# Ncert Solutions For Class 7 Maths Ex 11.2

## Ncert Solutions For Class 7 Maths Chapter 11 Ex 11.2

Q1. Determine the area of the parallelograms in the diagram given below

Ans:

(a) Area of a parallelogram = Height x Base

Base = 8 cm

Height = 5 cm

Area = 8 x 5 = 40 cm2

(b) Area of a parallelogram = Height x Base

Base = 6 cm

Height = 4 cm

Area = 6 x 4 = 24 cm2

(c) Area of a parallelogram = Height x Base

Base = 4.5 cm

Height = 7 cm

Area = 4.5 x 7 = 31.5 cm2

(d) Area of a parallelogram = Height x Base

Base = 9.6 cm

Height = 10 cm

Area = 9.6 x 10 = 96 cm2

(e) Area of a parallelogram = Height x Base

Base = 8.8 cm

Height = 4 cm

Area = 8.8 x 4 = 35.2 cm2

Q2. Determine the area of the triangles .

Ans:

(a) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height = 6 cm

Base = 8 cm

Area =12×8×6$= \frac{1}{2}\times 8 \times 6$ = 48 cm2

(b) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height = 6.4 cm

Base = 10 cm

Area =12×10×6.4$= \frac{1}{2}\times 10 \times 6.4$ = 64 cm2

(c) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height = 6 cm

Base = 8 cm

Area =12×8×6$= \frac{1}{2}\times 8 \times 6$ = 48 cm2

(d) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height = 4 cm

Base = 6 cm

Area =12×4×6$= \frac{1}{2}\times 4 \times 6$ = 24 cm2

Q3. Fill the empty cells:

 S. No Base Height Area of parallelogram A 10 cm 124 cm2 B 12 cm 96 cm2 C 7 cm 91 cm2 D 11 cm 121 cm2

Ans:

(a) Area of a parallelogram = Height x Base

B = 10 cm

H =?

Area = 124 cm2

10 x h = 124

h=12410=12.4$h = \frac{124}{10}=12.4$

So, the height of the parallelogram is 12.4 cm

(b) Area of a parallelogram = Height x Base

B =?

H = 12 cm

Area = 96 cm2

b x 12 = 96

b=9612=8$b = \frac{96}{12}=8$

So, the base of the parallelogram is 8 cm

(c) Area of a parallelogram = Height x Base

B = 7 cm

H =?

Area = 91 cm2

7 x h = 91

h=917=13$h = \frac{91}{7}=13$

So, the height of the parallelogram is 13 cm

(d) Area of a parallelogram = Height x Base

B = 11 cm

H =?

Area = 121 cm2

11 x h = 121

h=12111=11$h = \frac{121}{11}=11$

So, the height of the parallelogram is 11 cm

Q4. Fill in the blanks:

 Base Height Area of triangle 12 24 cm2 30 60 cm2 22 66 cm2

Ans:

(a) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height = ?

Base = 12 cm

Area =12×Base×height$= \frac{1}{2}\times Base \times height$ = 24 cm2

12×12×h=24$\frac{1}{2}\times 12 \times h = 24$ h=24×212=4cm$h = \frac{24\times 2}{12} = 4 cm$

Therefore, the height of the triangle is 4 cm

(b) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height = 30 cm

Base = ?

Area =12×Base×height$= \frac{1}{2}\times Base \times height$ = 60 cm2

12×b×30=60$\frac{1}{2}\times b \times 30 = 60$ h=60×230=4cm$h = \frac{60\times 2}{30} = 4 cm$

Therefore, the base of the triangle is 4 cm

(c) Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$

Height =?

Base = 22 cm

Area =12×Base×height$= \frac{1}{2}\times Base \times height$ = 66 cm2

12×22×h=24$\frac{1}{2}\times 22 \times h = 24$ h=66×222=6cm$h = \frac{66\times 2}{22} = 6 cm$

Therefore, the height of the triangle is 6 cm.

Q5. ABCD is a parallelogram (in the given figure). BX is the height from B to CD and BY is the height from B to AD. If CD = 12 cm and BX = 7.6 cm.

Find:      (a) the area of the parallelogram ABCD

(b) BY, if AD = 8 cm.

Ans:

(a) Area of parallelogram = Base x Height = CD x BX

= 7.6 x 12 = 91.2 cm2

(b) Area of parallelogram = Base x Height = AD x BY = 91.2 cm2

BY x 8 = 91.2

BY = 91.2/8 =11.4 cm

Q6. SL and QM are the heights on sides PQ and PS respectively of parallelogram PQRS (in the given figure). If area of parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

`Ans:

Area of a parallelogram = Height x Base = PQ x SL

1470 = 35 x SL

DL = 147035$\frac{1470}{35}$ = 42 cm

Also, PS x QM = 1470

1470 = 49 x QM

QM=147035=30 cm$QM = \frac{1470}{35} = 30\ cm$

Q7. ABC is right angled at A (in the given figure). AD Is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, determine the area of triangle ABC. Also find the length of AD.

Ans:

Area of a triangle =12×Base×height$= \frac{1}{2}\times Base \times height$  =12×5×12$= \frac{1}{2}\times 5 \times 12$ = 30 cm2

Also, area of triangle =12×AD×BC$= \frac{1}{2}\times AD \times BC$

30=12×AD×13$30 = \frac{1}{2}\times AD \times 13$ 30×213=AD$\frac{30\times 2}{13}= AD$

AD = 4.6 cm

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

Ans:

Area of triangle ABC =12×Base×height$= \frac{1}{2}\times Base \times height$ =12×BC×AD$= \frac{1}{2}\times BC \times AD$

$$= \frac{1}{2}\times 9 \times 6 = 27 cm2$$

Area of triangle ABC =12×Base×height$= \frac{1}{2}\times Base \times height$ =12×AB×CE$= \frac{1}{2}\times AB \times CE$

27=12×7.5×CE$27= \frac{1}{2}\times 7.5 \times CE$

CE = 7.2