Question

A piece of wire 32 cm long is bent to form the
figure given below. APD is a semicircle and
AB = BC = CD. Find the radius and area of the
figure.

​

Answer 1

Answer:

let AB=BC=CD=AD= s

so side of squar is s

and AD is also diameter of semi circle

then diameter of semicircle is also s

if diameter is s then radius of semicircle

will be s/2

perimeter of square = 4xside = 4 x s

circumference of semi circle will be = pie x r

In pic I find that side of square is 7

and rdius willbe 7/2

Answer 2

Answer:

Radius or Semicircle is x/2

Area of figure is [$x^{2}$(π + 4)]/4

Step-by-step explanation:

In the given figure,

AB = BC = CD = AC

So, lets consider.

AB = BC = CD = AC = x

Also,

Here APD is a semicircle and AD is the diameter of the semicircle.

So, the radius of semicircle will be diameter/2

i.e. AD/2

i.e. x/2

Therefore,

Radius of semicircle is x/2

Now, For calculating area of figure,

Divide figure into square ABCD and semicircle APD

Here,

Area of Figure = Area of square ABCD + Area of semicircle APD

We know that, Area of Square is $Side^{2}$

i.e. Area of square is $x^{2}$

Now, Area of semicircle is $(\pi r^{2} )/2$

i.e. $(\pi x^{2}) /2$

Therefore,

Area of figure = $x^{2}$ + $(\pi x^{2}) /2$

Area of figure = $[x^{2} (\pi + 4)] / 4$

SPJ2