Question

10. Each side of a rhombus is 10 cm. If one of its
diagonals is 16 cm, find the length of the
other diagonal.​

Answer 1

Answer:96 cm²

Step-by-step explanation:

a²=(d/2)² + (D/2)²,

10²=(16/2)² + (D/2)²,

100=8² + (D/2)²,

100-64=(D/2)²,

36=(D/2)²,

then

D/2=6,

then

D=6×2=12 cm,

now area of rhombus=1/2 × d ×D,  

=1/2×16×12,

=8×12=96 cm²

Answer 2

Answer:

hello

Step-by-step explanation:

we know that,

diagonals of a rhombus bisect each other at 90 degree

so, lets take abcd a rhombus where

ab=bc=cd=ad=10cm

also,

ac=16cm

In ∆AOB,

AB² = OA² + OB²

[By using Pythagoras theorem]

10² = OA² + 8²

100 = OA² + 64

100 − 64 = OA²

OA² = 36

OA = √36  

OA = 6 cm

AC = OC + OA

AC = 6 + 6  

[diagonal of a rhombus bisect each other at 90°, AO = OC]

AC =  12 cm

Hence, the length of the other diagonal is 12 cm.

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