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Maharashtra State BoardSSC (English Medium) 9th Standard

Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that □ABCD is a rhombus. - Geometry

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Question

Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that `square`ABCD is a rhombus. 

Sum
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Solution

Given: AO = 5, BO = 12 and AB = 13 

To Prove: `square`ABCD is a rhombus.

Proof: 

AO2 + BO2 

= 52 + 12

= 25 + 144

= 169   ...(i)

∴ AB2 = 132 = 169    ...(ii)

∴ AB2 = AO2 + BO2      ...[From (i) and (ii)]

∴ ∆AOB is a right-angled triangle.     ...[Converse of Pythagoras theorem]

∴ ∠AOB = 90°

∴ seg AC ⊥ seg BD    ...(iii) [A-O-C]

∴ In parallelogram ABCD,

∴ seg AC ⊥ seg BD     ...[From (iii)]

∴ `square`ABCD is a rhombus.      ...[A parallelogram is a rhombus perpendicular to each other]

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Chapter 5: Quadrilaterals - Practice Set 5.1 [Page 62]

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Balbharati Mathematics 2 [English] Standard 9 Maharashtra State Board
Chapter 5 Quadrilaterals
Practice Set 5.1 | Q 5 | Page 62

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