Question

Find the side of a rhombus whose diagonals are 10 cm and 24 cm

Answer 1

Diagonals of rhombus bisect each other perpendicularly....(1)

Therefore if we consider quadrilateral ABCD is rhombus with point of intersection as O when AC and BD are diagonals of length 10cm and 24cm

then here AO=1/2AC=1/2×10=5cm

similarly,

BO=12cm

From (1)

Angle AOB=90°

Therefore by Pythagoras theorem,

AO^2+BO^2=AB^2

5^2+12^2=AB^2

25+144=AB^2

√169=AB

AB=13cm

Therefore , Side of the given rhombus is 13cm.

Answer 2

there is a rhombus ABCD

O is the center where

AC is 10cm

BD is 24 cm

OA=OC=5cm

OB=OD= 12 cm

In OCD ,

CD is the side of the rhombus

CD^2= OD^2+OC^2

CD= 12^2+5^2

= 144 + 25

= root of 169 = 13 cm

so the answer is 13cm