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

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

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)

Question 8: In this figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AC⊥DE meets BC at Y. Show that

(a) ΔMBC ≅ ΔABD

**Answer:** MB = AB (Sides of a square)

BC = BD (Sides of a square)

∠MBA = ∠DBC = 90° (angles of squares)

So, ∠MBA + ∠ABC = ∠DBC + ∠ABC

So, ∠MBC = ∠ABD

So, ΔMBC ≅ ΔABD (SAS Theorem)

(b) ar(BYXD) = 2 ar(MBC)

**Answer:** As BYXD and ΔABD are on the same base BD and between same parallels BD and AX

So, ar(ΔABD) = `1/2` ar(BYXD)

In previous question we proved ΔMBC ≅ ΔABD

So, ar(ΔMBC) = `1/2` ar(BYXD)

Or, ar(BYXD) = 2 ar(ΔMBC)

(c) ΔFCB ≅ ΔACE

**Answer:** FC = AC (sides of a square)

CB = CE (sides of a square)

∠ACF = ∠BCE = 90°

So, ∠ACF + ∠ACB = ∠BCE + ∠ACB

Or, ∠FCB = ∠ACE

So, ΔFCB ≅ ΔACE

(d) ar(CYXE) = 2 ar(FCB)

**Answer:** CYXE and ΔACE are on same base CE and between same parallels AX and CE

So, ar(CYXE) = 2 ar(ACE)

In previous question we proved ΔACE ≅ ΔFCB

So, ar(CYXE) = 2 ar(FCB)

(e) ar(CYXE) = ar(ACFG)

**Answer:** ACFG and Δ FCB are on same base FC and between same parallels GB and FC

So, ar(ACFG) = 2 ar(FCB)

In previous question, we proved ar(CYXE) = 2 ar(FCB)

So, ar(CYXE) = ar(ACFG)

(f) ar(BCED) = ar(ABMN) + ar(ACFG)

**Answer:** Ar(BCED) = BC^{2}

Ar(ACFG) = AC^{2}

Ar(ABMN) = BA^{2}

Here, BC is hypotenuse, and AC and BC are remaining two sides of right angle triangle ABC

According to Pythagoras theorem,

`h^2=p^2+b^2`

So, ar(BCED) = ar(ABMN) + ar(ACFG)

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