ABCD is a trapezium in which AB and CD are the parallel sides. The diagonals meet at point 0. If area of triangle ABO is p and area of triangle COD is q, prove that area of the trapezium is (√p+√q)²

Question

Arun · Answer

Let h and g are the heights of the triangles AOB and COD respectively.Then, 1/2 x AB x h = p and 1/2 x CD x g = q so, AB = 2p/h and 2q/g also, p/q = AB²/CD² so, AB/CD = vp/vq

Area of trapezium = 1/2(sum of parallel sides) x distance between them

=1/2 (AB + CD) x (h + g)

= 1/2 (vp/vqCD + CD) x (2p/AB +2q/CD)

Now replacing AB by 2p x vq/vp x CD and then solving the equation we get the area as (vp +vq)² .

where v stands for squareroot

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

ABCD is a trapezium in which AB and CD are the parallel sides. The diagonals meet at point 0. If area of triangle ABO is p and area of triangle COD is q, prove that area of the trapezium is (√p+√q)²

ABCD is a trapezium in which AB and CD are the parallel sides. The diagonals meet at point 0. If area of triangle ABO is p and area of triangle COD is q, prove that area of the trapezium is (√p+√q)²

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on 10 grade maths

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship