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:

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

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

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

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:

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

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

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:

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 .

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.