The perimeter of a rhombus is 100,cm and one | vedclass

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(D) Let the side of the rhombus be $a$. Since the perimeter is $100\,cm$,we have $4a = 100\,cm$,which gives $a = 25\,cm$.Let the diagonals of the rhom

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(D) Let the side of the rhombus be $a$. Since the perimeter is $100\,cm$,we have $4a = 100\,cm$,which gives $a = 25\,cm$.Let the diagonals of the rhombus be $d_1 = 48\,cm$ and $d_2$. The diagonals of a rhombus bisect each other at right angles.This forms a right-angled triangle with the side of the rhombus as the hypotenuse and half of the diagonals as the other two sides.Using the Pythagorean theorem: $(a)^2 = (d_1/2)^2 + (d_2/2)^2$.Substituting the values: $(25)^2 = (48/2)^2 + (d_2/2)^2$.$625 = (24)^2 + (d_2/2)^2$.$625 = 576 + (d_2/2)^2$.$(d_2/2)^2 = 625 - 576 = 49$.$d_2/2 = \sqrt{49} = 7$.Therefore,$d_2 = 7 	imes 2 = 14\,cm$.

The perimeter of a rhombus is $100\,cm$ and one of its diagonals measures $48\,cm$. Then,the length of the other diagonal is ....... $cm.$

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