Question

The side of rhombus is 10cm and the length of one of the diagonals is 12cm .Find the length of the other diagonal

Answer 1

Answer:

${\sf{\therefore 2nd \: diagonal = 16cm}}$

Step-by-step explanation:

$\huge{\underline{\sf{SOLUTION-}}}$

• According to the given question:

• We know the properties of rhombus

• According to properties of rhombus:

$\implies sides \: of \: rhombus \: are \: equal \\\\ \implies diagonals \: bisects \: each \: other \: at \: 90 \degree \\ \\ \implies diagonals \: are \: unequal \: \\ \\ \implies rhombus \: form \: 4 \: equal \: right \: angled \: triangle$

• According to these properties:

$\bold{In \: a \: right \: angled \triangle} \\\\ \implies h= side \: of \: rhombus = 10 cm\\ \\\implies b = half \: of \: one \: diagonal = 6cm \\ \\ \bold{According \: to \: phythagoras \: theoram} \\ \\\implies {h}^{2} = {p}^{2} + {b}^{2} \\\\ \implies {10}^{2} = {p}^{2} + {6}^{2} \\ \\\implies 100 - 36 = {p}^{2} \\ \\ \implies 64 = {p}^{2} \\ \\ \implies p = \sqrt{64} \\ \\ { \bold{\implies p = 8cm}} \\ \\ \implies d_{2} = 2 \times p \\ \\ \implies d _{2} = 2 \times 8 \\\\ \bold{ \therefore 2nd \: diagonal = 16cm}$

Answer 2

Solution:-

let, Rhombus ABCD.

Given:-

•The sides of a rhombus are 10cm and one diagonal is 12 cm .

let, DO = OB = ? , BD = ?

and AO = OC = 6cm, and AC = 12 cm

1) we know diagonal of rhombus are equally bisect each other and they are perpendicular to each other.

2) All sides of rhombus are equal.

so,

by Pythagoras theorem.

=> (AB)² = ( AO )² + ( OB )²

=> (10)² = (6)² + (OB)²

=> 100 = 36 + (OB)²

=> 100 - 36 = (OB)²

=> 64 = (OB)²

i.e.

=> (OB)² = 64

=> OB = √64

=> OB = 8 cm

so, we know

DO = OB = 8 cm

hence, BD = 16 cm

Area of rhombus = [(AC)×(BD)]/2

Area of rhombus = [ 12 × 16 ]/2

Area of rhombus = [192]/2

Area of rhombus= 96 cm²

Hence length of diagonal rhombus is 16 cm and area of rhombus is

96 cm².

i hope it helps you.