Question

hey guys,

P and Q are two points lying on the sides of DC and AD respectively of a parallelogram
ABCD. show that a (APB) = a(BQC).
plz: ❌no spam ❌

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Answer 1

Given :

P and Q are two points lying on the sides of DC and AD respectively of parallelogram ABCD

To proof :

ar(Triangle ABP) is equal to ar (Triangle BQC)

Solution :

First refer to the attachment for labelling. This will help in understanding further process

Let's head to the question now

Given ABCD is a parallelogram and we know that opposite sides of parallelogram are parallel.

Therefore,

AB||CD and AD||BC

Now,

It is clear from attachment that parallelogram ABCD and and triangle APB are having common base i.e. AB and they are between the same parallel lines AB and DC.

Property :

If parallelogram is having same base and parallels the area of triangle is half of parallelogram.

$\star \sf \: Area \: of \: \triangle \: APB = \dfrac{1}{2} (Area \: of \: parallelogram ABCD)$

We will consider this as equation(i).....

Next,

Parallelogram ABCD and triangle QBC are having common base i.e. BC and they are between the same parallel lines AD and BC.

Property :

If parallelogram is having same base and parallels the area of triangle is half of parallelogram.

$\star \sf \: Area \: of \: \triangle \: QCB \: = \dfrac{1}{2} (Area \: of \: parallelogram ABCD)$

We will consider this as equation(ii).....

From equation(i)..... and equation(ii).....

Area of Triangle ABP is equal to Area of Triangle BQC

Hence,Proved!!

Answer 2

❁❁ Refer To Image First .. ❁❁

Given:--- In parallelogram ABCD, P & Q any two points lying on the sides DC and AD.

To Prove:--- ar (APB) = ar (BQC).

Concept Used :--- If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.

Proof :----

Here, ΔAPB and ||gm ABCD stands on the same base AB and lie between same parallel AB and DC.

Therefore,

ar(ΔAPB) = 1/2 ar(||gm ABCD) —-------------- Equation(1)

Similarly,

Parallelogram ABCD and ∆BQC stand on the same base BC and lie between the same parallel BC and AD.

ar(ΔBQC) = 1/2 ar(||gm ABCD) —-------------- Equation(2)

From eq (1) and (2), we can say that :---

ar(ΔAPB) = ar(ΔBQC)

✪✪ Hence Proved ✪✪

______________________________

★★Extra Brainly Knowledge★★

✯✯ Some Properties of Isosceles Trapezium ✯✯

→ Opposite sides are parallel by definition.

→ Opposite sides are congruent.

→ Opposite angles are congruent.

→ Consecutive angles are supplementary.

→ The diagonals bisect each other.