Question

What will be its area if a wire of 66 cm is made by bending the following figure, if the figure is (i) square (ii) circle (iii) equilateral triangle (iv) semicircle?​

Answer 1

Solution :-

(i)

One side of the square = 64/4 = 32/2cm

∴ Area of square = (33/2)^2 = 1089/4

= 272.25 sq.cm

(ii)

Circumference of circle 2πr = 66

2 × 22/7 × r = 66

r = 10.5 cm

∴ Area of the circle = πr²

= 22/7 × ( 10.5 )^2 = 346.5 sq.cm

(iii)

One arm of equilateral ∆ = 66/3 = 22cm

∴ Area of equilateral ∆ = (√3/4)a^2

= √3/4 ( 22) ^2

= 121 √3 = 121 × 1.73

= 209.33 sq.cm

(iv)

Circumference of the semicircle,

πr + 2r = 66

⇒ 22/7 r + 2r = 33

⇒ ( 22r + 14r )/7 = 33

⇒ r = ( 33 × 7 )/36 = 77/12

∴ Area of semicircle

= 1/2 × 22/7 × (77/12)^2

= 9317/144 sq.cm

Answer 2

Before, finding the answer. Let's find out how we can find the answer.

• In this question, we are asked to find the area if a wire of 66 cm is bent by different shapes.
• So, first, we have to divide the length of the wire with the no of sides in each shape.
• And after that, we have to use different formulas to find the area.

___________________

Given :

• Length of wire = 66 cm

To find :

• area if a wire of 66 cm is made by bending the following figure :

(i) square

(ii) circle

(iii) equilateral triangle

(iv) semicircle?​

Solution :

(i) No of sides in Square = 4

Each side = 66/4

                = 16.5 cm

Area of Square = a × a

                         = 16.5 × 16.5

                         = 272.25

Therefore, the Area of Square is 272.25 cm².

(ii) Circumference $\tt = 2 \pi r$

                 $\tt 66 cm = 2 \times \dfrac{22}{7} \times r$

                          $\tt = \dfrac{c}{ 2 \pi }$

                          $\tt = \dfrac{66}{ 2 \times 3.14}$

                         $\tt = \dfrac{66}{6.28}$

                         $\tt = 10.5 cm$                          

Area of Circle $\tt = \pi r^2$

                       $\tt = \dfrac{22}{7} \times 10.5^2$

                       $\tt = \dfrac{22}{7} \times 110.25$

                       $\tt = \dfrac{2425.5}{7}$

                       $\tt = 346.5$

Therefore, the Area of the Circle is 346.5 cm².

                       

(iii) No of sides in an equilateral triangle = 3

Each side = 66/3

                = 22 cm

Area of Equilateral Trinagle  $\tt = \dfrac{\sqrt{3} }{4} a^2$

                                                $\tt = \dfrac{\sqrt{3} }{4} (22)^2$

                                                $\tt = \dfrac{\sqrt{3} }{4} \times 484$

                                                $\tt = \dfrac{1.73 }{4} \times 484$

                                                $\tt = \dfrac{837.32 }{4}$

                                                $\tt = 209.33$

Therefore, the Area of the Equilateral triangle is 209.33 cm².

(iv) Circumference of Semi-Circle = πr + 2r = 66

                                                  = 22/7r + 2r = 66

                                                  = 22r + 14r = 66

                                                  = 36r/7 = 66

                                     Radius = 66 × 7/36

                                                  = 11 × 7/6

                                                  = 77/6

                                                 = 12.83 cm                                                                    

Area of Semi-Circle $\tt = \dfrac{1}{2} ( \pi r^2 )$

                                 $\tt = \dfrac{1}{2} ( \dfrac{22}{7} \times 12.83^2 )$

                                 $\tt = \dfrac{1}{2} ( \dfrac{22}{7} \times 164.60 )$

                                 $\tt = \dfrac{1}{2} \times ( \dfrac{3621.2}{7} )$

                                 $\tt = \dfrac{1}{2} \times 517.31$

                                 $\tt = 258.65 cm^2$

Therefore, the Area of the Semi-Circle is 258.65 cm².