Question

Find the area of a rhombus having one side of which measures to 15 cm and one of whose diagonals is 24 cm

Answer 1

hello friend!

given :-

side of the rhombus = 15cm

diagonal of the rhombus = 24cm

by pythagoras theorem :-

hyponetuse = 15cm

height = 24/2 = 12cm

base = a

15^2 = 12^2 + a^2

225 = 144 + a^2

225 - 144 = a^2

81 = a^2

a = 81^1/2

a = 9cm

hence, base is 9cm therefore another diagonal of the rhombus = 9 * 2 = 18cm

now area of the rhombus = 1/2 * d1 * d2

= 1/2 * 15 * 18

= 15 * 9

= 135cm^2

i hope it will help you..

Answer 2

Heya !!!

Let ABCD is a rhombus.

In which,

AB = AD = BC = DC = 15 cm.

And,

AC , BD are the Diagonals of ABCD.

Let,

AC = 24 cm.

We know that,

The diagonals of a rhombus bisect each other at right Angles.

Therefore,

OA = OC = 12 cm

Therefore,

By pythagoras theroem , we have :

AB² = (OA)² + (OB)²

(15)² = (12)² + (OB)²

(OB)² = (15)² - (12)²

(OB)² = 225 - 144

(OB)² = 81

OB = ✓81 = 9 cm

Therefore,

BD = 2 × OB = 2 × 9 = 18

************************************************

Area of rhombus = 1/2 × BD × AC

=> 1/2 × 18 × 24

=> 18 × 12

=> 216 cm².

HOPE IT WILL HELP YOU........ :-)