Question

Exercise 18.2
1. Each side of a rhombus is 13 cm and one diagonal is 10 cm. Find
(1) the length of its other diagonal
(ii) the area of the rhombus.​

Answer 1

From the question,

Rhombus ABCD with centre O,

side (S) = 13 cm

Shorter diagonal (P) = 10 cm

consider triangle OCB,

Applying pythagoras theorem,

$oc \: = \sqrt{ {(bc}^{2} } - {ob}^{2} ) \:$

$= \sqrt{(169 - 25)}$

$= \sqrt{144}$

$= 12$

OC = 12 cm

Longer diagonal, AC = 24 cm.

Area of rhombus ABCD = [(AC×BD)/2] = 10 × 24 /2

$area \: \: of \: \: rhombus \: = \: {120}^{2} cm$

Answer 2

✬ Other Diagonal = 24 cm ✬

✬ Area = 120 cm² ✬

Step-by-step explanation:

Given:

• Measure of each side of rhombus is 13 cm.
• Measure of diagonal is 10 cm.

To Find:

• What is the length of other diagonal and area of rhombus ?

Solution: As we know that diagonals of rhombus are equal to each other. Therefore,

➟ Other diagonal = 10 cm.

Also diagonals of rhombus bisect each other at 90°.

In rhombus ABCD we have

➭ AO = 1/2(AC)

➭ AO = 1/2(10) = 5 cm

➭ AB = 13 cm

In ∆AOB , by Pythagoras Theorem

$\implies{\rm }$ AB² = AO² + OB²

$\implies{\rm }$ 13² = 5² + OB²

$\implies{\rm }$ 169 = 25 + OB²

$\implies{\rm }$ 169 – 25 = OB²

$\implies{\rm }$ 144 = OB²

$\implies{\rm }$ √144 = OB

$\implies{\rm }$ 12 = OB

So,

➭ BD = 2(OB)

➭ BD = 2(12) = 24 cm

★ Ar. of Rhombus = 1/2(Diagonal¹)(Diagonal²) ★

Here,

• Two diagonals are 10 and 24 cm

$\implies{\rm }$ Area = 1/2(10)(24) cm²

$\implies{\rm }$ Area = 5(24) cm²

$\implies{\rm }$ Area = 120 cm²

Hence, the area of rhombus will be 120 cm².

_____________________

• Opposite sides of a rhombus are parallel.

• All sides of a rhombus are equal to each other.

• Perimeter of rhombus = 4(Side)