Question

find sides of rectangle if its diagonal is 50m and larger side is 10m more than its smaller side​

Answer 1

There you go! Hope this will be helpful!

Answer 2

Answer:



Step-by-step explanation:

In a rectangle, the opposite sides are equal.

So, we can break/ divide it into two Right Angled-Triangles.

Let's assume the shorter leg as x

By Pythagoras Theorem,

$x^{2} +(x+10)^{2} = 50^2$

=> 2$x^2+20x+100$= 2500

=> $x^2+10x+50$= 1250

=> $x^2+10x=1200$

=> $x^2+10x-1200$=0

(By the equation (x+a)(x+b)= $x^2+ax+bx+ab$,

we have to find a and b such that a+b=20 and ab=-1200).

a= 40

b=-30

Hence, (x+40)(x-30)=0

(x can be -40 or can be 30 so that product becomes 0)

I chose x= 30 because side length can't be negative.

So, smaller leg is 30 cm and bigger leg is 40 cm.

The sides therefore are, 30 cm, 40 cm, 30 cm and 40 cm.

Formulas used-

$a^2+b^2=c^2$ in right angled-triange

(x+a)(x+b)= $x^2+ax+bx+ab$

Variables used-

a, b and x