Question

ABCD is a parallelogram with AB || DC and P is any point inside the parallelogram.If the ABCD is 120 sq cm, then the sum of the area of triangle PAB
and triangle PCD is
(A) 120 sq cm
(B) 30 sq cm
(C) 60 sq cm
(D) 90 sq cm​

Answer 1

If ABCD is 120 cm², then the sum of the area of ΔPAB and ΔPCD is 60 cm².

Option (C) 60 cm² is correct.

Concept:

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

Given:

ABCD is a parallelogram with AB || DC and an area of 120 cm². P is a point inside the parallelogram

To find:

ar ΔPAB + ar ΔPCD

Construction :

Draw a line passing through P such that EF || AB.

Solution:

Step1: Proof that ABFE and CDEF a parallelogram:

AB || CD ……....(1)    [Given ]

AB || EF …....... (2)    [By construction]

From eq. (1) & (2)

EF || CD ….. (3)

[Lines parallel to the same line are parallel to each other.]

AD || BC      [Given ABCD is a parallelogram]

Therefore EA || FB  and DE || CF ….. (4)

From eq 2. and 4,  we have

AB || EF and  EA || BF

Hence, ABFE is a parallelogram.

Similarly, From eq 3. and 4,  we can prove that CDEF is a parallelogram.

Step2: Proof that the area of the triangle is half the area of the parallelogram in ΔPAB and || gm ABFE:

ΔPAB and || gm ABFE stands on the same base AB and lies between the same parallel AB and FE.

Therefore,

ar(ΔPAB) = $\frac{1}{2}$ ar( ||gm ABFE) …….. (1)

Step3: Proof that the area of the triangle is half the area of the parallelogram in ΔPCD and ||gm CDEF:

Similarly,

Parallelogram CDEF and ∆PCD stand on the same base CD and lie between the same parallel EF and CD.

ar(∆PCD) =  $\frac{1}{2}$ ar(||gm CDEF) …….. (2)

Step4: Adding eq.1 and 2:

ar(ΔPAB) + ar(∆PCD) =  $\frac{1}{2}$ ar ( || gm ABFE) +  $\frac{1}{2}$ ar (|| gm CDEF)

ar(ΔPAB) + ar(∆PCD) =  $\frac{1}{2}$ [ar ( || gm ABFE) + ar (|| gm CDEF ]

ar(ΔPAB) + ar(∆PCD) =  $\frac{1}{2}$ ar ( || gm ABCD)

ar(ΔPAB) + ar(∆PCD) =  $\frac{120}{2}$cm²

[Given: ar( ||gm ABCD) = 120 cm²]

ar(ΔPAB) + ar(∆PCD) = 60 cm²

Hence, the sum of the area of ΔPAB and ΔPCD is 60 cm².

Option (C) 60 cm² is correct.

On solving we get the sum of the area of ΔPAB and ΔPCD is 60 cm².

Hence, all the 3 options (A) 120 sq cm, (B) 30 sq cm and (D) 90 sq cm​ is incorrect.

Learn more on Brainly:

Question 3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

https://brainly.in/question/1426052

Question 14 In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

https://brainly.in/question/1427787

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