Question

In the given figure, ABCD is a parallelogram. If ar(∆BAP) = 10cm² & are(∆CPD) = 30cm² then find ar(//gm ABCD)​

Answer 1

Answer:

Area of this parallelogram is 80cm^2.

Step-by-step explanation:

Draw a line on BC parallel to AB & DC from point P. And let the drawn line touches the line BC at O.

= > area of ∆POB = area of ∆PAB { since both are congruent }

= > area of ∆POB = 10 cm^2

= > area of ∆POC = area of ∆PDC { since both are congruent }

= > area of ∆POC =30 cm^2

Thus,

= > Area of ∆BPC = area of POB + area of POC = 30 cm^2 + 10 cm^2 = 40 cm^2

= > Area of the parallelogram = area of ∆PAB + area of ∆PDC + area of BPC

= > Area of the parallelogram = 30 cm^2 + 10 cm^2 + 40 cm^2 = 80 cm^2

Hence, area of the parallelogram is 80 cm^2.

Answer 2

$\mathbb \pink{given}$

ar (∆ BAP) = 10 cm ²

ar (∆CPD) = 30 cm²

Let draw a line PO parallel to AB and DC

As we know that if a triangle and parallelogram are on same base and between same parallels that the area of triangle will be half of the area of parallelogram.

2 × ar(∆PBC) = ar( ABCD )

Now in ar( ∆ ABP) = ar( ∆ BPO )

and ar(∆DCP) = ar(∆POC)

Because these are congurent

ar(∆POB ) = 10 cm²

ar (∆POC) = 30 cm²

Area of∆ PBC = ar(∆POB) + ar(∆POC)

ar(∆PBC) = 10+30 = 40 cm²

Area of ABCD = 2{ar(∆PBC)

Area of ABCD = 2 × 40

Area of ||gm ABCD = 80 cm²