Question

find the area of Rhombus having each side equal to 15 cm and one of those diagonals is 24 cm​

Answer 1

Answer:

A=216cm²

Step-by-step explanation:

ar of rhombus = 1/2 x d1 x d2

Let ABCD is a rhombus with diagonals AC and BD which intersect each other at O.

AC = 24 ⇒ AO = 12

Let BO = x and AB = 15 cm (given)

By Pythagorean theorem

c2 = a2 + b2

225 = 144 + x²

x2 = 225 – 144

x2 = 81

x = 9 cm

BO = 9 cm

Diagonal BD = 2 x 9= 18cm.

Area = ½ x [ product of diagonals]

= ½ x 24 x 18

Area = 216sq.cm

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Answer 2

Answer:

$\sf \large \underline \green{given} \\ \\ \sf \: two \: diagonals \: of \: rhombus\\ \\ \sf \large \underline \pink{to \: find} \\ \\ \sf \: area \: of \: rhombus \\ \\ \sf \large \underline \blue{solution} \\ \\ \sf \: area \: of \: rhombus \\ \\ = \sf \: \frac{1 \: diagonal \times 2 \: diagonal}{2} \\ \\ \sf \: \frac{15 \times 24}{2} = 15 \times 12 = 180 {cm}^{2}$