Question

Find the area of rhombus having each side 20cm and one of it's diagonal is equal to 24cm​

Answer 1

Answer:

According to the question,

AB BC = CD = AD = 20 cmn

Consider, DB 24 cm

As diagonals of rhombus bisect each

other.

DO OB 12 cmn

In triangle AOB,

A0² +OB² = AB²

A0²+122 = 20²

AO² 400 - 144 = 256

AO 16 cm

Diagonal AC = 16 x 2 32 cm

Area of rhombus

$\tt{ \frac{1}{2} \times ac \times bd = \frac{1}{2} \times 32 \times 24 = {384cm}^{2} }$

Answer 2

Answer:

The area of the rhombus is 384cm

Given:

One side of a rhombus = 20cm

One diagonal = 24 cm

To find: The area of a rhombus

Solution:

In the rhombus ABCD,

Let, AB = 20 cm

And AC = 24 cm

From the diagram,

AO + CO = 24 cm

AO = 12 cm [AO = CO]

In ΔAOB,

AB = 20 cm and AO= 12 cm

Using Pythagoras Theorem,

BO= √20²-12² cm

= √400−144 cm

= √256 cm

= 16 cm

Hence, BO = DO = 16 cm

Therefore, BD = (16 + 16) cm = 32 cm

Area of a rhombus = ½ × Diagonal 1 × Diagonal 2

= ½ × AC × BD

= ½ × 24 cm × 32 cm

= 384 cm²

(Not my own answer, but hope it helps you)....