Mathematics – Class 7 – Chapter 11 – Perimeter and Area – Exercise 11.2 – NCERT Exercise Solution

1. Find the area of each of the following parallelograms:

1. Find the area of each of the following parallelograms:

(a)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Height of parallelogram = 4 cm

Base of parallelogram = 7 cm

Now,

Area of parallelogram = base × height

                                       = 7 × 4

                                       = 28 cm2

(b)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Height of parallelogram = 3 cm

Base of parallelogram = 5 cm

Now,

Area of parallelogram = base × height

                                        = 5 × 3

                                        = 15 cm2

(c)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Height of parallelogram = 3.5 cm

Base of parallelogram = 2.5 cm

Now,

Area of parallelogram = base × height

                                        = 2.5 × 3.5

                                        = 8.75 cm2

(d)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Height of parallelogram = 4.8 cm

Base of parallelogram = 5 cm

Now,

Area of parallelogram = base × height

                                        = 5 × 4.8

                                        = 24 cm2

(e)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Height of parallelogram = 4.4 cm

Base of parallelogram = 2 cm

Now,

Area of parallelogram = base × height

                                        = 2 × 4.4

                                        = 8.8 cm2

2. Find the area of each of the following triangles:

(a)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Base of triangle = 4 cm

Height of height = 3 cm

Now,

Area of triangle = ½ × base × height

                             = ½ × 4 × 3

                             = 1 × 2 × 3

                             = 6 cm2

(b)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Base of triangle = 3.2 cm

Height of height = 5 cm

Now,

Area of triangle = ½ × base × height

                             = ½ × 3.2 × 5

                             = 1 × 1.6 × 5

                            = 8 cm2

(c)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Base of triangle = 3 cm

Height of height = 4 cm

Now,

Area of triangle = ½ × base × height

                             = ½ × 3 × 4

                              = 1 × 3 × 2

                               = 6 cm2

(d)

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the figure,

Base of triangle = 3 cm

Height of height = 2 cm

Now,

Area of triangle = ½ × base × height

                             = ½ × 3 × 2

                             = 1 × 3 × 1

                             = 3 cm2

3. Find the missing values:

S.No.BaseHeightArea of the Parallelogram
a.20 cm 246 cm2
b. 15 cm154.5 cm2
c. 8.4 cm48.72 cm2
d.15.6 cm 16.38 cm2

Solution:

(a)

According to the table,

Given,

Base of parallelogram = 20 cm

Height of parallelogram =?

Area of the parallelogram = 246 cm2

Now,

Area of parallelogram = base × height

246 = 20 × height

Height = 246/20

Height = 12.3 cm

Hence, Height of the parallelogram is 12.3 cm.

(b)

According to the table,

Given,

Base of parallelogram =?

Height of parallelogram =15 cm

Area of the parallelogram = 154.5 cm2

Now,

Area of parallelogram = base × height

154.5 = base × 15

Base = 154.5/15

Base = 10.3 cm

Hence, Base of the parallelogram is 10.3 cm.

(c)

According to the table,

Given,

Base of parallelogram =?

Height of parallelogram =8.4 cm

Area of the parallelogram = 48.72 cm2

Now,

Area of parallelogram = base × height

48.72 = base × 8.4

Base = 48.72/8.4

Base = 5.8 cm

Hence, Base of the parallelogram is 5.8 cm.

(d)

According to the table,

Given,

Base of parallelogram = 15.6 cm

Height of parallelogram =?

Area of the parallelogram = 16.38 cm2

Now,

Area of parallelogram = base × height

16.38 = 15.6 × height

Height = 16.38/15.6

Height = 1.05 cm

Hence, Height of the parallelogram is 1.05 cm.

S.No.BaseHeightArea of the Parallelogram
a.20 cm12.3 cm246 cm2
b.10.3 cm15 cm154.5 cm2
c.5.8 cm8.4 cm48.72 cm2
d.15.6 cm1.0516.38 cm2

4. Find the missing values:

BaseHeightArea of Triangle
15 cm 87 cm2
 31.4 mm1256 mm2
22 cm 170.5 cm2

Solution:

(1)

According to the table,

Given,

Height of triangle =?

Base of triangle = 15 cm

Area of the triangle = 16.38 cm2

Then,

Area of triangle = ½ × base × height

87 = ½ × 15 × height

Height = (87 × 2)/15

Height = 174/15

Height = 11.6 cm

Hence, Height of the triangle is 11.6 cm.

(2)

According to the table,

Given,

Height of triangle =31.4 mm

Base of triangle =?

Area of the triangle = 1256 mm2

Then,

Area of triangle = ½ × base × height

1256 = ½ × base × 31.4

Base = (1256 × 2)/31.4

Base = 2512/31.4

Base = 80 mm = 8 cm

Hence, Base of the triangle is 80 mm or 8 cm.

(3)

According to the table,

Given,

Height of triangle =?

Base of triangle = 22 cm

Area of the triangle = 170.5 cm2

Then,

Area of triangle = ½ × base × height

170.5 = ½ × 22 × height

170.5 = 1 × 11 × height

Height = 170.5/11

Height = 15.5 cm

Hence, Height of the triangle is 15.5 cm.

5. PQRS is a parallelogram (Fig 11.23). QM is the height from 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

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

(a) The area of the parallelogram PQRS

According to the question,

Given,

SR = 12 cm

QM = 7.6 cm

Now,

Area of the parallelogram = base × height

= SR × QM

= 12 × 7.6

= 91.2 cm2

(b)

Area of the parallelogram = base × height

91.2 = PS × QN

91.2 = 8 × QN

QN = 91.2/8

QN = 11.4 cm

6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Mathematics claass 7 chapter 11

Solution:

According to the question,

Given,

Area of the parallelogram = 1470 cm2

AB = 35 cm

AD = 49 cm

Now,

Area of the parallelogram = base × height

1470 = AB × BM

1470 = 35 × DL

DL = 1470/35

DL = 42 cm

And,

Area of the parallelogram = base × height

1470 = AD × BM

1470 = 49 × BM

BM = 1470/49

BM = 30 cm

7. ΔABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. Also find the length of AD.

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the question,

Given,

AB = 5 cm

BC = 13 cm

AC = 12 cm

Now,

Area of the ΔABC = ½ × base × height

= ½ × AB × AC

= ½ × 5 × 12

= 1 × 5 × 6

= 30 cm2

Now,

Area of ΔABC = ½ × base × height

30 = ½ × AD × BC

30 = ½ × AD × 13

(30 × 2)/13 = AD

AD = 60/13

AD = 4.6 cm

8. ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?

Mathematics - Class 7 - Chapter 11 - Perimeter and Area - Exercise 11.2 - NCERT Exercise Solution

Solution:

According to the question it is given that,

AB = AC = 7.5 cm

BC = 9 cm

AD = 6cm

Now,

Area of ΔABC = ½ × base × height

= ½ × BC × AD

= ½ × 9 × 6

= 1 × 9 × 3

= 27 cm2

Now,

Area of ΔABC = ½ × base × height

27 = ½ × AB × CE

27 = ½ × 7.5 × CE

(27 × 2)/7.5 = CE

CE = 54/7.5

CE = 7.2 cm

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