Question

a rectangle abcd of maximum area is inscribed in a circle with center o. the length of the adjacent sides of the rectangle are 6 cm and 8cm. what is the area of the circle in which the rectangle is inscribed​

Answer 1

Step-by-step explanation:

Given :-

A rectangle ABCD of maximum area is inscribed in a circle with center o. The length of the adjacent sides of the rectangle are 6 cm and 8cm.

To find:-

What is the area of the circle in which the rectangle is inscribed ?

Solution :-

Given that

ABCD is a rectangle .

It is inscribed in a circle

Centre of the circle is O

Adjacent sides of the rectangle = 6 cm and 8 cm

Let the length of the rectangle = 8 cm

Let the breadth of the rectangle = 6 cm

AC and BD are the diagonals of ABCD rectangle

=> AC and BD are the diameters of the circle.

∆ ABD is a right angled triangle

By Pythagoras Theorem,

BD² = AB²+AD²

=> BD² = 6²+8²

=> BD² = 36+64

=> BD² = 100

=> BD = √100

=> BD = 10 cm

The diameter of the circle = BD = 10 cm

The radius of the circle = BO = OD

=> Diameter /2

=> BD/2

=> 10/2

=> 5 Cm

=>Radius of the circle BO = OD = 5 cm

We know that

The area of a circle whose radius is r units is

πr² sq.units

Area of the given circle =(22/7)×(5)² Cm²

=> (22/7)×25 sq.cm

=> (22×25)/7 sq.cm

=> 550/7 sq.cm

=> 78.57 sq.cm (approximately)

Therefore, Area = 78.57 sq.cm

Answer:-

The area of the given circle is 78.57 sq.cm

Used formulae:-

Pythagoras Theorem :-

" In a right angled triangle, The square of the hypotenuse is equal to the sum of the squares of the other two sides ".

→ The area of a circle whose radius is r units is

πr² sq.units

→ π = 22/7

→ r = radius

→ r = d/2

→ d = diameter