### NCERT Solutions Class 7 Maths Chapter 9 Perimeter and Area Exercise 9.1

## Introduction:

In this exercise 9.1 of chapter Area and Perimeter, we will learn about how to find the area, height and base of shapes other than square and rectangle like parallelogram , triangle. By using certain formulas given in this chapter we will solve the following Questions.

NCERT Class 7 Maths Chapter 9 Perimeter and Area Exercise 9.1

NCERT Class 7 Maths Chapter 9 Perimeter and Area Exercise 9.2

**Class 7 Maths Exercise 9.1 (Page-151)**

**Q1. **Find the area of each of the following parallelograms:

**Solutions: **

**a) **Given: Base = 7cm

Height = 4cm

So, the area of Parallelogram = base × height

= 7 × 4

= 28 cm²

**b) **Given: Base = 5cm

Height = 3cm

So, the area of Parallelogram = base × height

= 5 × 3

= 15 cm²

**c) **Given: Base = 2.5cm

Height = 3.5cm

So, the area of Parallelogram = base × height

= 2.5 × 3.5

= 8.75 cm²

**d) **Given: Base = 5cm

Height = 4.8cm

So, the area of Parallelogram = base × height

= 5 × 4.8

= 24 cm²

**e) **Given: Base = 2cm

Height = 4.4cm

So, the area of Parallelogram = base × height

= 2 × 4.4

= 8.8 cm²

**Q2. **Find the area of each of the following triangles:

**Solutions:**

**a) **Given: Base = 7cm

Height = 4cm

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

= \(\displaystyle \frac{{1}}{2}\) × 4 × 3

= \(\displaystyle \frac{{12}}{2}\)

= 6 cm²

**b) **Given: Base = 5cm

Height = 3.2cm

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

= \(\displaystyle \frac{{1}}{2}\) × 5 × 3.2

= \(\displaystyle \frac{{16}}{2}\)

= 8 cm²

**c) **Given: Base = 3cm

Height = 4cm

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

= \(\displaystyle \frac{{1}}{2}\) × 3 × 4

= \(\displaystyle \frac{{12}}{2}\)

= 6 cm²

**d) **Given: Base = 3cm

Height = 2cm

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

= \(\displaystyle \frac{{1}}{2}\) × 3 × 2

= \(\displaystyle \frac{{6}}{2}\)

= 3 cm²

**Q3. **Find the missing values:

**Solutions: **

**a) **

** **Given: Base = 20 cm

Area of Parallelogram = 246cm²

So, the area of Parallelogram = base × height

246 = 20 × h

h = \(\displaystyle \frac{{246}}{20}\)

= 12.3 cm

**b) **

** **Given: height = 15 cm

Area of Parallelogram = 154.5 cm²

So, the area of Parallelogram = base × height

154.5 cm² = base × 15

base = \(\displaystyle \frac{{154.5}}{15}\)

base = 10.3 cm

**c) **

Given: height = 8.4 cm

Area of Parallelogram = 48.72 cm²

So, the area of Parallelogram = base × height

48.72 cm² = base × 8.4

base = \(\displaystyle \frac{{48.72}}{8.4}\)

base = 5.8 cm

**d) **

Given: Base = 15.6 cm

Area of Parallelogram = 16.38cm²

So, the area of Parallelogram = base × height

16.38 = 15.6 × h

h = \(\displaystyle \frac{{16.38}}{15.6}\)

= 1.05 cm

**Q4. **** **Find the missing values:

**Solutions:**

**a) **

Given: Base = 15cm

Area of Triangle = 87 cm²

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

87 = \(\displaystyle \frac{{1}}{2}\) × 15 × h

87 × 2= 15 × h

h = \(\displaystyle \frac{{174}}{15}\)

= 11.6 cm

**b) **

Given: height = 31.4 mm

Area of Triangle = 1256 mm²

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

1256 = \(\displaystyle \frac{{1}}{2}\) × b × 31.4

1256 × 2 = 31.4× b

b = \(\displaystyle \frac{{2512}}{31.4}\)

= 80 mm

**c)**

Given: Base = 22cm

Area of Triangle = 170.5 cm²

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

170.5 = \(\displaystyle \frac{{1}}{2}\) × 22 × h

170.5 × 2 = 22 × h

h = \(\displaystyle \frac{{341}}{22}\)

= 15.5 cm

**Q5. **PQRS is a parallelogram(fig 9.14). QM is the height of Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:

a) the area of the parallelogram PQRS

b) QN, if PS = 8 cm

**Solution: **

**a) **

Given: SR (Base) = 12cm

QM (Height) = 7.6cm

So, the area of Parallelogram = base × height

= 12 × 7.6

= 91.2 cm²

So, the area of parallelogram PQRS is 91.2 cm² .

**b) **

Given: PS (Base) = 8cm

To Find : QN (height)

Area of parallelogram = 91.2 cm²

So, the area of Parallelogram = base × height

91.2 = 8 × h

h = \(\displaystyle \frac{{91.2}}{8}\)

h = 11.4 cm

So, the QN is 11.4 cm

**Q6. **DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (fig 9.15). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

**Solutions:**

Given :

Area of parallelogram = 1470 cm²

AB = 35 cm

AD = 49 cm

the area of Parallelogram = base × height

1470 = AB × DL

1470 = 35 × DL

DL = \(\displaystyle \frac{{1470}}{35}\)

DL = 42cm

the area of Parallelogram = base × height

1470 = DA × BM

1470 = 49 × BM

DL = \(\displaystyle \frac{{1470}}{49}\)

DL = 30 cm

So, the length of BM is 30 cm and DL is 42 cm.

**Q7. **∆ABC is right angled at A (fig 9.16). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.

**Solutions: **ATQ , given triangle is right angle. so, AB will be the base and AC will be the height.

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

= \(\displaystyle \frac{{1}}{2}\) × AB × AC

= \(\displaystyle \frac{{1}}{2}\) × 5 × 12

= \(\displaystyle \frac{{1}}{2}\) × 60

= 30 cm²

So, the area of ΔABC is 30 cm²

the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

30 = \(\displaystyle \frac{{1}}{2}\) × BC × AD

30 = \(\displaystyle \frac{{1}}{2}\) × 13 × AD

AD = \(\displaystyle \frac{{30 × 2}}{13}\)

AD = \(\displaystyle \frac{{60}}{13}\)

Hence, the length of AD is \(\displaystyle \frac{{60}}{13}\) cm.

**Q8. **∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (fig 9.17). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?

**Solution: **In ΔABC, AD(height) is 6cm and BC ( base) is 9cm.

So, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

= \(\displaystyle \frac{{1}}{2}\) × BC × AD

= \(\displaystyle \frac{{1}}{2}\) × 9 × 6

= \(\displaystyle \frac{{1}}{2}\) × 54

= \(\displaystyle \frac{{54}}{2}\)

= 27cm² .

Now, the area of Triangle = \(\displaystyle \frac{{1}}{2}\) × base × height

27 = \(\displaystyle \frac{{1}}{2}\) × AB × CE

27 = \(\displaystyle \frac{{1}}{2}\) × 7.5 × CE

CE= \(\displaystyle \frac{{27 × 2}}{7.5}\)

CE = \(\displaystyle \frac{{54}}{7.5}\)

CE = 7.2cm

Hene the area of ∆ABC is 27cm² and the height from C to AB i.e., CE is 7.2cm.