Given
A rectangle is placed inside a circle whose diameter is 13unit all corners of it on the circle.
area of the rectangle = 60 square unit
Find the dimensions of the rectangle.
To proof
As given in the question
rectangle is placed inside a circle whose diameter is 13unit .
now by using the diagram given below
the diagonal of the rectangle is equal to the diameter of the circle.
Thus DB = 13 unit
FORMULA
Area of the rectangle = Length × Breadth
area of the rectangle = 60 square unit
let the value of the length be ( Say AB ) = x
put the value in the above equation
we get
60 = x × Breadth
[tex]Breadth( say AD)= \frac{60}{x}[/tex]
now in ABD
Using the pythagorus theorem
we get
BD²= AB² + AD²
[tex]13^{2} = x^{2}+ (\frac{60}{x}) ^{2}[/tex]
solving the above equation
we get
[tex]169x^{2}= x^{4} + 3600[/tex]
[tex]x^{4} - 169x^{2}+3600 =0[/tex]
[tex]x^{4} - 144x^{2} -25x^{2} + 3600 =0[/tex]
[tex]x^{2} (x^{2} - 144) - 25 (x^{2} - 144) =0[/tex]
(x² - 25) (x² -144) =0
The roots are
x = -12 ,-5, 5, 12
-12 and -5 are neglected because of negative terms .
( Length and breadth of recangle cannot be negative )
thus x = 12 and 5
Two case arise
First case
when
Length of rectangle= 12unit
[tex]Breadth = \frac{60}{12}[/tex]
Breadth of rectangle = 5 unit
Second case
Length of rectangle= 5unit
[tex]Breadth = \frac{60}{5}[/tex]
Breadth of rectangle = 12 unit
Hence proved
