Class 9 Maths

Parallelograms

Exercise 9.4

Part 3

Question 6: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) × ar(CPD) = ar(APD) × ar(BPC)

(Hint: From A and C, draw perpendiculars to BD)

Answer: This figure shows a quadrilateral ABCD in which diagonals AC and BD intersect at P. Let us draw AM⊥BD and CN⊥BD

parallelogram

ar(ΔAPB) = `1/2` × BP × AM

ar(ΔCDP) = `1/2` × DP × CN

Or, ar(ΔAPB) × ar(ΔCDP) = (`1/2` × BP × AM) × (`1/2` × DP × CN)

`=1/4` × BP × DP × AM × CN ……………..(1)

Similarly, ar(ΔAPD) × ar(ΔBPC)

= (`1/2` × DP × AM) × (`1/2` × BP × CN)

`=1/4` × BP × DP × AM × CN …………….(2)

From equations (1) and (2)

ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar(ΔBPC)

Question 7: P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

(a) ar(PRQ) = `1/2` ar(ARC)

Answer: In ΔAPQ, R is the midpoint of AP

parallelogram

So, RQ is a median of ΔAPQ

So, ar(ΔPRQ) = `1/2` ar(ΔAPQ) ………. (1)

In ΔABQ, P is the midpoint of AB

So, QP is a median of ΔABQ

So, ar(ΔAPQ) = `1/2` ar(ΔABQ) ………….(2)

From equations (1) and (2)

ar(ΔPRQ) = `1/2xx1/2` ar(ΔABQ)

= `1/4` ar(ΔABQ) = `1/4xx1/2` ar(ΔABC)

(Since AQ is a median of ΔABC)

So, ar(ΔPRQ) = `1/8` ar(ΔABC) …………..(3)

Now, ar(ΔARC) = `1/2` ar(ΔAPC)

(Since CR is a median of ΔAPC)

= `1/2xx1/2` ar(ΔABC)

(Since CP is a median of ΔABC)

So, ar(ΔARC) = `1/4` ar(ΔABC) ……………(4)

From equation (3)

ar(ΔPRQ) = `1/8` ar(ΔABC)

= `1/2` × `1/4` ar(ΔABC))

= `1/2` ar(ΔARC) (From equation (4))

Hence, ar(ΔPRQ) = `1/2` ar(ΔARC)

(b) ar(RQC) = `3/8` ar(ABC)

Answer: PQ is a median in ΔRBC

So, ar(RQC) = ar(RBQ)

= ar(PRQ) + ar(BPQ)

= `1/8` ar(ABC) + ar(BPQ) (From equation (3) of question (a))

= `1/8` ar(ABC) + `1/2` ar(PBC) (Since PQ is a median of BPC)

`=1/8` ar(ABC) + `1/2xx1/2` ar(ABC) (since CP is a median of ABC)

`=1/8` ar(ABC) + `1/4` ar(ABC)

Thus, ar(RQC) = `3/8` ar(ABC)

(c) ar(PBQ) = ar(ARC)

Answer: QP is a median of ABQ

So, ar(PBQ) = `1/2` ar(ABQ)

`=1/2xx1/2` ar(ABC) (since AQ is a medina of ABC)

`=1/4` ar(ABC) = ar(ARC) (from equation (4) of question (a))

Thus, ar(PBQ) = ar(ARC)