Question

area of a rectangle is 24cm2 and perimeter 20cm.length of the diagonal

Answer 1

if length will of 6 then breadth will be of 4 and vice versa

Answer 2

Answer:

Diagonal of the rectangle =  $\sqrt{52}$

Step-by-step explanation:

Given : Area of rectangle = $24 cm^2$

Perimeter of rectangle = 20 cm

To find : The diagonal of rectangle

Solution : Area of rectangle = $length \times breadth=l\timesb=24cm^2$ ......[1]

Perimeter of rectangle = $2(length+breadth)=2(l+b)=20cm$ ......[2]

From [1] we take $l=\frac{24}{b}$ and put in [2]

$2(l+b)=20cm$

$2(\frac{24}{b}+b)=20cm$

$\frac{24+b^2}{b}=10cm$  (Take LCM)

$b^2-10b+24$

$b^2-6b-4b+24$ (Middle term split)

$b(b-6)-4(b-6)$

$(b-6)(b-4)$

b=6, b=4

if b=4 then l=24/4 = 6

if b=6 then l=24/6=4

There are two cases,

If b=4 and length = 6

Diagonal of the rectangle = $\sqrt{l^2+b^2}$

                                           = $\sqrt{6^2+4^2}$

                                           = $\sqrt{36+16}=\sqrt{52}$

If b=6 and length = 4

Diagonal of the rectangle = $\sqrt{l^2+b^2}$

                                           = $\sqrt{4^2+6^2}$

                                           = $\sqrt{16+36}=\sqrt{52}$

Therefore, Diagonal of the rectangle =  $\sqrt{52}$