Question

find the area of a rhombus having each side equal to 13cm one of whose diagonals is 24cm

Answer 1

let the other diagonal be x
then

diagonal of the rhombus = (1/2×13)^2 +(x/2)^2 =24^2

=56.25 + x^2/4 = 576

56.25+px^2=576/4

Answer 2

Answer: 120 cm²

Step-by-step explanation:

Let us name the rhombus ABCD. (Refer to the image.)

All sides of a rhombus are equal.

Side of the rhombus = AD = 13 cm

Diagonals of a rhombus perpendicularly bisect each other. Let the point of intersection of the diagonals be = O. (Refer to the image.)

Let us assume that the diagonal AC = 24 cm. (Given in the question)

AO = OC (∵ Diagonals bisect each other.)

AO = OC = 1/2 of AC

∴ AO = 12 cm

ΔAOD is a right-angled triangle. (∵ Diagonals are perpendicular to each other.)

According to pythagoras theorem,

AO² + OD² = AD²

(12cm)² + OD² = (13cm)²

144 cm² + OD² = 169 cm²

OD² = 169 cm² - 144 cm²

OD² = 25 cm²

OD = √25 cm²

∴ OD = 5 cm

Diagonal BD = 2 × OD (∵ Diagonals bisect each other.)

∴ BD = 2 × 5 cm = 10 cm

Area of rhobmus = 1/2 × Product of the diagonals

Area of ABCD = 1/2 × 24 cm × 10 cm

∴ Area of ABCD = 120 cm²