Question

answer in detail with steps

Answer 1

Answer: 30 cm.

Step by step explanation:

Refer the attachment for the rhombus ABCD.

Given :

Perimeter of rhombus ABCD = 100 cm.

AC = 40 cm.

[Perimeter of rhombus = 4 × Side]

So, one side of rhombus = 100/4 = 25 cm.

AB = BC = CD = DA = 25 cm.

We know that,

Diagonals of rhombus bisect each other at 90°.

.°. Angle AOB = 90°

AO = 1/2 × AC

AO = 1/2 × 40 = 20 cm.

Thus, In ∆ BOA,

(AB)² = (BO)² + (AO)²

(BO)² = (AB)² - (AO)²

BO = √[(AB)² - (AO)²]

BO = √[(25)² - (20)²]

BO = √[625 - 400]

BO = √225

BO = 15 cm.

Length of other diagonal (BD) = 2 × BO

= 2 × 15

BD = 30 cm.

Thus, the length of other diagonal is 30 cm.

Answer 2

GIVEN:

• Perimeter of rhombus = 100cm

• One of diagonal $\sf{ [d_1(AO)]}$ = 40cm

TO FIND:

• Other diagonal $\sf{[d_2(BO)]}$ = ?

SOLUTION:

We know that

★ Area$\sf{_{Rhombus}}$ = 4(side)

→ 4s = 100

→ s = 100/4

→ side(AB) = 25

★ Diagonal of rhombus bisect perpendicularly each other

$\therefore\angle$ AOB = 90°

AO = ½ × AC

→ AO = ½ × 40

→ AO = 20

Taking ∆BOA

Using Pythagoras theorem

AB² = BO² + AO²

→ 25² = BO² + 20²

→ 625 - 400 = BO²

→ 225 = BO²

→ BO = $\sqrt{\sf{225}}$

→ BO = 15cm

Length of Diagonal (BD)

→ BO = ½ × BD

→ BD = 15 × 2

→ BD = 30

$\therefore$ Hence, Other diagonal = 30cm