Question

Each side of a rhombus is 10 cm long and one of its diagonals measures
16 cm. Find the length of the other diagonal and hence find the area of
the rhombus.
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Answer 1

Answer:

30 cm......................

Answer 2

Answer:

Diagonals of a rhombus bisect each other at right angles.

Let ABCD be the rhombus, Diagonal AC=16 cm and side AB=10 cm.

In right △AOB, AB=10 cm, AO=8 cm

By Pythagoras theorem,

( AB)^2=(AO)^2+(BO)^2

=> (10)^2= (8)^2+(BO)^2

=> 100 = 64 +(BO)^2

=>100−64= (BO)^2

=>BO =√36

∴BO= 6 cm.

Diagonal BD=2×6=12cm.

Area of Rhombus=

$( \frac{1}{2}\times \: product \: of \: the \: diagonals) \\ = \frac{1}{2} \times 16 \times 12 \\ = 8 \times 12 \\ = 96 \: {cm}^{2} .$