Question

ABCD is a parallelogram points p and q on bc trisect bc on three equal parts prove that ar of triangle APQ=ar of triangle PQD=1\6ar of parallelogram ABCD

Answer 1

Answer:

Step-by-step explanation:

Student name Varun Jogi

ABCD is a parallelogram.P and Q on BC trisects it.Prove that ar(APQ) =ar(DPQ)=1/6ar(ABCD)

Given, ABCD is a parallelogram in which points P and Q trisects the BC.

⇒ BP = PQ = QC

To prove:

Construction: Through P and Q, draw PR and QS parallel to AB and CD.

Proof:

Now, ΔAPD and ΔAQD lie on the same base AD and between the same parallel AD and BC.

∴ ar (ΔAPD) = ar (ΔAQD)

⇒ ar (ΔAPD) - ar (ΔAOD) = ar (ΔAQD) - ar (ΔAOD) [on subtracting ar (ΔAOD) from both sides]

⇒ ar (ΔAPO) = ar (ΔOQD) ..... (1)

⇒ ar (ΔAPO) + ar (ΔOPQ) = ar (ΔOQD) + ar (ΔOPQ) [on adding ar (ΔOPQ) on both sides]

⇒ ar (ΔAPQ) = ar (ΔDPQ) ..... (2)

Again, ΔAPQ and parallelogram PQSR are on the same base PQ and between same parallels PQ and AD

.... (3)

Now,

Using (2), (3) and (4), we get

[Hence proved]

Answer 2

As ∆ APQ and ∆ DPQ lie on same base and between same ||

area of ∆APQ=area of∆DPQ- 1

construction-extend P and Q to AD to point Z and L.

area of ||gm PZLQ= 1/3area of ||gm ABCD -2

area of ∆ APQ= 1/2area of ||gm ABCD -2(Iie on same base and between same||) -3

from 2 and 3

area of∆APQ =1/3 * 1/2= 1/6 -4

FROM 4 and 1

area of ∆APQ=area of ∆PDQ= 1/6 area of ||gm

hope it helps