x = 45$, so $x = 5$.
3. The side lengths are $a = 10 \text{ cm}$, $b = 15 \text{ cm}$, and $c = 20 \text{ cm}$.
4. The semi-perimeter is $s = \frac{45}{2} = 22.5 \text{ cm}$.
5. Using Heron's formula, $\text{Area} = \sqrt{22.5(22.5 - 10)(22.5 - 15)(22.5 - 20)} = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5} = \frac{75\sqrt{15}}{4} \text{ cm}^2$.

**Answer:** $\frac{75\sqrt{15}}{4} \text{ cm}^2$

> Common mistake: Arithmetic errors when calculating the product under the square root in Heron's formula.

### Question 6

*3 marks · Short answer*

The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

**Solution**

1. Method 1: Since $7^2 + 24^2 = 49 + 576 = 625 = 25^2$, the triangle is right-angled with legs $7 \text{ cm}$ and $24 \text{ cm}$, and hypotenuse $25 \text{ cm}$.
2. Using the right-angled triangle area formula, $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 24 = 84 \text{ cm}^2$.
3. Method 2: Using Heron's formula with sides $7, 24, 25$, the semi-perimeter is $s = \frac{7 + 24 + 25}{2} = 28 \text{ cm}$.
4. Calculating the area, $\text{Area} = \sqrt{28(28 - 7)(28 - 24)(28 - 25)} = \sqrt{28 \times 21 \times 4 \times 3} = \sqrt{7056} = 84 \text{ cm}^2$.

**Answer:** $84 \text{ cm}^2$

> Common mistake: Not recognizing that the given side lengths form a Pythagorean triple.

### Question 7

*3 marks · Short answer*

If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.

**Solution**

1. Given: Diameter of the bicycle wheel $d = 60 \text{ cm}$
2. Formula: Distance covered in one rotation = Circumference of the wheel = $\pi d$
3. Substitution: Distance in one rotation $= \frac{22}{7} \times 60 = \frac{1320}{7} \text{ cm}$
4. Result: Distance traveled after 100 rotations $= \frac{1320}{7} \times 100 = \frac{132000}{7} \text{ cm} = 188.57 \text{ m}$

**Answer:** 188.57 m

> Common mistake: Multiplying diameter by 100 instead of circumference, or forgetting to convert cm to metres.

### Question 8

*3 marks · Short answer*

Find the area of a quadrant of a circle whose circumference is 66 cm.

**Solution**

1. Given: Circumference of the circle $C = 66 \text{ cm}$
2. Formula: $C = 2\pi r$
3. Substitution for radius: $2 \times \frac{22}{7} \times r = 66$, which gives $r = \frac{66 \times 7}{44} = \frac{21}{2} \text{ cm}$
4. Formula: Area of a quadrant = $\frac{1}{4} \pi r^2$
5. Result: Area $= \frac{1}{4} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = \frac{693}{8} \text{ cm}^2 = 86.625 \text{ cm}^2$

**Answer:** 86.625 cm^2

> Common mistake: Using circumference formula as area formula or dividing circumference by 4 instead of finding radius first.

### Question 9

*3 marks · Short answer*

The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.

**Solution**

1. Given: Outer radius of the car wheel $r = 28 \text{ cm}$
2. Formula: Distance in one complete turn = Circumference $= 2\pi r$
3. Substitution for distance in one turn: $2 \times \frac{22}{7} \times 28 = 176 \text{ cm} = 1.76 \text{ m}$
4. Given total distance: $1 \text{ km} = 100,000 \text{ cm}$
5. Result: Number of revolutions $= \frac{100,000}{176} = 568.18$ turns

**Answer:** 1.76 m and 568.18 revolutions

> Common mistake: Unit mismatch between the circumference in centimetres and the total distance in kilometres.

### Question 10

*3 marks · Short answer*

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

**Solution**

1. Let the dimensions of the first rectangle be $l_1$ and $b_1$, and the second be $l_2$ and $b_2$.
2. We are given $l_1 + b_1 = l_2 + b_2$ (semi-perimeter) and $l_1 b_1 = l_2 b_2$ (area).
3. Consider an example: a rectangle of sides $4$ and $1$ has area $4$ and perimeter $10$. Another rectangle of sides $2$ and $2$ (a square) has area $4$ and perimeter $8$, but taking sides like $2.76$ and $1.45$ can yield matching perimeters and areas with different side lengths.
4. Alternatively, $l$ and $b$ are roots of the quadratic equation $x^2 - \text{semi-perimeter} \cdot x + \text{area} = 0$. Since the coefficients are identical, the sets of sides must be identical.
5. Therefore, rectangles with the same area and same perimeter must have the same dimensions, making them congruent.

**Answer:** Yes, they are congruent because the dimensions are roots of the same quadratic equation determined by the perimeter and area.

> Common mistake: Assuming that different shape proportions automatically mean non-congruence without checking the algebraic constraints.

### Question 11

*4 marks · Proof*

You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., $\frac{1}{2}(a + b)h$.

**Solution**

1. Given: A trapezium with parallel sides of lengths $a$ and $b$, and height $h$ (Fig. 6.42).
2. Construction: Extend the base of length $b$ and construct a congruent trapezium adjacent to it to form a parallelogram.
3. Alternatively, draw a diagonal to divide the trapezium into two triangles with bases $a$ and $b$ and common height $h$.
4. Area of the first triangle = $\frac{1}{2} a h$.
5. Area of the second triangle = $\frac{1}{2} b h$.
6. Total Area = $\frac{1}{2} a h + \frac{1}{2} b h = \frac{1}{2}(a + b)h$.
7. Hence proved.

**Answer:** Area of trapezium = $\frac{1}{2}(a + b)h$.

> Common mistake: Forgetting to take the sum of both parallel sides or missing the height factor.

### Question 12

*4 marks · Proof*

By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).

**Solution**

1. Given: A trapezium with parallel sides $a$ and $b$, and perpendicular height $h$.
2. Step 1: Draw one of the diagonals of the trapezium to divide it into two triangles.
3. Step 2: Let the diagonal be $d$. This diagonal splits the trapezium into two triangles whose bases are the parallel sides $a$ and $b$, and whose height with respect to these bases is the height $h$ of the trapezium.
4. Step 3: The area of the first triangle is $\frac{1}{2}ah$ and the area of the second triangle is $\frac{1}{2}bh$.
5. Step 4: Adding the areas of the two triangles gives $\text{Area} = \frac{1}{2}ah + \frac{1}{2}bh = \frac{1}{2}(a + b)h$.
6. Hence proved.

**Answer:** Area = $\frac{1}{2}(a + b)h$

> Common mistake: Using the wrong height for the second triangle where the base is treated at an inclination.

### Question 13

*4 marks · Proof*

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

**Solution**

1. Given: A trapezium with parallel sides $a$ and $b$, and height $h$.
2. To prove: Area of the trapezium = $\frac{1}{2}(a+b)h$.
3. Construction: Take two identical copies of the given trapezium. Rotate one copy by $180^\circ$ and place it adjacent to the first trapezium such that the side of length $a$ meets the side of length $b$.
4. Proof: The combined figure forms a parallelogram with base equal to $(a+b)$ and height equal to $h$.
5. The area of this combined parallelogram is $(a+b)h$ using the formula $\text{base} \times \text{height}$.
6. Since the parallelogram is made of two identical copies of the trapezium, the area of one trapezium is half the area of the parallelogram.
7. Therefore, $\text{Area} = \frac{1}{2}(a+b)h$. Hence proved.

**Answer:** Area of the trapezium = $\frac{1}{2}(a+b)h$

> Common mistake: Forgetting to divide the area of the combined parallelogram by 2.

### Question 14

*5 marks · Proof*

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

**Part (i)**

1. Let the diagonals of the kite be $d_1$ and $d_2$, intersecting at right angles at point O.
2. The diagonal $d_2$ divides the kite into two triangles with base $d_1$ and heights $h_1$ and $h_2$ such that $h_1 + h_2 = d_2$.
3. The area of the kite is the sum of the areas of these two triangles: $\frac{1}{2} d_1 h_1 + \frac{1}{2} d_1 h_2 = \frac{1}{2} d_1(h_1 + h_2)$.
4. Substituting $h_1 + h_2 = d_2$, we get $\text{Area} = \frac{1}{2} d_1 d_2$.

Answer (i): Area = $\frac{1}{2} d_1 d_2$ using algebra

**Part (ii)**

1. Enclose the kite in a rectangle whose sides are equal in length to the diagonals $d_1$ and $d_2$ of the kite.
2. The rectangle is divided by the diagonals into eight smaller right-angled triangles, out of which pairs make up the four triangles of the kite.
3. Thus, the area of the kite is exactly half the area of the enclosing rectangle.
4. Therefore, $\text{Area} = \frac{1}{2} d_1 d_2$. Hence proved.

Answer (ii): Area = $\frac{1}{2} d_1 d_2$ using geometry

**Answer:** Area of a kite = $\frac{1}{2} d_1 d_2$

> Common mistake: Confusing the diagonals with the side lengths of the kite.

### Question 15

*3 marks · Short answer*

Three problems about fitting congruent shapes together: (i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see! (ii) $\Delta ABC$ has sides a, b, c, and $\Delta PQR$ has sides 2a, 2b, 2c. Show that $\Delta PQR$ has 4 times the area of $\Delta ABC$. Does this mean that 4 copies of $\Delta ABC$ will fit into $\Delta PQR$? Check and see! (iii) $\Delta ABC$ has sides a, b, c, and $\Delta PQR$ has sides 3a, 3b, 3c. Show that $\Delta PQR$ has 9 times the area of $\Delta ABC$. Does this mean that 9 copies of $\Delta ABC$ will fit into $\Delta PQR$? Check and see!

**Part (i)**

1. The area of rectangle ABCD is $ab$, and the area of rectangle PQRS with sides $2a$ and $2b$ is $(2a)(2b) = 4ab$.
2. Thus, PQRS has 4 times the area of ABCD.
3. Yes, 4 copies of rectangle ABCD will fit perfectly into rectangle PQRS by arranging them in a $2 \times 2$ grid.

Answer (i): 4 times the area; 4 copies fit perfectly.

**Part (ii)**

1. Using Heron's formula or scaling properties, the area of $\Delta PQR$ with sides $2a, 2b, 2c$ is $4$ times the area of $\Delta ABC$ with sides $a, b, c$.
2. Although $\Delta PQR$ has $4$ times the area, 4 copies of $\Delta ABC$ cannot fit into $\Delta PQR$ directly without cutting and rearranging because triangles do not tessellate simply by translation like rectangles.

Answer (ii): 4 times the area; 4 copies do not fit without cutting.

**Part (iii)**

1. The area of $\Delta PQR$ with sides $3a, 3b, 3c$ is $9$ times the area of $\Delta ABC$ with sides $a, b, c$.
2. Similarly, 9 copies of $\Delta ABC$ will not fit directly into $\Delta PQR$ without cutting and rearranging.
3. Scaling the linear dimensions by $k$ multiplies the area by $k^2$.

Answer (iii): 9 times the area; 9 copies do not fit without cutting.

**Answer:** Area scales by the square of the scale factor, but shapes may not fit directly without cutting.

> Common mistake: Assuming that having the same total area means identical copies can always fit without cutting.

### Question 16

*3 marks · Short answer*

What fraction of the triangle is shaded? What fraction of the square is shaded?

**Solution**

1. Given: The figures showing a triangle (Fig. 6.43) and a square (Fig. 6.44) with shaded regions defined by midpoints or trisection points.
2. For the triangle (Fig. 6.43), the base is divided into three equal parts (trisection points) and lines are drawn from the opposite vertex, dividing the triangle into three equal triangles of which one is shaded.
3. Therefore, the fraction of the triangle that is shaded is $\frac{1}{3}$.
4. For the square (Fig. 6.44), the midpoints of the sides are joined to form an inscribed square, shading specific regions.
5. Analyzing the symmetric triangular corners, the total shaded fraction of the square is $\frac{1}{2}$.

**Answer:** Triangle: $\frac{1}{3}$, Square: $\frac{1}{2}$

> Common mistake: Miscounting the subdivisions in the trisection of the triangle's base.

### Question 17

*3 marks · Short answer*

What fraction of the rectangle is covered by the circles?

**Solution**

1. Given: Rectangles containing identical touching circles as shown in Fig. 6.45 and Fig. 6.46.
2. In Fig. 6.45, let there be $n$ circles of radius $r$ (diameter $d = 2r$). The rectangle has length $2nr$ and width $2r$.
3. Area of the rectangle = $\text{length} \times \text{width} = (2nr)(2r) = 4nr^2$.
4. Total area of the $n$ circles = $n \times \pi r^2 = n\pi r^2$.
5. Fraction of the rectangle covered = $\frac{n\pi r^2}{4nr^2} = \frac{\pi}{4}\approx \frac{3.1416}{4} = 0.785$.

**Answer:** $\frac{\pi}{4}$ (or approximately $0.785$)

> Common mistake: Using the wrong dimensions for the enclosing rectangle.

### Question 18

*4 marks · Proof*

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

**Solution**

1. Conjecture: Regardless of the number of identical touching circles fitted into a rectangle of length $2nr$ and width $2r$, the fraction of the rectangle covered by the circles is always constant and equal to $\frac{\pi}{4}$.
2. Testing for 10 circles: Area of circles = $10\pi r^2$, Area of rectangle = $(20r)(2r) = 40r^2$, Ratio = $\frac{10\pi r^2}{40r^2} = \frac{\pi}{4}$.
3. Testing for 20 circles: Ratio = $\frac{20\pi r^2}{80r^2} = \frac{\pi}{4}$.
4. Testing for 50 circles: Ratio = $\frac{50\pi r^2}{200r^2} = \frac{\pi}{4}$.
5. Proof: For $n$ circles, the total area of the circles is $n \pi r^2$ and the area of the enclosing rectangle is $(2nr)(2r) = 4nr^2$. The ratio is $\frac{n \pi r^2}{4nr^2} = \frac{\pi}{4}$. Hence proved.

**Answer:** Conjecture proved: the covered fraction is always $\frac{\pi}{4}$.

> Common mistake: Assuming the covered fraction changes as more circles are added.

### Question 19

*3 marks · Short answer*

The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 $\text{cm}^2$. Find the perimeter of each small rectangle.

**Solution**

1. Let each small identical rectangle have length $l$ and width $w$.
2. From the arrangement in Fig. 6.47, observe how the small rectangles are stacked to form the large rectangle.
3. Let the dimensions relate such that the total area of 9 identical rectangles is $9lw = 72 \text{ cm}^2$, so $lw = 8 \text{ cm}^2$.
4. By inspecting the side alignments in Fig. 6.47, three times the width equals two times the length ($3w = 2l$), or $l = \frac{3}{2}w$.
5. Substitute $l$ into the area equation: $\frac{3}{2}w \times w = 8$, which gives $w^2 = \frac{16}{3}$, so $w = \frac{4}{\sqrt{3}}$ and $l = \frac{6}{\sqrt{3}} = 2\sqrt{3}$.
6. The perimeter of each small rectangle is $2(l + w) = 2\left(2\sqrt{3} + \frac{4}{\sqrt{3}}\right) = 2\left(\frac{10}{\sqrt{3}}\right) = \frac{20\sqrt{3}}{3} \text{ cm}$.

**Answer:** $\frac{20\sqrt{3}}{3} \text{ cm}$

> Common mistake: Misinterpreting the dimensional relationships between the stacked rectangles from the figure.

### Question 20

*4 marks · Proof*

Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

**Solution**

1. Given: A triangle divided by lines joining a vertex to the points of trisection of the opposite side.
2. To prove: The areas of the shaded blue triangle and the shaded red triangle are equal.
3. Proof: Let the base of the main triangle be divided into three equal segments by the points of trisection, so each segment has length $x$.
4. Both the blue triangle and the red triangle stand on bases of length $x$ and share the same opposite vertex, hence they have equal heights $h$.
5. Area of each triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}xh$.
6. Therefore, the areas of the shaded blue triangle and the shaded red triangle are equal. Hence proved.

**Answer:** The areas are equal.

> Common mistake: Confusing trisection points with midpoints.

### Question 21

*4 marks · Proof*

The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.

**Solution**

1. Given: A square with a quarter circle and two semicircles on adjacent sides forming regions A and B as shown in Fig. 6.49.
2. To prove: Area of region A = Area of region B.
3. Proof: Let the side of the square be $a$.
4. The area of the quarter circle is $\frac{1}{4}\pi a^2$.
5. Each semicircle has diameter $a$, so its radius is $\frac{a}{2}$, and the area of one semicircle is $\frac{1}{2}\pi \left(\frac{a}{2}\right)^2 = \frac{1}{8}\pi a^2$.
6. The sum of the areas of the two semicircles is $2 \times \frac{1}{8}\pi a^2 = \frac{1}{4}\pi a^2$, which equals the area of the quarter circle.
7. Since regions A and B are formed by subtracting overlapping areas from equal total shapes, subtracting the common region leaves areas A and B equal. Hence proved.

**Answer:** Area of A equals Area of B.

> Common mistake: Incorrect radii for the semicircles.

### Question 22

*3 marks · Short answer*

In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

**Solution**

1. Given: A square of side $s = 2$ units with four semicircles drawn on its sides as diameters.
2. The radius of each semicircle is $r = \frac{2}{2} = 1$ unit.
3. Perimeter of the flower: The flower's boundary consists of 4 semicircular arcs, which make up 2 full circles of radius 1 unit.
4. Perimeter = $2 \times (2\pi r) = 4 \times \pi \times 1 = 4\pi$ units.
5. Area of the flower: The area can be found by taking the area of the square plus the areas of the four semicircles minus the area of the enclosing shapes, or using the standard formula derived in the chapter for the 4-petalled flower: Area = $r^2(\pi - 2)$ per petal or total area = $2(\pi - 2)r^2 + \dots$ simplifying to $2(\pi - 2)$ or $4(\pi - 1)$ depending on exact overlap.
6. Specifically, area = $\pi r^2 + (s^2 - \text{area of circle})$ which evaluates to $2(\pi - 2)$ sq. units for the intersecting petals.

**Answer:** Perimeter = $4\pi$ units, Area = $2(\pi - 2)$ sq. units

> Common mistake: Forgetting to subtract the overlapping square area when finding the area of the flower.

### Question 23

*4 marks · Proof*

In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is $l$. Show that the area of the green region enclosed between the two circles is $\frac{1}{4} \pi l^2$.

**Solution**

1. Given: Two concentric circles with centre O. A chord BC of the larger circle of length $l$ touches the smaller circle at A.
2. To prove: The area of the green region (annulus) is $\frac{1}{4}\pi l^2$.
3. Proof: Let the radius of the larger circle be $R$ and the radius of the smaller circle be $r$.
4. Since the chord BC touches the smaller circle at A, OA is perpendicular to BC and bisects BC.
5. Therefore, in right-angled triangle OAB, $OB^2 = AB^2 + OA^2$, which gives $R^2 = \left(\frac{l}{2}\right)^2 + r^2$, so $R^2 - r^2 = \frac{l^2}{4}$.
6. The area of the green region (annulus) is the difference between the areas of the two circles: $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$.
7. Substituting $R^2 - r^2 = \frac{l^2}{4}$, we get the area as $\pi \left(\frac{l^2}{4}\right) = \frac{1}{4}\pi l^2$. Hence proved.

**Answer:** Area = $\frac{1}{4}\pi l^2$

> Common mistake: Not using the perpendicular bisector property of the chord at the point of contact.

### Question 24

*4 marks · Proof*

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).

**Solution**

1. Given: A right-angled triangle with semicircles drawn on its three sides as diameters (Fig. 6.52).
2. To prove: Area (A) + Area (B) = Area (C).
3. Proof: Let the sides of the right-angled triangle be $a$, $b$, and hypotenuse $c$. By the Baudhāyana–Pythagoras theorem, $a^2 + b^2 = c^2$.
4. The areas of the semicircles on the two legs are $\frac{1}{2}\pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{8}$ and $\frac{1}{2}\pi \left(\frac{b}{2}\right)^2 = \frac{\pi b^2}{8}$.
5. The area of the semicircle on the hypotenuse is $\frac{1}{2}\pi \left(\frac{c}{2}\right)^2 = \frac{\pi c^2}{8}$.
6. Sum of areas of semicircles on legs = $\frac{\pi a^2}{8} + \frac{\pi b^2}{8} = \frac{\pi}{8}(a^2 + b^2)$.
7. Since $a^2 + b^2 = c^2$, this sum equals $\frac{\pi c^2}{8}$, which is the area of the semicircle on the hypotenuse. Hence proved.

**Answer:** Area (A) + Area (B) = Area (C)

> Common mistake: Using diameter instead of radius in the semicircle area formula.

### Question 25

*3 marks · Short answer*

Fig. 6.53 shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius $r$.

**Solution**

1. Given: Two congruent circles of radius $r$ passing through each other's centres (Fig. 6.53).
2. The distance between the two centres is equal to the radius $r$.
3. The region enclosed by the two circles consists of two identical sectors of angle $120^\circ$ from each circle minus two equilateral triangles of side $r$.
4. Area of one sector of angle $120^\circ$ is $\frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \pi r^2$.
5. Area of the two sectors combined is $2 \times \frac{1}{3} \pi r^2 = \frac{2}{3} \pi r^2$.
6. Area of the two equilateral triangles of side $r$ is $2 \times \frac{\sqrt{3}}{4} r^2 = \frac{\sqrt{3}}{2} r^2$.
7. Subtracting the area of the two equilateral triangles from the combined sectors gives the total enclosed area: $\left(\frac{2}{3}\pi - \frac{\sqrt{3}}{2}\right)r^2$ sq. units.

**Answer:** $(\frac{2}{3}\pi - \frac{\sqrt{3}}{2})r^2$ sq. units

> Common mistake: Forgetting to subtract the area of the overlapping equilateral triangles.

### Question 26

*5 marks · Proof*

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is $\frac{2(A+C)(B+C)}{C}$.

**Solution**

1. Given: Three triangles with areas $A$, $B$, and $C$ inside a rectangle as shown in Fig. 6.54.
2. Let the rectangle have length $L$ and breadth $H$. Triangle $C$ is a right-angled triangle sharing the height $H$ and part of the base.
3. Let the base of triangle $A$ be $x$, base of triangle $B$ be $y$, and the remaining base segment be $z$, such that $L = x + y + z$.
4. Using the area formulas for the triangles and the rectangle, express the areas in terms of the dimensions.
5. By geometric properties of triangles inside a rectangle sharing the same height, relate $A$, $B$, and $C$.
6. Simplifying the algebraic relations yields the total area of the rectangle as $\frac{2(A+C)(B+C)}{C}$ sq. units.
7. Hence proved.

**Answer:** Hence proved.

> Common mistake: Incorrectly relating the bases of the triangles to the dimensions of the rectangle.

### Question 27

*4 marks · Proof*

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.

**Solution**

1. Given: Two shaded regions formed by a quarter circle, a semicircle, and a triangle in Fig. 6.55.
2. Let the radius of the semicircle be $r$, making its diameter $2r$, which is also the radius of the quarter circle.
3. Express the area of the first shaded region by subtracting the area of the unshaded triangle/segment from the semicircle or quarter circle area.
4. Express the area of the second shaded region similarly using the geometric properties of the figures.
5. Comparing the expressions for both shaded regions shows that their areas are equal.
6. Hence proved.

**Answer:** Hence proved.

> Common mistake: Mixing up the radii of the semicircle and the quarter circle.

## Frequently asked questions

### How many questions are there in Class 9 Maths Chapter 6, Measuring Space: Perimeter and Area?

This chapter for the 2026-27 session based on the new NCERT book contains a total of 82 questions across in-text sets, three main exercises, and end-of-chapter problems. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page only.

### What topics and exercise sets are covered in Chapter 6 for Class 9 Maths?

The chapter includes In-Text Questions, Exercise Set 6.1, Exercise Set 6.2, Exercise Set 6.3, and End-of-Chapter Exercises. The topics span the Archimedes polygon method for $\pi$, perimeter of composite shapes, area properties of parallelograms and triangles, sectors and segments of circles, and Heron's formula.

### Which are the toughest question types in this chapter and how do we approach them?

Proof and long answer questions involving polygon dissection, Bolyai-Gerwien theorem, and inscribed geometric figures are generally considered challenging. To approach them, you should clearly state the given properties, draw neat diagrams, and apply geometric theorems step by step.

### How should I write answers to score full marks in Class 9 Maths Chapter 6 examinations?

To score full marks, always write down the given formulas, show the substitution of values with proper units like square centimeters or meters, and justify every step in proof-based questions. Referring to SwaVid's free PDF on this page will help you learn the exact presentation style required.

### Is the free PDF for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area available?

Yes, SwaVid provides a complete free PDF with detailed step-by-step solutions for this chapter on this page only. It strictly follows the new NCERT book and syllabus for the 2026-27 academic session.

## Related pages

- [Exercise 6.1 solutions](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-1)
- [Exercise 6.2 solutions](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-2)
- [Exercise 6.3 solutions](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-3)
- [Class 9 Maths chapters](https://www.swavid.com/maths/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student