Question

1.
Diagonals of a parallelogram intersect each other at point 0. If AO = 5,
BO = 12 and AB = 13 then show that DABCD is a rhombus.​

Answer 1

Step-by-step explanation:

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. [A parallelogram is a rhombus perpendicular to each other]Read more on Sarthaks.com - https://www.sarthaks.com/849977/diagonals-of-parallelogram-intersect-each-other-at-point-if-ao-bo-show-that-abcd-is-rhombus

Answer 2

Answer:

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