Question

the perimeter of a rhombus is 40cm and length of one of the diagonal is 16cm the area of rhombus is ​

Answer 1

Step-by-step explanation:

Perimeter of a rhombus=40cm

Therefore its side 40/4=10cm

one diagnol=16cm

Now find other diagnol by dividing rhombus in four triangle and using pythagoras theorem

lets try to find

keep other diagnol's half as x

10square=8square+x square

100-64=x square

root under 36=x

6 cm=x

Therefore other diagnol will be 2*6=12cm

Now lets find area

Area of rhombus= 1/2* Diagnol 1*Diagnol 2

Area = 1/2*16*12

There fore,

Area =96 sqcm.

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Answer 2

OB = √AB² - OA²

= √(10)² - (8)²

= √100-64

= √36

= 6cm

Diagonal BD = 6×2 = 12cm

Area of rhombus = 1/2 × product of diagonals

= 1/2 × 12× 16

= 96 cm²

Hope it helps✔.