Question

find area of rhombus whose one side is 10 cm and one of the diagonal is equal to 12 cm.

Answer 1

The diagonals in a rhombus are perpendicular , and bisect each other

Given length of side is 10 cm and one diagonal as 12 cm we can find the length of other diagonal

From the  above figure let DB be the 12 cm diagonal,now OB and OD are 6 cm

Now consider triangle AOB ,this is a right andled triangle and we know AB=10cm and OB=6cm ,using Pythagoras theorem we can find out lenth of OA as 8cm ,so length of diagonal AC is 2*8=16cm

Area of rhombus is product of diagonal divided  by 2, 16*12/2=96 sq.cm

Answer 2

Answer:

96 cm ^2

Step-by-step explanation:

Let the diagnals be d1, d2.

Given: d1 = 12cm

Since diagonals bisect each other at right angles,

By Pythagoras theorem,

( d1 / 2 ) ^2 + ( d2 / 2 ) ^2 = 10 ^2

( 12 / 2 ) ^2 + ( d2 / 2 ) ^2 = 100

36 + ( d2 / 2 ) ^2 = 100

( d2 / 2 ) ^2 = 64

d2 / 2 = 8

d2 = 16 cm

Area of a rhombus =

( 1 / 2 ) × product of diagonals

= ( 1 / 2 ) × d1 × d2

= ( 1 / 2 ) × 12 × 16

= 96 cm ^2