NCERT Solutions Class 7th Math Chapter – 9 Perimeter and Area Example

Example 1. One of the sides and the corresponding height of a parallelogram are 4 cm and 3 cm respectively. Find the area of the parallelogram (Fig 9.8).

Solution –
Given that length of base (b) = 4 cm, height (h) = 3 cm
Area of the parallelogram = b × h
= 4 cm × 3 cm = 12 cm2

Example 2. Find the height ‘x’ if the area of the parallelogram is 24 cm2 and the base is 4 cm.

Solution – Area of parallelogram = b × h Therefore,24 = 4 × x (Fig 9.9)
or 24/4 = x or x = 6 cm
So, the height of the parallelogram is 6 cm.

Example 3. The two sides of the parallelogram ABCD are 6 cm and 4 cm. The height corresponding to the base CD is 3 cm (Fig 9.10). Find the
(i) area of the parallelogram.
(ii) the height corresponding to the base AD.

Solution –
(i) Area of parallelogram = b × h
= 6 cm × 3 cm = 18 cm2
(ii) base (b) = 4 cm, height = x (say),
Area = 18 cm2
Area of parallelogram = b × x
18 = 4 × x
18/4 = x
Therefore, x = 4.5 cm
Thus, the height corresponding to base AD is 4.5 cm.

Example 4. Find the area of the following triangles (Fig 9.11).

Solution –

(i) Area of triangle = 1/2bh = 1/2 × QR × PS

=1/2 × 4cm x 2cm = 4 cm2

(ii) Area of triangle = 1/2 bh = 1/2 × MN × LO

= 1/2 × 3cm × 2cm = 3 cm

Example 5. Find BC, if the area of the triangle ABC is 36 cm2 and the height AD is 3 cm (Fig 9.12).

Solution – Height = 3 cm, Area = 36 cm2 Area of the triangle ABC = 1/2 bh
or 36 = 1/2 × b × 3 i.e. b = 36 x 2/3 = 24 cm
So, BC = 24 cm

Example 6. In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm (Fig 9.13).
Find:
(i) the area of the ∆PQR
(ii) QM

Solution –
(i) QR = base =4 cm, PL = height = 5 cm
Area of the triangle PQR = 1/2 bh
=1/2 x 4cm × 5cm = 10 cm2
(ii) PR = base = 8 cm QM = height = ? Area = 10 cm2
Area of triangle = 1/2 × ×b h i.e., 10 = 1/2 × 8 × h
h = 10/4 = 5/2 = 2.5. So, QM = 2.5 cm

Example 7. What is the circumference of a circle of diameter 10 cm (Take π = 3.14)?

Solution – Diameter of the circle (d) = 10 cm
Circumference of circle = πd
= 3.14 × 10 cm = 31.4 cm
So, the circumference of the circle of diameter 10 cm is 31.4 cm

Example 8. What is the circumference of a circular disc of radius 14 cm? (Use π = 22/7)

Solution – Radius of circular disc (r) = 14 cm
Circumference of disc = 2πr

= 2 x 22/7 × 14 cm = 88 cm
So, the circumference of the circular disc is 88 cm.

Example 9. The radius of a circular pipe is 10 cm. What length of a tape is required to wrap once around the pipe (π = 3.14)?

Solution –
Radius of the pipe (r) = 10 cm
Length of tape required is equal to the circumference of the pipe.
Circumference of the pipe = 2πr
= 2 × 3.14 × 10 cm
= 62.8 cm
Therefore, length of the tape needed to wrap once around the pipe is 62.8 cm

Example 10. Find the perimeter of the given shape (Fig 9.23) (Take π = 22/7).

Solution – In this shape we need to find the circumference of semicircles on each side of the square. Do you need to find the perimeter of the square also? No. he outer boundary, of this figure is made up of semicircles. Diameter of each semicircle is 14 cm.
We know that:
Circumference of the circle = πd

Circumference of the semicircle = 1/2πd

= 1/2 x 22/7 × 14 cm = 22 cm
Circumference of each of the semicircles is 22 cm
Therefore, perimeter of the given figure = 4 × 22 cm = 88 cm

Example 11. Sudhanshu divides a circular disc of radius 7 cm in two equal parts. What is the perimeter of each semicircular shape disc? (Use π = 22/7)

Solution – To find the perimeter of the semicircular disc (Fig 9.24), we need to find
(i) Circumference of semicircular shape
(ii) Diameter Given that radius (r) = 7 cm. We know that the circumference of circle = 2πr

So, the circumference of the semicircle = × 1/2 πr = πr
= 22/7 x 7 cm = 22 cm
So, the diameter of the circle = 2r = 2 × 7 cm = 14 cm
Thus, perimeter of each semicircular disc = 22 cm + 14 cm = 36 cm

Example 12. Find the area of a circle of radius 30 cm (use π = 3.14).

Solution – Radius, r = 30 cm
Area of the circle = πr2 = 3.14 × 302 = 2,826 cm2

Example 13. Diameter of a circular garden is 9.8 m. Find its area.

Solution – Diameter, d = 9.8 m. Therefore, radius r = 9.8 ÷ 2 = 4.9 m
Area of the circle = πr2 = 22/7 × (4 .9 )2 m2 = 22/7 × 4 9 4 9 . . m2 = 75.46 m2

Example 14. The adjoining figure shows two circles with the same centre. The radius of the larger circle is 10 cm and the radius of the smaller circle is 4 cm.

Find: (a) the area of the larger circle
(b) the area of the smaller circle
(c) the shaded area between the two circles. (π = 3.14)

Solution –
(a) Radius of the larger circle = 10 cm
So, area of the larger circle = πr2
= 3.14 × 10 × 10 = 314 cm2
(b) Radius of the smaller circle = 4 cm
Area of the smaller circle = πr2
= 3.14 × 4 × 4 = 50.24 cm2
(c) Area of the shaded region = (314 – 50.24) cm2 = 263.76 cm2