Question

the diagonals of a rhombus are 30 cm and 16 cm find its perimeter​

Answer 1

Answer:

Hope this will help you

Answer 2

Given:

• The diagonals of a rhombus are 30 cm and 16 cm find its perimeter.

Diagram:

• Refer to the attachment please

Solution:

• We know that diagonals of a rhombus bisect each other at right angles.

Let ABCD be a rhombus and AC, BD be it's two diagonals, where:

• AC = 30 cm

• BD = 16 cm

Let the two diagonals bisect each other at E.

Then,

• ∠BEC = 90°

• EC = 15 cm

• EB = 8 cm

In right angled BEC,

$\\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: BC {}^{2} = BE {}^{2} + EC {}^{2} \\ \implies \: \: \: \: \: \: \: \: \sf \: BC {}^{2} = (8 + 15) {}^{2} cm \\ \: \: \: \: \: \: \: \: \: \: \implies \ \sf \: \: \: \: \: \: \: \: \: \: \: BC {}^{2} = (64+225) {}^{2} cm {}^{2} \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: BC {}^{2} = 289cm {}^{2} \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \:BC {}^{2} = (17) {}^{2} cm \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \:BC {}^{} = 17 cm$

In a rhombus, all sides are if equal lengths.

$\\ \therefore \: \: \: \: \: \: \: \: \: \: \rm \: Perimeter \: of \: rhombus = 4 × side \: of \: rhombus \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = (4 \times 17) cm\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \ \sf \: = 68 \: cm$

${ \underline{ \underline{ \boldsymbol{ \bigstar{ \pink{\:Hence, \: The \: perimeter \: of \: rhombus \: is \: 68 cm.}}}}}}$