Question

If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

Answer 1

Given : P is any point in the interior of a parallelogram ABCD .

 

To Prove :

Area of the triangle APB is less than the area of the parallelogram.

Construction: Draw DN⊥ AB and PM⊥ AB.

 

Proof :  

ar (|| gm  ABCD) = AB × DN ……(1)

ar (Δ APB) = ½ (AB × PM) ……(2)

Now,

PM < DN

AB × PM < AB × DN

[Multiplying by AB on both sides]

= ½ (AB × PM) < ½ (AB × DN)

[Multiplying by ½ on both sides]

= ar (Δ APB) < ½ ar(|| gm ABCD

[From eq 1 and 2]  

Hence proved that the area of the ∆ APB is less than the area of the parallelogram.

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Answer 2

Answer:-

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Given :

P is any point in the interior of a parallelogram ABCD .

 

To Prove :

Area of the triangle APB is less than the area of the parallelogram.

Construction: Draw DN⊥ AB and PM⊥ AB.

 

Proof :  

ar (|| gm  ABCD) = AB × DN ……(1)

ar (Δ APB) = ½ (AB × PM) ……(2)

Now,

PM < DN

AB × PM < AB × DN

[Multiplying by AB on both sides]

= ½ (AB × PM) < ½ (AB × DN)

[Multiplying by ½ on both sides]

= ar (Δ APB) < ½ ar(|| gm ABCD

[From eq 1 and 2]  

Hence proved that the area of the ∆ APB is less than the area of the parallelogram.

$\:$

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