Question

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Answer 1

Answer:

(A) 24 cm²

Step-by-step explanation:

Let us take a rhombus ABCD with O as the point of intersection of both of its diagonals and AC as its diagonal of length 8cm

We know that all the sides of a rhombus are of the same length. So,

$4 \times side = perimeter$

We are given the value of the perimeter to be 20cm. So,

$4 \times side = 20cm \\ side = \frac{20cm}{4} \\ side = 5cm$

We also know that a parallelogram's diagonals bisect each other. Then BD bisects AC at O and AC bisects BD at O. Which leads to AO and CO with length 4cm.

Now in triangle AOB, angle AOB is 90° since the diagonals intersect each other at right angles.

Then,

AO² + BO² = AB²

(4cm) + BO² = (5cm)²

BO² = 25cm² - 16cm²

BO² = 9cm²

BO = 3cm

Since BO is half the length of BD, hence BD is 6cm

Now can find the area of the rhombus, which is given by half the product of its diagonals

$\frac{1}{2} \times ac \times bd \\ \frac{1}{2} \times 8cm \times 6cm \\ {24cm}^{2}$

Hence option A is correct