Question

find the area of a rhombus one side of which measures 20cm and one of whose diagonals is 24cm .​

Answer 1

Answer:

Let ABCD be the rhombus where AC = 10 cm and BD = 24 cm.

Let AC and BD intersect each other at O. Now, diagonals of rhombus bisect each other at right angles. Thus, we have.

Since AOB is a right angled triangle, by pythagoras theorem, we have.

Hence, AB = 13. Thus, length of each side of rhombus is 13 cm.

Answer 2

Answer:

384cm²

Explanation:

area of a rhombus =

$\frac{1}{2} p \sqrt{4 {a}^{2} - p^{2} }$

where,

p = diagonal = 24cm

a = side = 20cm

$12 \sqrt{1600 - 576} \\ = 12 \sqrt{1024} \\ = 12 \times 32 \\ = 384 \: cm^{2}$

Please mark me as the brainliest...