Question

3. The shape of a garden is rectangular in the middle and semi circular
at the ends as shown in the diagram. Find the area and the perimeter
T
of this garden (Length of rectangle is
7 m 20-(3.5 +3.5) metres).
I
- 20 m
DI​

Answer 1

Answer:

48M

Step-by-step explanation:

Ncert solutions

Grade 8

Mathematics

Science

Chapters in NCERT Solutions - Mathematics , Class 8

Exercises in Mensuration

Question 3

Q3) The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].

Solution 3:

We know that,

length of the garden=20 m

diameter of semicircle=7m,r=3.5m

(breadth of the rectangle)

Length of rectangle=20m-3.5m-3.5m20m−3.5m−3.5m

=13 m

Area of rectangle=length\times breadthlength×breadth

=13m\times7m=13m×7m

=91\ m^2=91 m

2

Area of semi-circle =\frac{1\ }{2}\times\pi\times\left(3.5\right)^2

2

1

×π×(3.5)

2

==\frac{1}{2}\times\frac{22}{7}\times\frac{35}{10\ }\times\frac{35}{10}=

2

1

×

7

22

×

10

35

×

10

35

=19.25\ m^2=19.25 m

2

Area of garden=91+19.25+19.2591+19.25+19.25

=129.5\ m^2=129.5 m

2

Perimeter of garden=perimeter of rectangle+2(Perimeter of semicircle)

Perimeter of rectangle==2\times length=2×length

=2\times26=2×26

=26\ m=26 m

Perimeter of semicircle==\pi r=πr

=\frac{22}{7}\times3.5=

7

22

×3.5

=\frac{22}{7}\times\frac{35}{10}=

7

22

×

10

35

=11m

Perimeter of Garden=26m+11m+11

=48m