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Maharashtra State BoardSSC (English Medium) 9th Standard

Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.

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Question

Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.

Sum
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Solution

Let `square`ABCD be the rhombus.

AC = 20 cm, BD = 21 cm

AO = `1/2` AC        ...[Diagonals of a rhombus bisect each other]

= `1/2 xx 20`

= 10 cm       ...(i)

BO = `1/2` BD         ...[Diagonals of a rhombus bisect each other]

= `1/2 xx21`

= `21/2`

= 10.5 cm      ...(ii)

In ∆AOB,

∠AOB = 90°       ...[Diagonals of a rhombus are perpendicular to each other]

By Pythagoras theorem,

∴ AB2 = AO2 + BO2

∴ AB2 = 102 + 10.52

∴ AB2 = 100 + 110.25

∴ AB2 = 210.25

∴ AB = `sqrt(210.25)`

∴ AB = 14.5 cm

Perimeter of `square`ABCD = 4 × AB

= 4 × 14.5

= 58 cm

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Chapter 5: Quadrilaterals - Practice Set 5.3 [Page 69]

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Balbharati Mathematics 2 [English] Standard 9 Maharashtra State Board
Chapter 5 Quadrilaterals
Practice Set 5.3 | Q 4 | Page 69

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