---
title: "NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2"
url: https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-2
dateModified: 2026-10-07T15:48:27+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2

Chapter 6: Measuring Space: Perimeter and Area. Every question from Exercise 6.2, with full working and the final answer.

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## 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.

## Related pages

- [All Chapter 6 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions)
- [Exercise 6.1](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-1)
- [Exercise 6.3](https://www.swavid.com/maths/class/9/chapter/measuring-space-perimeter-and-area/ncert-solutions/exercise-6-3)

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