Q25 of 77 Page 15

In Fig. 15.87, X and Y are the mid-point of AC and AB respectively, QP||BC and CYQ and BXP are straight lines. Prove that:

ar(Δ ABP) = ar(Δ ACQ).


In a ΔAXP and ΔCXB,

∠PAX = XCB (Alternative angles AP || BC)


AX = CX (Given)


∠AXP = ∠CXB (Vertically opposite angles)


ΔAXP ≅ ΔCXB (By ASA rule)


AP = BC (By c.p.c.t) (i)


Similarly,


QA = BC (ii)


From (i) and (ii), we get


AP = QA


Now,


AP || BC


And,


AP = QA


Area (APB) = Area (ACQ) (Therefore, Triangles having equal bases and between the same parallels QP and BC)


More from this chapter

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23

In Fig. 15.86, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that


(i) ar(Δ BDE) = ar(Δ ABC)


(ii) ar(Δ BDE) = ar(Δ BAE)


(iii) ar(Δ BFE) = ar(Δ AFD)


(iv) ar(Δ ABC) = ar(Δ BEC)


(v) ar(Δ FED) = ar(Δ AFC)


(vi) ar(Δ BFE) = 2ar(Δ EFD)

 24

D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(ΔBOE) = ar(Δ ABC)

 26

In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that:


(i) PE=FQ


(ii) ar(ΔAPE):ar(Δ PFA)= ar(Δ QFD)=ar(Δ PFD)


(iii) ar(Δ PEA) = ar(Δ QFD)

 27

In Fig. 15.89, ABCD is a ||gm. O is any point on AC. PQ||AB and LM||AD. Prove that:


ar(||gm DLOP) = ar(||gm BMOQ)