Question

The perimeter of a rectangle is 46 cm, the length of a diagonal of this rectangle is 17 cm, find the area of this rectangle

Answer 1

.l equals length
b equals breadth
perimeter,p= 2(L+b)

2(L+b)=46
L+b=23

diagonal^2=length^2+breadth^2

(l+b)^2= l^2+b^2+2*l*b

substitute l+b and diagonal^2 value in the above equation
solve for l*b
Area of rectangle equals l*b :)

Answer 2

Answer:

$\pink{120 \: {cm}^{2} }$

Solution:

• Let the length of rectangle be l
• And the breadth be b

Then Diagonal of rectangle

$: \implies \sqrt{ {l}^{2} } + {b}^{2} = 17$

And, Perimeter of rectangle

$: \implies \: 2(l + b) = 46$

$: \implies \: l + b = 23$

$: \implies ({l + b})^{2} = 529$

$: \implies \: {l}^{2} + {b}^{2} + 2lb = 529$

$: \implies \: 2lb = 529 - 289$

$: \implies \: lb \: = 120$

$\sf \: { \therefore \: Area \: of \: Rectangle = \: \: 120 {cm}^{2}}$