Question

If the sides of a rhombus are 10 cm each and one
diagonal is 16 cm, find the area of the rhombus.​

Answer 1

Answer :-

Area of the rhombus is 96 cm²

Explanation :-

For solving this problem first let us draw draw rhombus ABCD and let the the intersection of diagonals be 'O' (Refer attachment )

Given

• All sides are of 10 cm i.e AB = BC = CD = AD = 10 cm

• One of the diagonal is 16 cm i.e AC = 16 cm

Diagonals in a rhombus bisect each other perpendicularly.

• AO = AC/2 = 16/2 = 8 cm

• ∠AOB = 90 °

Consider Δ AOB

As one of the angle in Δ AOB is 90° it is a right angled triangle

By Pythagoras theorem

AO² + OB² = AB²

⇒ 8² + OB² = 10²

⇒ 64 + OB² = 100

⇒ OB² = 100 - 64

⇒ OB² = 36

⇒ OB² = 6²

⇒ OB = 6

Diagonals in a rhombus bisect each other perpendicularly

∴ BD = 2OB = 2(6) = 12 cm

Finding the area of the rhombus ABCD

One of the diagonal (d₁) = 16 cm

Another diagonal (d₂) = 12 cm

We know that

Area of the rhombus = (d₁ * d₂)/2

= (16 * 12)/2

= 192/2

= 96 cm²

∴ Area of the rhombus is 96 cm².

Answer 2

$\large{\mathfrak{\red{\underline{\underline{........Solution........}}}}}$

Given:

=> AB = BC = CD = AD = 10

=> One side of Diagonal is 16 i.e BD = 16 cm.

To Find:

=> Area of rhombus

Formula used:

$\sf{\implies Area\;of\;rhombus=\dfrac{1}{2}\times d_{1}\times d_{2}}$

So, as we know the diagonals of a rhombus bisect each other at right angles.

=> OB = 1/2 BD

=> OB = 1/2 × 16 = 8cm

And,

=> ∠AOB = 90°

Now, using Pythagoras in right triangle ΔAOB,

=> OA² + OB² = AB²

=> OA² + 8² = 10²

=> OA² = 100 - 64

=> OA² = 36

=> OA = 6 cm

∴ AC = 2 OA

=> AC = 12 cm

Thus, the length of other diagonal is 12 cm.

Now, we will find the area of rhombus.

$\sf{\implies Area\;of\;rhombus=\dfrac{1}{2}\times d_{1}\times d_{2}}$

$\sf{\implies \dfrac{1}{2}\times 12\times 16}$

$\sf{\implies 96\;cm^{2}}$

∴ Hence, Area of the rhombus is 96 cm².