Question

the length of a rectangle is three times its width.If the length of the diagnol is 15√2cm,then find perimeter of rectangle......plz solve it fast

Answer 1

Heya!!

hope it will help u

let the width of the rectangle be 'x' cm.

So,its length will be '3x' cm.

consider a rectangle ABCD in which AB is length,BC is breadth and AC is diagonal.

We have,


l² + b² = d² (where,d is diagonal).

=>(3x)² + (x)² =( 15√2)²

=>9x² + x² = 450

=>10 x² = 450

=> x² = 45

thus, x = √45= 3√5 cm

and,3x = 9√5.

so,perimeter of rectangle = 2(l + b)

=2(3√5 + 9√3)

=(12√5) × 2

=24√5 cm Ans.

thanks

Answer 2

Let the width be x,  and then the length will be  (3*x) = 3x

Now, 

As we know that all angles of a rectangle is at 90

So, By Pythagoras Theorem
 

length^2  + breadth^2 = diagonal^2



$(3 x)^{2} + x^{2} = (15 \sqrt{2}) ^{2}$

$9 x^{2} + x^{2} = (15 \sqrt{2}) ^{2}$



= $10 x^{2} = 225*2$

= $x^{2} = \frac{450}{10}$

= $x^{2} = 45$

$\sqrt{45}$ = x


Now,

breadth = $\sqrt{45}$

and length = 3$\sqrt{45}$

Now,,,

Perimeter  = 2(length+breadth )


perimeter = 2($\sqrt{45} + 3 \sqrt{45}$)


Perimeter= 2*4$\sqrt{45}$ 


perimeter of rectangle = 8$\sqrt{45}$cm

perimeter of rectangle =8*3($\sqrt{5}$)

perimeter of rectangle = 24$\sqrt{5}$


i hope this will help you


-by ABHAY