Question

the area of the rhombus ABCD shown in adjoining figure..​

Answer 1

Answer:

HERE'S your answer mate,

hope it helps you...

Answer 2

Recall:

$\text {Area of rhombus} = \dfrac{1}{2} \times \text{diagonal 1} \times \text{diagonal 2}$

$\text {pythagoras theorem state that: }a^2 + b^ = c^2$

Note that:

The diagonals of a rhombus intersect at a right angle.

⇒ ∠AOB is a right angle

Find half of the other diagonal, AO:

$a^2 + b^2 = c^2$

$AO^2 + OB^2 = AB^2$

$AO^2 + 63^2 = 65^2$

$AO^2 = 65^2 - 63^2$

$AO^2 =256$

$AO = 16 \text { cm}$

Find the diagonal AC:

$AO = 16 \text { cm}$

$AC = 16 \times 2$

$AC = 32 \text { cm}$

Find the diagonal AC:

$BO = 63 \text { cm}$

$BD = 63 \times 2$

$AC = 126 \text { cm}$

Find the area:

$\text {Area of rhombus} = \dfrac{1}{2} \times \text{diagonal 1} \times \text{diagonal 2}$

$\text {Area of rhombus} = \dfrac{1}{2} \times 32 } \times 126$

$\text {Area of rhombus} = 2016 \text { cm}^2$

Answer: The area is 2016 cm²