Question

what is the area of a rhombus with side 10 cm and length of the diogonal 16 cm?​

Answer 1

Answer:

96 cm square is the answer

Answer 2

Answer:

96cm²

Step-by-step explanation:

Let's take a rhombus ABCD with a diagonal BC 16cm, length of each side 10cm and centre as O.

[All the sides of a rhombus are equal in length, ∴ AC = CD = BD = AB]

First of all, we will find the length another diagonal in order to find the area of the rhombus.

Since diagonals of a rhombus bisect each other at 90°, therefore,

BO = OC = $\sf\dfrac{BC}{2} = \dfrac{16cm}{2} = 8cm$

Now, in right-angled ∆ BOD [Using Pythagoras theorem]

$\rm \dashrightarrow BO^2 + OD^2 = BD^2$

$\rm \dashrightarrow (8cm)^2 + OD^2 = (10cm)^2$

$\rm \dashrightarrow OD^2 = 100cm^2 - 64cm^2$

$\rm \dashrightarrow OD^2 = 36cm^2$

$\rm \dashrightarrow OD = 6cm \qquad ..[on \ taking \ the \ square \ root]$

Now, OD = OA

$\therefore\rm \quad AD = OD + OA$

$\dashrightarrow \rm AD = 6cm + 6cm$

$\dashrightarrow\rm AD = 12cm$

Now, Area of the rhombus = $\rm \dfrac{1}{2} \times AD \times BC$

$\rm \dashrightarrow\dfrac{1}{2} \times 12cm \times 16cm$

$\bf \dashrightarrow 96cm^2$

Hence, the area of the rhombus is 96cm².