Question

5. Diagonals of a parallelogrum intersect each other at point O. ICAO - 5. BO - 12 and
AB - 13 then show that ABCD is a rhombus.​

Answer 1

Answer:

5, 12 and 13 form a Pythagorean triplet. Thus, Δ AOB is a right angle triangle, right angled at O. ... Thus, in parallelogram ABCD, diagonals bisect at right angles. Hence, ABCD is a rhombus.

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hope this helps you mate:)

Answer 2

answer :-

Given: AO = 5, BO = 12 and AB = 13.  To prove: ABCD is a rhombus.

Proof: AO = 5, BO = 12, AB = 13 [Given] 

AO2 + BO2 = 52 + 122 = 25 + 144 

∴ AO2 + BO2 = 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)] 

∴ ABCD is a rhombus.