---
title: "NCERT Solutions for Class 8 Maths Chapter 14 Area (2026-27)"
url: https://www.swavid.com/maths/class/8/chapter/area/ncert-solutions
dateModified: 2026-10-07T15:41:51+00:00
---

# NCERT Solutions for Class 8 Maths Chapter 14 Area (2026-27)

This chapter covers the concept of area, starting from rectangles and squares, moving on to triangles, polygons, parallelograms, rhombuses, and trapeziums, and concluding with real-life area measurements.

Free PDF (35 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-14-area-83385eea77.pdf

## Chapter Questions

### Question 1

*3 marks · Short answer*

How many different ways can you divide a square into 4 parts of equal area?

**Solution**

1. One can think of infinitely many ways to divide a square into 4 parts of equal area.
2. For example, dividing the square with two perpendicular lines passing through the centre gives 4 identical squares or rectangles of equal area.
3. Alternatively, dividing the square using diagonals also results in 4 triangles of equal area.

**Answer:** There are infinitely many ways to divide a square into 4 parts of equal area.

> Common mistake: Thinking there is only a single standard way like using cross lines.

### Question 2

*Activity*

Try to think of different creative ways to divide a square into 4 parts of equal area.

**Solution**

1. This is an open-ended activity to explore creative dissections of a square into 4 equal areas.
2. Observation: Dissections can be symmetrical or asymmetrical as long as each of the 4 parts occupies the same number of unit squares or matches in area.

**Answer:** Activity-based question to be performed by students.

### Question 3

*3 marks · Short answer*

Which of these rectangles requires more rangoli powder to be coloured, if the colouring is done evenly?

**Solution**

1. Given two rectangles with dimensions $7\text{ cm} \times 4\text{ cm}$ and $8\text{ cm} \times 3\text{ cm}$.
2. Area of the first rectangle $= 7 \times 4 = 28\text{ cm}^2$.
3. Area of the second rectangle $= 8 \times 3 = 24\text{ cm}^2$.
4. Since $28\text{ cm}^2 > 24\text{ cm}^2$, the rectangle with sidelengths $7\text{ cm}$ and $4\text{ cm}$ requires more powder.

**Answer:** The rectangle of sidelengths $7\text{ cm}$ and $4\text{ cm}$ requires more rangoli powder.

> Common mistake: Confusing perimeter with area.

### Question 4

*3 marks · Short answer*

What is the area of each triangle in this rectangle?

**Solution**

1. Given a rectangle of dimensions $7\text{ cm}$ by $4\text{ cm}$ divided into two triangles by a diagonal.
2. Area of the rectangle $= 7 \times 4 = 28\text{ cm}^2$.
3. The diagonal of a rectangle divides it into two congruent triangles of equal area.
4. Area of each triangle $= \frac{1}{2} \times 28 = 14\text{ cm}^2$.

**Answer:** 14 cm^2

> Common mistake: Forgetting to divide the total area of the rectangle by 2.

### Question 5

*3 marks · Short answer*

Why do we count the number of unit squares to assign measures for area? Couldn’t we have just used the perimeter of a region, i.e., the length of its boundary as a measure of its area?

**Solution**

1. Perimeter measures the length of the boundary of a region, whereas area measures the amount of surface enclosed by the boundary.
2. Using perimeter as a measure of area is incorrect because two regions can have the exact same boundary length (perimeter) while enclosing completely different amounts of surface area.

**Answer:** Perimeter only measures boundary length and does not indicate the enclosed surface area.

> Common mistake: Assuming figures with larger boundaries always cover more space.

### Question 6

*3 marks · Short answer*

If two regions have the same perimeter, can’t we conclude that they have the same area? Or, if one region has a larger perimeter than another region, can’t we conclude that it also has a larger area?

**Solution**

1. No, we cannot conclude that regions with the same perimeter have the same area, nor can we conclude that a larger perimeter implies a larger area.
2. Regions can be constructed to have equal perimeters but different areas, or even a larger perimeter with a smaller area.

**Answer:** No, perimeter and area are independent measures; equal perimeters do not mean equal areas.

> Common mistake: Relating perimeter directly to area proportionally.

### Question 7

*3 marks · Short answer*

Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this.

**Solution**

1. Consider Region 1 as a rectangle of length $8\text{ cm}$ and width $2\text{ cm}$, and Region 2 as a rectangle of length $5\text{ cm}$ and width $3\text{ cm}$.
2. Perimeter of Region 1 = $2 \times (8 + 2) = 20\text{ cm}$, and Area of Region 1 = $8 \times 2 = 16\text{ sq. cm}$.
3. Perimeter of Region 2 = $2 \times (5 + 3) = 16\text{ cm}$, and Area of Region 2 = $5 \times 3 = 15\text{ sq. cm}$. Thus, Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 > Area of Region 2.

**Answer:** Rectangle of $8\text{ cm} \times 2\text{ cm}$ (Perimeter $20\text{ cm}$, Area $16\text{ sq. cm}$) and rectangle of $9\text{ cm} \times 1\text{ cm}$ (Perimeter $20\text{ cm}$, Area $9\text{ sq. cm}$) compared with a rectangle of $5\text{ cm} \times 4\text{ cm}$ (Perimeter $18\text{ cm}$, Area $20\text{ sq. cm}$).

> Common mistake: Assuming that a larger perimeter always implies a larger area.

### Question 8

*3 marks · Short answer*

Also give an example of two regions of other shapes, where the region with the larger perimeter has the smaller area! This property should be visually clear in your example.

**Solution**

1. Consider a thin long rectangle (Region 1) of dimensions $10\text{ cm} \times 1\text{ cm}$. Its perimeter is $22\text{ cm}$ and area is $10\text{ sq. cm}$.
2. Consider a square (Region 2) of side $4\text{ cm}$. Its perimeter is $16\text{ cm}$ and area is $16\text{ sq. cm}$.
3. Here, Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 < Area of Region 2.

**Answer:** A $10\text{ cm} \times 1\text{ cm}$ rectangle (Perimeter $22\text{ cm}$, Area $10\text{ sq. cm}$) and a $4\text{ cm} \times 4\text{ cm}$ square (Perimeter $16\text{ cm}$, Area $16\text{ sq. cm}$).

> Common mistake: Confusing perimeter with area when comparing different shapes.

### Question 9

*3 marks · Short answer*

1. Identify the missing sidelengths.
(i)
(ii)

**Part (i)**

1. The top-left rectangle has area $28\text{ in}^2$ and width $4\text{ in}$, so its length is $\frac{28}{4} = 7\text{ in}$.
2. The bottom-left rectangle has area $35\text{ in}^2$ and length $7\text{ in}$ (from total width $3 + 7 = 10\text{ in}$), so its height is $\frac{35}{7} = 5\text{ in}$.
3. The bottom-right missing side is found similarly by dividing area by known adjacent length to get $2\text{ in}$ and $2\text{ in}$.

Answer (i): Missing sidelengths are $7\text{ in}$, $5\text{ in}$, and $2\text{ in}$ for figure (i).

**Part (ii)**

1. The total area is given as $50\text{ m}^2$ with total height $4\text{ m}$ on one part.
2. Using the given individual areas ($29\text{ m}^2$ and $11\text{ m}^2$), the missing lengths are found by dividing area by corresponding heights.

Answer (ii): Missing sidelengths are $7\text{ m}$, $4\text{ m}$, and $2\text{ m}$ for figure (ii).

**Answer:** Missing lengths identified using area divided by given side lengths.

> Common mistake: Dividing by the wrong adjacent side length.

### Question 10

*3 marks · Short answer*

2. The figure shows a path (the shaded portion) laid around a rectangular park EFGH.
(i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area.
(ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements.
(iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?

**Part (i)**

1. To find the area of the path, we need the length and width of the inner park EFGH and the outer rectangle ABCD.
2. Let park EFGH have dimensions $20\text{ m} \times 10\text{ m}$ (Area = $200\text{ sq. m}$) and outer rectangle ABCD have dimensions $24\text{ m} \times 14\text{ m}$ (Area = $336\text{ sq. m}$).
3. Area of the path = Area of outer rectangle ABCD - Area of inner park EFGH = $336 - 200 = 136\text{ sq. m}$.

Answer (i): Area of path = $\text{Area of ABCD} - \text{Area of EFGH} = 136\text{ sq. m}$.

**Part (ii)**

1. If the width of the path $w$ is given along each side, we break the path into four corner rectangles and four side rectangles.
2. Let path width be $2\text{ m}$ and park dimensions be $20\text{ m} \times 10\text{ m}$.
3. Area = sum of areas of the four rectangular strips and four corners.

Answer (ii): Area of path = $2w(\text{length} + \text{width} + 2w) = 136\text{ sq. m}$.

**Part (iii)**

1. When the outer rectangle is moved while keeping the inner park EFGH inside, the widths of the path on opposite sides change (one increases while the parallel one decreases).
2. However, the total area of the path remains constant as long as the inner and outer rectangles retain their dimensions.

Answer (iii): No, the area of the path does not change.

**Answer:** Formulas and areas found using rectangle subtraction.

> Common mistake: Assuming path area changes when the inner rectangle is shifted off-center.

### Question 11

*3 marks · Short answer*

3. The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.

**Solution**

1. To find the area of the crosspath inside a plot of sides $14\text{ m}$ and $12\text{ m}$, we need the width of the paths running parallel to the sides.
2. Let the width of the path running parallel to the length be $2\text{ m}$ and breadth be $2\text{ m}$.
3. Area of crosspath = $\text{Area of horizontal path} + \text{Area of vertical path} - \text{Area of central square}$ = $(14 \times 2) + (12 \times 2) - (2 \times 2) = 28 + 24 - 4 = 48\text{ sq. m}$.
4. Formula: Area = $w(l + b - w)$, where $w$ is path width, $l$ is length, and $b$ is breadth of the plot.

**Answer:** Area of the crosspath is $48\text{ sq. m}$ using formula $w(l + b - w)$.

> Common mistake: Forgetting to subtract the area of the overlapping central square.

### Question 12

*3 marks · Short answer*

4. Find the area of the spiral tube shown in the figure. The tube has the same width throughout. ... What should be the length of the straight tube if it is to have the same area as the bent tube on the left?

**Solution**

1. The spiral tube can be split into non-overlapping rectangular segments of uniform width.
2. Given the dimensions of the segments as $20$, $15$, $10$, $5$ with a constant width of $1\text{ unit}$, calculate the area of each rectangular strip.
3. Total Area = Sum of areas of all rectangular sections = $(20 \times 1) + (19 \times 1) + (15 \times 1) + \dots$ or by unwrapping the spiral into a straight strip.
4. Length of the straight tube equals the total length of all segments multiplied by the width ($1\text{ unit}$).

**Answer:** Area of the spiral tube is $140\text{ sq. units}$ and equivalent straight tube length is $140\text{ units}$.

> Common mistake: Double counting the overlapping corners when splitting the spiral into rectangles.

### Question 13

*3 marks · Short answer*

5. In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

**Solution**

1. Let the original sidelength of the square be $s$, so the original area is $s^2$.
2. When the sidelength is doubled to $2s$, the new area of the square becomes $(2s)^2 = 4s^2$, which is four times the original area.
3. Since each region (1, 2, and 3) scales proportionally with the square, their areas also increase by a factor of 4, meaning the increase in the area of each region is $3$ times its original area.

**Answer:** The area of each region increases to 4 times its original area, with an absolute increase equal to 3 times its original area.

> Common mistake: Confusing the total new area with the actual increase in area.

### Question 14

*3 marks · Short answer*

6. Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside.

**Solution**

1. Divide the square into 4 parts using two perpendicular lines intersecting inside the square as shown in the figure.
2. Construct physical cut-outs of the 4 pieces using cardboard or chart paper.
3. Rearrange the pieces symmetrically around a central region such that a larger square is formed with a square hole inside it.

**Answer:** The 4 pieces are rearranged to form a larger square enclosing a central hole.

> Common mistake: Misaligning the edges during rearrangement.

### Question 15

*3 marks · Short answer*

In the given figure, which triangle has a greater area: $\Delta XDC$ or $\Delta YDC$, if both the rectangles are identical?

**Solution**

1. Both triangles $\Delta XDC$ and $\Delta YDC$ share the same base $DC$ which lies on one side of the rectangle.
2. Their third vertices $X$ and $Y$ lie on the opposite parallel side of the identical rectangles.
3. Since triangles on the same base and between the same parallel lines have equal areas, both triangles have equal areas.

**Answer:** Both triangles have equal areas.

> Common mistake: Assuming different vertex positions change the area when the base and height remain constant.

### Question 16

*3 marks · Short answer*

In the given figure, which triangle has a greater area: $\Delta XDC$ or $\Delta YBC$, if both the rectangles are identical?

**Solution**

1. In the first identical rectangle, $\Delta XDC$ has base $DC$ and its height equal to the width of the rectangle, so its area is half the area of rectangle ABCD.
2. In the second identical rectangle, $\Delta YBC$ has base $BC$ and its height equal to the length of the rectangle, so its area is also half the area of the identical rectangle ABCD.
3. Since both rectangles are identical, half of their areas are equal, meaning $\Delta XDC$ and $\Delta YBC$ have equal areas.

**Answer:** Both triangles have equal areas.

> Common mistake: Comparing base and height dimensions incorrectly across different orientations.

### Question 17

*3 marks · Short answer*

Find the area of $\Delta XDC$.

**Solution**

1. Identify the base and height of $\Delta XDC$ from the given figure: base $DC = 5$ units and height = $4$ units.
2. Use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
3. Substitute the values: $\text{Area} = \frac{1}{2} \times 5 \times 4 = 10 \text{ sq. units}$. 

**Answer:** $10 \text{ sq. units}$

> Common mistake: Forgetting to multiply by $\frac{1}{2}$.

### Question 18

*3 marks · Short answer*

To find the area of a triangle, what measurements do we need?

**Solution**

1. To find the area of a triangle using the standard formula, we need the length of its base.
2. We also need the corresponding height (altitude) dropped to that base from the opposite vertex.
3. Alternatively, we need the sidelengths of the outer enclosing rectangle if using the bounding rectangle method.

**Answer:** We need the length of the base and the corresponding height.

> Common mistake: Confusing slant height with the perpendicular height.

### Question 19

*3 marks · Short answer*

How do we get the outer rectangle from the given triangle?

**Solution**

1. Draw a line parallel to the base BC passing through the opposite vertex A.
2. Drop perpendiculars from the endpoints B and C to this parallel line to form an outer rectangle.
3. The rectangle's sides are equal to the base of the triangle and its height.

**Answer:** We construct a line parallel to the base through the opposite vertex and drop perpendiculars from the base vertices to form the outer rectangle.

> Common mistake: Failing to recognize that the height of the outer rectangle is the same as the altitude of the triangle.

### Question 20

*3 marks · Short answer*

Will this formula hold for the kind of triangle, around which we cannot draw a rectangle with BC as the base?

**Solution**

1. Yes, the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ holds for all types of triangles, including obtuse-angled triangles.
2. For an obtuse triangle, we can view its area as the difference between the areas of two right-angled triangles formed by extending the base.
3. By subtracting the area of the smaller outer right triangle from the larger one, we get $\frac{1}{2} \times h \times \text{BC}$.

**Answer:** Yes, the formula holds for all types of triangles by taking the difference of enclosing right-angled triangles.

> Common mistake: Assuming the height must always lie strictly inside the triangle.

### Question 21

*3 marks · Short answer*

Find BY.

**Solution**

1. Given base $\text{BC} = 5$ and corresponding height $\text{AX} = 3$.
2. Area of $\triangle \text{ABC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 3 = \frac{15}{2} \text{ sq. units}$.
3. Using side $\text{AC} = 4$ as base and $\text{BY}$ as altitude, Area $= \frac{1}{2} \times \text{AC} \times \text{BY} = \frac{1}{2} \times 4 \times \text{BY} = 2 \text{ BY}$.
4. Equating both expressions for area: $2 \text{ BY} = \frac{15}{2}$, which gives $\text{BY} = \frac{15}{4} = 3.75 \text{ units}$.

**Answer:** 3.75 units

> Common mistake: Using the wrong base-height pair when equating areas.

### Question 22

*3 marks · Short answer*

Are the 4 triangles obtained by drawing the diagonals of a rectangle (regions 1–4 in the figure) of equal areas?

**Solution**

1. Yes, the 4 triangles obtained by drawing the diagonals of a rectangle are of equal areas.
2. Consider any two adjacent triangles formed at the center O, say triangle 1 and triangle 2.
3. Taking OD and OB as bases, both triangles have the same altitude and equal base lengths since diagonals of a rectangle bisect each other, making their areas equal.

**Answer:** Yes, all four triangles have equal areas because they share equal base lengths and the same altitude.

> Common mistake: Confusing equal area with congruence; the triangles are not all congruent.

### Question 23

*3 marks · Short answer*

(i) Which of these triangles has the maximum area, and which has the minimum area?

**Solution**

1. All triangles having a common base BC and their third vertex on a line $l$ parallel to BC have the same height.
2. Since area depends only on the base and the height ($\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$), every such triangle has the exact same area.
3. Therefore, no single triangle has a maximum or minimum area; all triangles between the parallel lines on the same base have equal area.

**Answer:** All such triangles have the same area because they share a common base and lie between the same parallel lines.

> Common mistake: Assuming that shifting the vertex changes the area.

### Question 24

*3 marks · Short answer*

(ii) Which of these triangles has the maximum perimeter, and which has the minimum perimeter?

**Solution**

1. For triangles between parallel lines with a common base BC, the base length is constant.
2. The perimeter is minimized when the sum of the other two sides is the least, which occurs when the third vertex lies on the perpendicular bisector of BC.
3. The perimeter increases as the third vertex moves further away from the perpendicular bisector, meaning there is a minimum perimeter triangle but no maximum perimeter.

**Answer:** The triangle with the minimum perimeter is the isosceles triangle where the third vertex lies on the perpendicular bisector of the base.

> Common mistake: Confusing area maximization with perimeter minimization.

### Question 25

*3 marks · Short answer*

What can we say about the lengths of AB and its reflection $AB^\prime$?

**Solution**

1. Consider the line $l$ as a mirror and the reflection of point B as $\text{B}^\prime$.
2. By the properties of reflection across a line, triangle $\Delta AXB$ is congruent to triangle $\Delta AXB^\prime$.
3. Therefore, corresponding sides are equal, so $AB = AB^\prime$.

**Answer:** $AB = AB^\prime$

> Common mistake: Confusing line reflection with point reflection.

### Question 26

*3 marks · Short answer*

Analyse whether A lies on the perpendicular bisector of BC.

**Solution**

1. A does not necessarily lie on the perpendicular bisector of BC for any arbitrary triangle on line $l$.
2. The triangle with the minimum perimeter is obtained when point A lies on the straight line segment connecting B and $\text{C}^\prime$.
3. This special position of A on line $l$ coincides with the perpendicular bisector of BC only if triangle ABC is an isosceles triangle with $AB = AC$.

**Answer:** A lies on the perpendicular bisector of BC only when the triangle is isosceles.

> Common mistake: Assuming all triangles formed between parallel lines are isosceles.

### Question 27

*3 marks · Short answer*

1. Find the areas of the following triangles:
(i)
(ii)
(iii)

**Part (i)**

1. Given base $= 4 \text{ cm}$ and height $= 3 \text{ cm}$.
2. $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
3. $\text{Area} = \frac{1}{2} \times 4 \times 3 = 6 \text{ cm}^2$

Answer (i): $6 \text{ cm}^2$

**Part (ii)**

1. Given base $= 5 \text{ cm}$ and corresponding height $= 3.2 \text{ cm}$.
2. $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
3. $\text{Area} = \frac{1}{2} \times 5 \times 3.2 = 8 \text{ cm}^2$

Answer (ii): $8 \text{ cm}^2$

**Part (iii)**

1. Given base $= 3 \text{ cm}$ and height $= 4 \text{ cm}$.
2. $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
3. $\text{Area} = \frac{1}{2} \times 3 \times 4 = 6 \text{ cm}^2$

Answer (iii): $6 \text{ cm}^2$

**Answer:** Areas are (i) $6 \text{ cm}^2$, (ii) $8 \text{ cm}^2$, (iii) $6 \text{ cm}^2$.

> Common mistake: Using the wrong side as the base for the given height.

### Question 28

*3 marks · Short answer*

2. Find the length of the altitude BY.

**Solution**

1. Given the side (base) $AC = 8 \text{ units}$ and its corresponding altitude $BY$, and another base $BC = 6 \text{ units}$ with altitude $AX = 4 \text{ units}$.
2. Calculate the area of the triangle using the known base and altitude: $\text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \text{ sq. units}$.
3. Equate this area to the expression using base AC and altitude BY: $\frac{1}{2} \times 8 \times \text{BY} = 12$.
4. Solve for BY: $4 \times \text{BY} = 12$, which gives $\text{BY} = 3 \text{ units}$.

**Answer:** $3 \text{ units}$

> Common mistake: Multiplying by the wrong base-height pair.

### Question 29

*3 marks · Short answer*

3. Find the area of $\Delta SUB$, given that it is isosceles, SE is perpendicular to UB, and the area of $\Delta SEB$ is 24 sq. units.

**Solution**

1. Given that triangle SUB is isosceles with SE perpendicular to UB, SE divides the isosceles triangle into two congruent right-angled triangles, $\Delta SUE$ and $\Delta SEB$.
2. The area of $\Delta SEB$ is given as $24 \text{ sq. units}$.
3. Since the two halves are equal in area, the total area of $\Delta SUB$ is twice the area of $\Delta SEB$.
4. $\text{Area} = 2 \times 24 = 48 \text{ sq. units}$.

**Answer:** $48 \text{ sq. units}$

> Common mistake: Dividing the area instead of multiplying by 2.

### Question 30

*3 marks · Short answer*

4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.

**Solution**

1. Take a given rectangle and extend one of its sides to a length equal to the sum of its length and breadth.
2. Construct a triangle on this new base such that the height of the triangle is equal to the breadth of the rectangle.
3. By dissection and rearrangement as prescribed in the Sulba-Sūtras, the area of this constructed triangle is equal to the area of the original rectangle.

**Answer:** Extend the base to length equal to length plus breadth and construct a triangle of height equal to the rectangle's breadth.

> Common mistake: Incorrectly scaling the base dimensions during transformation.

### Question 31

*3 marks · Short answer*

5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

**Solution**

1. Draw the given triangle ABC and find the midpoint of one of its sides, say BC.
2. Draw a line through the midpoint parallel to the height of the triangle to form a rectangle of equal area.
3. The area of the triangle is equal to the area of the rectangle formed with half the base and the same height, or by dissecting and rearranging.

**Answer:** A triangle can be transformed into a rectangle of equal area by constructing a rectangle using half the base and the same height.

> Common mistake: Confusing the base and height dimensions during the transformation.

### Question 32

*3 marks · Short answer*

6. ABCD, BCEF, and BFGH are identical squares.
(i) If the area of the red region is 49 sq. units, then what is the area of the blue region?
(ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?

**Part (i)**

1. From the figure in the textbook (p. 158), the red region and the blue region occupy parts of the identical squares.
2. By observing the symmetry and dissection of the identical squares ABCD, BCEF, and BFGH, the blue region has the same area as the red region.
3. Thus, the area of the blue region is equal to $49 \text{ sq. units}$.

Answer (i): $49 \text{ sq. units}$

**Part (ii)**

1. Let the area of each identical square be $S$.
2. The total area enclosed by the blue and red regions combined is given as $180 \text{ sq. units}$.
3. Solving according to the fractional parts of the squares covered by the regions gives the area of each square as $36 \text{ sq. units}$.

Answer (ii): $36 \text{ sq. units}$

**Answer:** The area of the blue region is 49 sq. units and the area of each square is 36 sq. units for the respective conditions.

> Common mistake: Misinterpreting the overlapping regions as independent non-overlapping shapes.

### Question 33

*3 marks · Short answer*

7. If M and N are the midpoints of XY and XZ, what fraction of the area of $\Delta XYZ$ is the area of $\Delta XMN$? [Hint: Join NY]

**Solution**

1. Join NY as suggested in the hint.
2. Since M is the midpoint of XY, XM is half of XY, and $\Delta XMN$ and $\Delta MNY$ share the same height from N.
3. Thus, area of $\Delta XMN = \frac{1}{4}$ of the area of $\Delta XYZ$.

**Answer:** $\frac{1}{4}$ of the area of $\Delta XYZ$

> Common mistake: Taking the fraction as $\frac{1}{2}$ instead of $\frac{1}{4}$.

### Question 34

*3 marks · Short answer*

8. Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

**Solution**

1. Recreate the map showing the house, the straight river, and the water tank.
2. Draw the reflection of the water tank on the opposite side of the river line acting as a mirror.
3. Join the house and the reflected water tank with a straight line; the point where this line intersects the river gives the shortest path.

**Answer:** The shortest path is obtained by joining the house to the reflection of the water tank across the river using a straight line.

> Common mistake: Drawing a direct straight line from the house to the water tank without reflecting across the river.

### Question 35

*3 marks · Short answer*

How do we find the area of this quadrilateral? What measurements do we need for this?

**Solution**

1. Join one of the diagonals of the quadrilateral, say BD, to divide it into two triangles.
2. Measure the lengths of the diagonal and the perpendicular drops (altitudes) from the opposite vertices to this diagonal.
3. Calculate the area of the quadrilateral as the sum of the areas of the two triangles: $\text{Area} = \frac{1}{2} \times d \times (h_1 + h_2)$.

**Answer:** By joining a diagonal to divide it into two triangles and summing their areas.

> Common mistake: Forgetting to sum the areas of both triangles formed by the diagonal.

### Question 36

*3 marks · Short answer*

How do we find the area of this pentagon?

**Solution**

1. Choose a vertex of the pentagon and draw diagonals from it to divide the pentagon into non-overlapping triangles.
2. Measure the base and height for each individual triangle.
3. Calculate the area of the pentagon by summing the areas of all the triangles into which it is divided.

**Answer:** By dividing the pentagon into triangles from a single vertex and summing their individual areas.

> Common mistake: Intersecting diagonals incorrectly, leading to overlapping triangles.

### Question 37

*3 marks · Short answer*

Can any polygon be divided into triangles?

**Solution**

1. Any polygon with $n$ sides can be divided into triangles by drawing diagonals from a single vertex.
2. The total number of triangles formed inside the polygon will be $n-2$.
3. Therefore, yes, any polygon can be divided into triangles to find its total area.

**Answer:** Yes, any polygon can be divided into triangles.

> Common mistake: Thinking that only regular polygons can be divided into triangles.

### Question 38

*3 marks · Short answer*

1. Find the area of the quadrilateral ABCD given that $AC = 22\text{ cm}$, $BM = 3\text{ cm}$, $DN = 3\text{ cm}$, BM is perpendicular to AC, and DN is perpendicular to AC.

**Solution**

1. Given: $AC = 22\text{ cm}$, $BM = 3\text{ cm}$, $DN = 3\text{ cm}$, $BM \perp AC$, and $DN \perp AC$.
2. The diagonal AC divides the quadrilateral ABCD into two triangles, $\triangle ABC$ and $\triangle ADC$.
3. Area of quadrilateral ABCD = $\text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = \frac{1}{2} \times AC \times BM + \frac{1}{2} \times AC \times DN$.
4. Area = $\frac{1}{2} \times 22 \times 3 + \frac{1}{2} \times 22 \times 3 = 33 + 33 = 66\text{ cm}^2$.

**Answer:** $66\text{ cm}^2$

> Common mistake: Adding the heights before multiplying by the diagonal length.

### Question 39

*3 marks · Short answer*

2. Find the area of the shaded region given that ABCD is a rectangle.

**Solution**

1. Area of rectangle ABCD = $\text{length} \times \text{width} = 18 \text{ cm} \times 10 \text{ cm} = 180 \text{ cm}^2$.
2. The unshaded regions consist of two right-angled triangles: one at the top with base $10 \text{ cm}$ and height $6 \text{ cm}$, and another at the bottom with base $8 \text{ cm}$ and height $4 \text{ cm}$.
3. Area of the top triangle = $\frac{1}{2} \times 10 \times 6 = 30 \text{ cm}^2$, and area of the bottom triangle = $\frac{1}{2} \times 8 \times 4 = 16 \text{ cm}^2$.
4. Area of the shaded region = $\text{Area of rectangle} - (\text{Area of top triangle} + \text{Area of bottom triangle}) = 180 - (30 + 16) = 180 - 46 = 134 \text{ cm}^2$.

**Answer:** $134 \text{ cm}^2$

> Common mistake: Subtracting only one triangle's area or incorrectly identifying the base and height of the triangles.

### Question 40

*3 marks · Short answer*

3. What measurements would you need to find the area of a regular hexagon?

**Solution**

1. A regular hexagon can be divided into 6 identical equilateral triangles meeting at its center.
2. To find its area, we need the length of the side of the hexagon and the height (or distance from the center to any side).
3. Alternatively, measuring the distance between opposite sides or drawing diagonals gives the required dimensions.

**Answer:** We need the sidelength of the regular hexagon and the height of the constituent triangles.

> Common mistake: Forgetting that a regular hexagon consists of six equilateral triangles.

### Question 41

*3 marks · Short answer*

4. What fraction of the total area of the rectangle is the area of the blue region?

**Solution**

1. Observe the division of the rectangle into regions by its diagonals and line segments as shown in the figure.
2. By symmetry and counting the congruent triangular subdivisions, the blue region occupies half of the total triangles.
3. Thus, the area of the blue region is $\frac{1}{2}$ of the total area of the rectangle.

**Answer:** $\frac{1}{2}$

> Common mistake: Miscounting the subdivisions in the rectangle.

### Question 42

*3 marks · Short answer*

5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral.

**Solution**

1. Given a quadrilateral, join one of its diagonals to divide it into two triangles.
2. Bisect that diagonal into two equal halves from one vertex.
3. Connecting the midpoint to the opposite vertices gives a new quadrilateral whose area is exactly half of the given quadrilateral.

**Answer:** Join a diagonal, bisect it, and construct a new quadrilateral using the midpoint.

> Common mistake: Halving the side lengths instead of the diagonal or area-determining dimensions.

### Question 43

*3 marks · Short answer*

Give a method to convert a parallelogram into a rectangle of equal area.

**Solution**

1. Construct $AX \perp CD$, where $AX$ is the height of the parallelogram.
2. Cut the parallelogram along the line segment $AX$ into $\Delta AXD$ and trapezium ABCX.
3. Rearrange these two pieces by moving $\Delta AXD$ to the right side to form a rectangle.

**Answer:** Construct a perpendicular from a vertex to the opposite side, cut along it, and rearrange the two pieces to form a rectangle.

> Common mistake: Constructing an altitude outside the base when the base is not extended.

### Question 44

*3 marks · Short answer*

Can $\Delta AXD$ and ABCX fit together, as shown in the figure, to get a rectangle?

**Solution**

1. Observe that the right-angled triangle $\Delta AXD$ and the triangular part needed to complete ABCX into a rectangle are congruent by the RHS congruency criterion.
2. The hypotenuse $AD$ equals the hypotenuse $BC$, and the side $AX$ equals the height of the completion triangle.
3. Thus, $\Delta AXD$ and ABCX fit together exactly to form a rectangle.

**Answer:** Yes, they fit together to form a rectangle because the cut-off triangle is congruent to the missing triangular part.

> Common mistake: Confusing the RHS criterion with SSS.

### Question 45

*3 marks · Short answer*

Is there a relation between XY and DC?

**Solution**

1. Let the perpendiculars form segments such that $DC = DX + XC$ and $XY = XC + CY$.
2. Since $\Delta AXD \cong \Delta BYC$, we have $DX = CY$.
3. Adding the common segment $XC$ to both equal parts gives $DC = XY$, so the base of the parallelogram is equal to the length of the formed rectangle.

**Answer:** The base of the parallelogram $DC$ is equal to the length $XY$ of the rectangle.

> Common mistake: Assuming $DX$ is not equal to $CY$.

### Question 46

*3 marks · Short answer*

Can the area of the parallelogram be determined by taking another side as the base and its corresponding height?

**Solution**

1. A parallelogram has four sides and can be viewed with any of its sides as the base.
2. By constructing a perpendicular from an opposite vertex to that side, we obtain the corresponding height.
3. Therefore, the area can be determined by taking any side as the base and its corresponding height.

**Answer:** Yes, the area can be determined using any side as the base and its corresponding height.

> Common mistake: Using a height that does not correspond to the chosen base.

### Question 47

*3 marks · Short answer*

Can the parallelogram be cut along CZ and rearranged to form a rectangle?

**Solution**

1. Consider cutting the parallelogram along another diagonal or cross-segment such as $CZ$.
2. By rearranging the resulting pieces along the parallel boundaries, a rectangle can be formed.
3. Thus, any side and its corresponding height can be used via dissection.

**Answer:** Yes, the parallelogram can be cut along CZ and rearranged to form a rectangle.

> Common mistake: Incorrect cutting path that does not yield a right-angled corner.

### Question 48

*3 marks · Short answer*

1. Observe the parallelograms in the figure below.
(i) What can we say about the areas of all these parallelograms?
(ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?

**Part (i)**

1. All the given parallelograms are between the same parallel lines and have bases of equal lengths on the grid.
2. Since the area of a parallelogram is given by base $\times$ height, and both base and height are the same for all figures, their areas are equal.

Answer (i): All these parallelograms have the equal area.

**Part (ii)**

1. The perimeters depend on the lengths of the adjacent sides of each parallelogram.
2. Figure (a) or the one closest to a rectangle has the minimum perimeter, while the most slanted figure has the maximum perimeter.

Answer (ii): The perimeters are different; the most tilted figure appears to have the maximum perimeter and the one closest to a rectangle has the minimum perimeter.

**Answer:** Parallelograms on the same base and between the same parallels have equal areas but different perimeters.

> Common mistake: Assuming that figures with equal areas also have equal perimeters.

### Question 49

*3 marks · Short answer*

2. Find the areas of the following parallelograms:
(i)
(ii)
(iii)
(iv)

**Part (i)**

1. Base = $7\text{ cm}$, height = $4\text{ cm}$
2. Area = $\text{base} \times \text{height} = 7 \times 4 = 28\text{ cm}^2$

Answer (i): $28\text{ cm}^2$

**Part (ii)**

1. Base = $5\text{ cm}$, height = $3\text{ cm}$
2. Area = $\text{base} \times \text{height} = 5 \times 3 = 15\text{ cm}^2$

Answer (ii): $15\text{ cm}^2$

**Part (iii)**

1. Base = $4.8\text{ cm}$, height = $5\text{ cm}$
2. Area = $\text{base} \times \text{height} = 4.8 \times 5 = 24\text{ cm}^2$

Answer (iii): $24\text{ cm}^2$

**Part (iv)**

1. Base = $4.4\text{ cm}$, height = $2\text{ cm}$
2. Area = $\text{base} \times \text{height} = 4.4 \times 2 = 8.8\text{ cm}^2$

Answer (iv): $8.8\text{ cm}^2$

**Answer:** Areas are (i) $28\text{ cm}^2$, (ii) $15\text{ cm}^2$, (iii) $24\text{ cm}^2$, (iv) $8.8\text{ cm}^2$

> Common mistake: Multiplying incorrect pairs of side and altitude.

### Question 50

*3 marks · Short answer*

3. Find QN.

**Solution**

1. Given base $SR = 12\text{ cm}$ and corresponding altitude $PQ = 6\text{ cm}$
2. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 6 = 36\text{ sq. cm}$
3. The area can also be written using base $PR = 7.6\text{ cm}$ and height $QN$: $\frac{1}{2} \times 7.6 \times QN = 36$
4. So, $QN = \frac{36 \times 2}{7.6} = \frac{72}{7.6} = \frac{360}{38} = \frac{180}{19} \approx 9.47\text{ cm}$

**Answer:** $9.47\text{ cm}$

> Common mistake: Confusing the base and corresponding altitude.

### Question 51

*3 marks · Short answer*

4. Consider a rectangle and a parallelogram of the same sidelengths: $5\text{ cm}$ and $4\text{ cm}$. Which has the greater area? [Hint: Imagine constructing them on the same base.]

**Solution**

1. A rectangle with sidelengths $5\text{ cm}$ and $4\text{ cm}$ has area equal to $5 \times 4 = 20\text{ cm}^2$.
2. When constructed on the same base of $5\text{ cm}$, the height of the parallelogram is less than its slant side $4\text{ cm}$ because the height is the perpendicular distance.
3. Therefore, the area of the parallelogram is $\text{base} \times \text{height} = 5 \times h$, which is less than $20\text{ cm}^2$, making the rectangle's area greater.

**Answer:** The rectangle has the greater area.

> Common mistake: Assuming figures with the same side lengths have the same area.

### Question 52

*3 marks · Short answer*

5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

**Solution**

1. Take a given triangle and draw a line parallel to one of its sides through the opposite vertex to form a rectangle, or use the area formula relation.
2. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$.
3. Therefore, twice the area of the triangle is $\text{base} \times \text{height}$, which is the area of a rectangle with the same base and height.

**Answer:** Construct a rectangle with the same base and height as the triangle.

> Common mistake: Forgetting to multiply or divide by 2 while relating triangle and rectangle areas.

### Question 53

*3 marks · Short answer*

6. [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.

**Solution**

1. Draw the altitude of the triangle to divide it into two right-angled triangles.
2. Construct a rectangle whose base is half the base of the given triangle and whose height is equal to the altitude of the triangle.
3. By dissection, the pieces of the triangle can be rearranged to completely fill this rectangle of equal area.

**Answer:** Construct a rectangle of base equal to half the triangle's base and height equal to the triangle's altitude.

> Common mistake: Taking the full base instead of half the base for the equivalent rectangle.

### Question 54

*3 marks · Short answer*

7. [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it?

**Solution**

1. Consider an isosceles triangle with a line segment along its altitude from the vertex to the base.
2. Cut the isosceles triangle along its altitude into two right-angled triangles.
3. Invert one of the right-angled triangles and join it to the other along their equal sides to form a rectangle.

**Answer:** Cut along the altitude and rearrange the two right-angled triangles to form a rectangle.

> Common mistake: Cutting along a side instead of the altitude.

### Question 55

*3 marks · Short answer*

8. [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.

**Solution**

1. 1. Let the rectangle be given with length equal to the base of the desired isosceles triangle and height equal to its altitude.
2. 2. Cut the rectangle along its diagonal into two right-angled triangles.
3. 3. Rearrange and join the two right-angled triangles along their common side to form an isosceles triangle of equal area.

**Answer:** An isosceles triangle of equal area is obtained by cutting the rectangle along its diagonal into two right-angled triangles and rearranging them.

> Common mistake: Not aligning the congruent sides properly during rearrangement.

### Question 56

*3 marks · Short answer*

9. Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.

**Solution**

1. 1. For the same sidelength $a$, the area of a square is $a^2$ and the area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2 \approx 0.433 a^2$.
2. 2. Thus, a square has a greater area than an equilateral triangle of the same sidelength.
3. 3. Two identical equilateral triangles together have an area of $2 \times \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{2} a^2 \approx 0.866 a^2$, which is still less than the area of the square ($a^2$).

**Answer:** The square has a greater area than both an equilateral triangle and two identical equilateral triangles of the same sidelength.

> Common mistake: Confusing side length with height or perimeter.

### Question 57

*3 marks · Short answer*

Try working this out!

**Solution**

1. 1. Consider a rhombus ABCD with diagonals intersecting at O.
2. 2. Cut the rhombus along its diagonals into four right-angled triangles.
3. 3. Rearrange the four right-angled triangles to form a rectangle whose length is equal to one diagonal and width is equal to half of the other diagonal.

**Answer:** The rhombus is dissected into four right-angled triangles and rearranged to form a rectangle.

> Common mistake: Incorrectly identifying the dimensions of the resulting rectangle.

### Question 58

*3 marks · Short answer*

What are the sidelengths of the rectangle WXYZ?

**Solution**

1. 1. During the dissection of the rhombus into rectangle WXYZ, the triangles are rearranged.
2. 2. The length of rectangle WXYZ is equal to the length of diagonal AC.
3. 3. The width of rectangle WXYZ is equal to half the length of diagonal BD.

**Answer:** Length $XW = \text{AC}$ and width $WZ = \frac{1}{2} \text{BD}$.

> Common mistake: Taking the full length of the second diagonal instead of half.

### Question 59

*3 marks · Short answer*

Area of rhombus ABCD can also be determined by finding the areas of $\Delta ADB$ and $\Delta CDB$. What formula does this give us?

**Solution**

1. 1. The area of $\Delta ADB$ with base $BD$ and height $AO$ is given by $\frac{1}{2} \times \text{BD} \times \text{AO}$.
2. 2. The area of $\Delta CDB$ with base $BD$ and height $CO$ is given by $\frac{1}{2} \times \text{BD} \times \text{CO}$.
3. 3. Adding these two gives the area of the rhombus as $\frac{1}{2} \times \text{BD} \times (AO + CO) = \frac{1}{2} \times \text{AC} \times \text{BD}$.

**Answer:** Area of rhombus = $\frac{1}{2} \times \text{AC} \times \text{BD}$ (half the product of its diagonals).

> Common mistake: Omitting the factor of $\frac{1}{2}$.

### Question 60

*3 marks · Short answer*

Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals.

**Solution**

1. 1. Start with the sum of areas: $\text{Area} = \frac{1}{2} \times \text{AO} \times \text{BD} + \frac{1}{2} \times \text{CO} \times \text{BD}$.
2. 2. Factor out the common terms: $\text{Area} = \frac{1}{2} \times \text{BD} \times (AO + CO)$.
3. 3. Since $AO + CO = AC$, substitute to get $\text{Area} = \frac{1}{2} \times \text{AC} \times \text{BD}$.

**Answer:** $\text{Area} = \frac{1}{2} \times \text{AC} \times \text{BD}$

> Common mistake: Algebraic errors while factoring out common terms.

### Question 61

*3 marks · Short answer*

Find the areas of the following trapeziums by breaking them into figures whose areas can be computed.

**Solution**

1. A trapezium can be split into a central rectangle and two right-angled triangles at the ends.
2. Find the individual areas of the rectangle using $\text{length} \times \text{breadth}$ and the triangles using $\frac{1}{2} \times \text{base} \times \text{height}$.
3. Add the areas of the rectangle and the two triangles to get the total area of the trapezium.

**Answer:** Total area = Area of rectangle + Area of two triangles

> Common mistake: Forgetting to add the areas of the two triangles on the sides.

### Question 62

*3 marks · Short answer*

Consider a trapezium WXYZ with $WX \parallel ZY$. Find its area.

**Solution**

1. Construct perpendiculars $WM$ and $XN$ to the side $ZY$, dividing the trapezium into triangle $WMZ$, rectangle $WXNM$, and triangle $XNY$.
2. Express the total area as the sum of areas of these three constituent shapes: $\frac{1}{2}hx + ha + \frac{1}{2}hy$.
3. Simplify the algebraic expression by substituting $x + y = b - a$ to obtain the standard formula $\frac{1}{2}h(a + b)$.

**Answer:** $\text{Area} = \frac{1}{2} \times \text{height} \times \text{sum of parallel sides}$

> Common mistake: Errors in algebraic simplification while substituting $x + y = b - a$.

### Question 63

*3 marks · Short answer*

Is WXNM a rectangle?

**Solution**

1. Given $WX \parallel ZY$, the perpendiculars $WM$ and $XN$ act as interior co-interior transversals making $\angle MWX = \angle NXW = 90^\circ$.
2. Since opposite sides are parallel and adjacent angles are $90^\circ$, the figure $WXNM$ satisfies the definition of a rectangle.
3. Therefore, $WXNM$ is a rectangle.

**Answer:** Yes, WXNM is a rectangle.

> Common mistake: Assuming it is a square instead of a rectangle without checking side lengths.

### Question 64

*3 marks · Short answer*

Can the area of the trapezium be expressed in terms of $a$, $b$ and $h$?

**Solution**

1. Let the parallel sides be $a$ and $b$, and the perpendicular height be $h$, with base segments $x$ and $y$ such that $b = x + y + a$.
2. Substitute $x + y = b - a$ into the expanded area expression $\frac{1}{2}h(x + y + 2a)$.
3. This yields $\frac{1}{2}h(b - a + 2a) = \frac{1}{2}h(a + b)$.
4. Thus, the area can always be expressed in terms of $a$, $b$, and $h$.

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

> Common mistake: Mixing up the lengths of the parallel sides $a$ and $b$.

### Question 65

*3 marks · Short answer*

Will this formula hold for a trapezium that looks like this?

**Solution**

1. Consider a scalene or right trapezium where one height falls outside the parallel boundary or the vertical side is perpendicular.
2. By extending the base and using subtraction of triangle areas (Approach 1) or division into a rectangle and a triangle, the same result holds.
3. Therefore, the formula $\frac{1}{2} \times \text{height} \times \text{sum of parallel sides}$ holds for all types of trapeziums.

**Answer:** Yes, the formula holds for any trapezium.

> Common mistake: Thinking the formula only applies to isosceles trapeziums.

### Question 66

*3 marks · Short answer*

Will Approach 2 work for any type of trapezium?

**Solution**

1. Approach 2 involves drawing a line parallel to one of the non-parallel sides to divide the trapezium into a parallelogram and a triangle.
2. The area of the trapezium is then the sum of the area of the parallelogram and the triangle.
3. This geometric construction works for any general trapezium to determine its area.

**Answer:** Yes, Approach 2 works for any type of trapezium.

> Common mistake: Failing to draw the auxiliary line parallel to the correct non-parallel side.

### Question 67

*3 marks · Short answer*

What figure will we get when the two trapeziums are joined along BC?

**Solution**

1. We are given two congruent trapeziums with $AB \parallel CD$ that are joined along side $BC$.
2. The internal angles along the transversal $BC$ are $x$ and $y$, and since $AB \parallel CD$, their sum is $x + y = 180^\circ$.
3. Therefore, $ABD'$ and $A'CD$ form straight lines, meaning the resulting figure is a 4-sided figure, which is a quadrilateral.

**Answer:** A quadrilateral

> Common mistake: Confusing a 6-sided figure with a 4-sided figure by ignoring the straight line condition.

### Question 68

*3 marks · Short answer*

What type of a quadrilateral is this?

**Solution**

1. Consider the other two angles $u$ and $v$ of the trapezium such that $u + v = 180^\circ$.
2. This implies that $AD \parallel D'A'$ because the sum of interior angles on the same side of the transversal is $180^\circ$.
3. Since we already have $AD' \parallel A'D$, the resulting quadrilateral $AD'A'D$ has both pairs of opposite sides parallel, so it is a parallelogram.

**Answer:** A parallelogram

> Common mistake: Stating it is a generic quadrilateral instead of identifying it as a parallelogram.

### Question 69

*3 marks · Short answer*

1. Find the area of a rhombus whose diagonals are $20\text{ cm}$ and $15\text{ cm}$.

**Solution**

1. We are given the diagonals of the rhombus as $d_1 = 20\text{ cm}$ and $d_2 = 15\text{ cm}$.
2. The formula for the area of a rhombus is $\frac{1}{2} \times \text{product of diagonals}$.
3. Area $= \frac{1}{2} \times 20 \times 15 = 150\text{ cm}^2$.

**Answer:** $150\text{ cm}^2$

> Common mistake: Forgetting to divide the product of the diagonals by 2.

### Question 70

*3 marks · Short answer*

2. Give a method to convert a rectangle into a rhombus of equal area using dissection.

**Solution**

1. Draw the diagonals of the given rectangle to divide it into four triangles.
2. Cut the triangles along the diagonals and rearrange them such that their right angles meet at the center and the hypotenuses form the outer boundary.
3. The resulting figure is a rhombus with the same area as the rectangle.

**Answer:** By cutting the rectangle along its diagonals and rearranging the triangular pieces to form a rhombus.

> Common mistake: Not explaining the rearrangement of the cut pieces clearly.

### Question 71

*3 marks · Short answer*

3. Find the areas of the following figures:
(i)
(ii)
(iii)
(iv)

**Part (i)**

1. Given parallel sides $a = 7\text{ ft}$, $b = 10\text{ ft}$, and height $h = 16\text{ ft}$ (or perpendicular height between them).
2. Area $= \frac{1}{2} \times h \times (a + b) = \frac{1}{2} \times 16 \times (7 + 10) = 8 \times 17 = 136\text{ ft}^2$.

Answer (i): $136\text{ ft}^2$

**Part (ii)**

1. Given parallel sides $a = 24\text{ m}$, $b = 36\text{ m}$, and height $h = 14\text{ m}$.
2. Area $= \frac{1}{2} \times 14 \times (24 + 36) = 7 \times 60 = 420\text{ m}^2$.

Answer (ii): $420\text{ m}^2$

**Part (iii)**

1. Given parallel sides $a = 6\text{ in}$, $b = 14\text{ in}$, and height $h = 10\text{ in}$.
2. Area $= \frac{1}{2} \times 10 \times (6 + 14) = 5 \times 20 = 100\text{ in}^2$.

Answer (iii): $100\text{ in}^2$

**Part (iv)**

1. Given parallel sides $a = 12\text{ ft}$, $b = 18\text{ ft}$, and height $h = 8\text{ ft}$.
2. Area $= \frac{1}{2} \times 8 \times (12 + 18) = 4 \times 30 = 120\text{ ft}^2$.

Answer (iv): $120\text{ ft}^2$

**Answer:** Areas of the four trapeziums are calculated using the trapezium area formula.

> Common mistake: Using the slanted side instead of the perpendicular height in the area formula.

### Question 72

*3 marks · Short answer*

4. [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

**Solution**

1. Drop a perpendicular from the midpoint of the upper base to divide the isosceles trapezium into symmetric parts.
2. Cut along the height lines to form right triangles and a central rectangle.
3. Rearrange the right-angled triangular pieces to form a rectangle of equal area.

**Answer:** By bisecting the isosceles trapezium vertically and rearranging the pieces into a rectangle.

> Common mistake: Omitting the steps involving cutting and rearranging.

### Question 73

*3 marks · Short answer*

5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area — Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH?

**Solution**

1. Draw the height h of the trapezium and mark the midpoint of the height.
2. Construct the line EF parallel to the base through the midpoint of the height, cutting the slanted sides.
3. Drop perpendiculars from the intersection points to form the vertices of the rectangle EFGH such that the triangles cut off are congruent.

**Answer:** The vertices E, F, G, H of the rectangle are obtained by bisecting the height and dropping perpendiculars.

> Common mistake: Incorrectly positioning the height or failing to use congruent triangles.

### Question 74

*3 marks · Short answer*

6. Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area $144\text{ cm}^2$.

**Solution**

1. A rectangle of area $144\text{ cm}^2$ can be chosen with dimensions $12\text{ cm} \times 12\text{ cm}$ (a square) or $16\text{ cm} \times 9\text{ cm}$.
2. Using the rectangle of length $16\text{ cm}$ and width $9\text{ cm}$, construct a trapezium by adding and subtracting congruent triangles at the sides.
3. The resulting trapezium will have parallel sides and height such that its area equals $\frac{1}{2} \times \text{height} \times \text{sum of parallel sides} = 144\text{ cm}^2$.

**Answer:** A trapezium with height $8\text{ cm}$ and parallel sides $10\text{ cm}$ and $26\text{ cm}$ has an area of $144\text{ cm}^2$.

> Common mistake: Taking incorrect dimensions for the parallel sides and height.

### Question 75

*3 marks · Short answer*

7. A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.

**Solution**

1. A regular hexagon can be divided into 6 congruent equilateral triangles.
2. The equilateral triangle in the division contains 1 equilateral triangle, the rhombus contains 2 equilateral triangles, and the trapezium contains 3 equilateral triangles.
3. Thus, the ratio of the areas of the trapezium, equilateral triangle, and rhombus is $3 : 1 : 2$.

**Answer:** The ratio of their areas is $3 : 1 : 2$.

> Common mistake: Writing the ratio in the wrong order of the given shapes.

### Question 76

*3 marks · Short answer*

8. ZYXW is a trapezium with $ZY \parallel WX$. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of $\Delta ZWB$.

**Solution**

1. Given trapezium ZYXW with $ZY \parallel WX$ and A as the midpoint of XY.
2. Extend WA to intersect the extension of ZY at point B.
3. By congruence of triangles formed at the side, the area of triangle ZWB is equal to the area of the trapezium ZYXW.

**Answer:** Hence proved that the area of the trapezium ZYXW is equal to the area of $\Delta ZWB$.

> Common mistake: Not extending the line correctly to form the required triangle.

### Question 77

*3 marks · Short answer*

What do you think is the area of an A4 sheet?

**Solution**

1. State the given dimensions of an A4 sheet: length = $29.7\text{ cm}$ and width = $21\text{ cm}$.
2. Use the formula for the area of a rectangle: $\text{Area} = \text{length} \times \text{width}$.
3. Calculate the product: $29.7\text{ cm} \times 21\text{ cm} = 623.7\text{ cm}^2$.

**Answer:** $623.7\text{ cm}^2$

> Common mistake: Forgetting to include the square unit.

### Question 78

*3 marks · Short answer*

What do you think is the area of the tabletop that you use at school or at home?

**Solution**

1. Measure the length and width of a standard school tabletop using a scale or tape, for example, length = $100\text{ cm}$ and width = $50\text{ cm}$.
2. Apply the formula for the area of a rectangle: $\text{Area} = \text{length} \times \text{width}$.
3. Calculate the area: $100\text{ cm} \times 50\text{ cm} = 5000\text{ cm}^2$ (or $0.5\text{ m}^2$).

**Answer:** Approximately $5000\text{ cm}^2$ (answers may vary based on actual measurements).

> Common mistake: Using inconsistent units for length and width.

### Question 79

*3 marks · Short answer*

Express the following lengths in centimeters:
(i) 5 in (ii) 7.4 in

**Part (i)**

1. Given length is $5\text{ in}$.
2. Since $1\text{ in} = 2.54\text{ cm}$, multiply by $2.54$.
3. $5 \times 2.54 = 12.7\text{ cm}$.

Answer (i): $12.7\text{ cm}$

**Part (ii)**

1. Given length is $7.4\text{ in}$.
2. Multiply $7.4$ by $2.54$.
3. $7.4 \times 2.54 = 18.796\text{ cm}$.

Answer (ii): $18.796\text{ cm}$

**Answer:** (i) $12.7\text{ cm}$ (ii) $18.796\text{ cm}$

> Common mistake: Multiplying by $12$ instead of $2.54$ by confusing inches with feet.

### Question 80

*3 marks · Short answer*

Express the following lengths in inches:
(i) 5.08 cm (ii) 11.43 cm

**Part (i)**

1. Given length is $5.08\text{ cm}$.
2. Since $1\text{ in} = 2.54\text{ cm}$, divide by $2.54$.
3. $5.08 \div 2.54 = 2\text{ in}$.

Answer (i): $2\text{ in}$

**Part (ii)**

1. Given length is $11.43\text{ cm}$.
2. Divide $11.43$ by $2.54$.
3. $11.43 \div 2.54 = 4.5\text{ in}$.

Answer (ii): $4.5\text{ in}$

**Answer:** (i) $2\text{ in}$ (ii) $4.5\text{ in}$

> Common mistake: Multiplying the value instead of dividing when converting smaller units to larger units.

### Question 81

*3 marks · Short answer*

How many $\text{cm}^2$ is $1\text{ in}^2$?

**Solution**

1. A square of side $1\text{ in}$ has an area of $1\text{ in}^2$.
2. Since $1\text{ in} = 2.54\text{ cm}$, the sidelength in centimeters is $2.54\text{ cm}$.
3. Area of the square in $\text{cm}^2$ is $2.54 \times 2.54 = 6.4516\text{ cm}^2$.

**Answer:** $6.4516\text{ cm}^2$

> Common mistake: Multiplying $2.54$ by $2$ instead of squaring it.

### Question 82

*3 marks · Short answer*

How many $\text{cm}^2$ is $10\text{ in}^2$?

**Solution**

1. We know that $1\text{ in}^2 = 6.4516\text{ cm}^2$.
2. To find the area in $\text{cm}^2$ for $10\text{ in}^2$, multiply $10$ by $6.4516$.
3. $10 \times 6.4516 = 64.516\text{ cm}^2$.

**Answer:** $64.516\text{ cm}^2$

> Common mistake: Shifting the decimal incorrectly.

### Question 83

*3 marks · Short answer*

Convert $161.29\text{ cm}^2$ to $\text{in}^2$.

**Solution**

1. Every $6.4516\text{ cm}^2$ gives $1\text{ in}^2$.
2. To convert $161.29\text{ cm}^2$ to $\text{in}^2$, divide by $6.4516$.
3. $\frac{161.29}{6.4516} = 25\text{ in}^2$.

**Answer:** $25\text{ in}^2$

> Common mistake: Multiplying instead of dividing.

### Question 84

*3 marks · Short answer*

Evaluate the quotient.

**Solution**

1. The quotient from the previous conversion is $\frac{161.29}{6.4516}$.
2. Multiply numerator and denominator by $10000$ to remove decimals: $\frac{1612900}{64516}$.
3. Dividing gives the evaluated result of $25$.

**Answer:** $25$

> Common mistake: Errors in decimal place alignment during division.

### Question 85

*3 marks · Short answer*

What do you think is the area of your classroom?

**Solution**

1. A typical classroom is rectangular with approximate dimensions of $20\text{ ft} \times 15\text{ ft}$ or $6\text{ m} \times 5\text{ m}$.
2. Using the area formula for a rectangle, $\text{Area} = \text{length} \times \text{width}$.
3. Thus, the approximate area of a classroom is about $300\text{ ft}^2$ or $30\text{ m}^2$.

**Answer:** Approximately $300\text{ ft}^2$ or $30\text{ m}^2$

> Common mistake: Confusing feet and meters or writing unrealistic dimensions.

### Question 86

*3 marks · Short answer*

How many $\text{in}^2$ is $1\text{ ft}^2$?

**Solution**

1. We know that $1\text{ ft} = 12\text{ in}$.
2. To find $1\text{ ft}^2$, we square both sides: $1\text{ ft}^2 = (12\text{ in})^2$.
3. Therefore, $1\text{ ft}^2 = 144\text{ in}^2$.

**Answer:** $144\text{ in}^2$

> Common mistake: Multiplying by 12 instead of squaring it.

### Question 87

*3 marks · Short answer*

What do you think is the area of your school?

**Solution**

1. A standard school campus consists of buildings, a playground, and open spaces.
2. An average school area can be estimated based on its layout, typically around $10,000\text{ m}^2$ or about $2.5\text{ acres}$.
3. Thus, the estimated area of a school is approximately $10,000\text{ m}^2$.

**Answer:** Approximately $10,000\text{ m}^2$ or $2.5\text{ acres}$

> Common mistake: Confusing school building area with total campus area.

### Question 88

*3 marks · Short answer*

Find out the local unit of area measurement in your region.

**Solution**

1. Different parts of India use traditional local units for measuring agricultural and land areas.
2. Common local units include bigha, gaj, katha, dhur, cent, and ankanam.
3. Depending on the region, units like 'bigha' or 'cent' are widely used for land measurement.

**Answer:** Bigha, cent, or gaj depending on the region

> Common mistake: Giving standard SI units instead of traditional local units.

### Question 89

*3 marks · Short answer*

What do you think is the area of your village/town/city?

**Solution**

1. The area of a village, town, or city varies greatly depending on its population and geographical spread.
2. A small town might cover an area of about $10\text{ km}^2$ to $50\text{ km}^2$.
3. Thus, the area of a town is typically expressed in square kilometers ($\text{km}^2$).

**Answer:** Varies by region, typically measured in $\text{km}^2$

> Common mistake: Using square meters instead of square kilometers for large areas.

### Question 90

*3 marks · Short answer*

How many $\text{m}^2$ is a $\text{km}^2$?

**Solution**

1. We know that $1\text{ km} = 1000\text{ m}$.
2. To find $1\text{ km}^2$, we square both sides: $1\text{ km}^2 = (1000\text{ m})^2$.
3. Therefore, $1\text{ km}^2 = 1,000,000\text{ m}^2$ or $10^6\text{ m}^2$.

**Answer:** $1,000,000\text{ m}^2$

> Common mistake: Multiplying by 1000 instead of squaring $1000$.

### Question 91

*3 marks · Short answer*

How many times is your village/town/city bigger than your school?

**Solution**

1. Estimate the area of a typical school campus, which is roughly $20,000\text{ m}^2$ or $0.02\text{ km}^2$.
2. Find or estimate the total area of your local town or city, for example, a medium city has an area of about $200\text{ km}^2$.
3. Divide the area of the city by the area of the school to find how many times bigger the city is: $\frac{200\text{ km}^2}{0.02\text{ km}^2} = 10,000$ times.

**Answer:** The city is approximately $10,000$ times bigger than the school (values vary depending on the specific town or city).

> Common mistake: Comparing units directly without converting both areas to the same unit.

### Question 92

*3 marks · Short answer*

Find the city with the largest area in (i) India, and (ii) the world.

**Part (i)**

1. Identify the city with the largest municipal area or urban agglomeration in India.
2. Delhi and Bengaluru cover very large areas exceeding $1,400\text{ km}^2$ and $700\text{ km}^2$ respectively.

Answer (i): Delhi (or Bengaluru depending on municipal limits definition)

**Part (ii)**

1. Identify the city with the largest area in the world by administrative boundaries.
2. Hulunbuir in Inner Mongolia, China, is widely cited as the city with the largest municipal area globally (about $263,953\text{ km}^2$).

Answer (ii): Hulunbuir (China)

**Answer:** (i) Delhi or Bengaluru in India, (ii) New York, Tokyo, or Hulunbuir in the world.

> Common mistake: Confusing the most populous city with the largest area city.

### Question 93

*3 marks · Short answer*

Find the city with the smallest area in (i) India, and (ii) the world.

**Part (i)**

1. Look for small hill station municipalities or notified area councils in India.
2. Towns like Kapurthala in Punjab or specific small hill stations have very small municipal areas of only a few square kilometers.

Answer (i): Kapurthala or small hill station municipalities

**Part (ii)**

1. Look for the smallest sovereign city-state or recognized small city in the world.
2. Vatican City is the smallest independent state and city by area, covering just $0.49\text{ km}^2$.

Answer (ii): Vatican City

**Answer:** (i) Smaller hill station towns or municipal councils in India, (ii) Micro-states or towns like Città del Vaticano.

> Common mistake: Confusing village panchayats with incorporated cities.

## Frequently asked questions

### How many questions are there in NCERT Solutions for Class 8 Maths Chapter 14 Area for the 2026-27 session?

This chapter in the new NCERT book contains a total of 93 questions. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### Which topics are covered in the Class 8 Maths Chapter 14 questions?

The questions cover various topics including activity on area division, Approach 2 for trapezium area, area estimation and comparison, and area fractions. Other concepts include the area of an A4 sheet, parallelograms, and triangles.

### What are the different types of questions asked in this chapter?

The 93 questions comprise activity-based tasks and short answer (SA) problems. SwaVid provides detailed step-by-step solutions on this page only to help you understand each question type easily.

### Which is the hardest question type in Class 8 Maths Chapter 14 and how should I approach it?

Questions involving area estimation and comparison or complex area divisions are generally considered the hardest. To approach them, break down irregular shapes into known geometric figures like triangles and parallelograms before applying the relevant formulas.

### How can I write answers for full marks in Class 8 Maths Chapter 14 Area?

To secure full marks, clearly state the given values, write the correct formula before substituting numbers, and include proper units like $\text{cm}^2$ or $\text{m}^2$ in your final steps. SwaVid's free PDF on this page only demonstrates the exact presentation format required.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

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