(i) ar(PRQ) = \(\frac{1}{2} \) ar(ARC)

(ii) ar(PBQ) = ar(ARC)

(iii) ar(RQC) = \(\frac{3}{8} \) ar(ABC)

Answer :

We have a \(\triangle \)ABC such that P is the mid-point of AB and Q is the mid-point of BC.

Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.

(i) In \(\triangle \) APQ , R is the mid point of AP(Given )

\(\therefore \) RQ is the median of \(\triangle \) APQ

\(\Rightarrow \) ar(PRQ) = \(\frac{1}{2} \) ar (APQ) ..... (1 )

In triangle ABQ ,P is the mid point of AB

\(\therefore \) QP is the median of \(\triangle \) ABQ

\(\Rightarrow \) ar(APQ) = \(\frac{1}{2} \) ar (ABQ) ..... (2 )

from (1) and (2) we get

\(\Rightarrow \) ar(PRQ) = \(\frac{1}{2} × \frac{1}{2} \) ar (ABQ)

\(\Rightarrow \) ar(PRQ) = \(\frac{1}{4} \) ar (ABQ) = \(\frac{1}{4} × \frac{1}{2} \) ar (ABC)

[ \(\because \) AQ is the Median of \(\triangle \) ABC ]

\(\therefore \) ar(PRQ) = \(\frac{1}{8} \) × ar (ABC).....(3)

Now ar(ARC) = \(\frac{1}{2} \) ×ar(APC)

[ \(\because \)
CR is a Median of \(\triangle APC \) ]

\(\Rightarrow \) ar(ARC) = \(\frac{1}{2} \frac{1}{2} \) × ar(ABC)

[ \(\because \)
CP is a Median of \(\triangle ABC \) ]

\(\therefore \) ar(ARC) = \(\frac{1}{4} \) × ar(ABC) .....(4)

Now from (3) we get

ar(PRQ) = \(\frac{1}{8} \) (ABC)

ar(PRQ) = \(\frac{1}{2} \frac{1}{4} \) ar(ABC)

ar(PRQ) = \(\frac{1}{2} \) (ARC)

(iii) In triangle RBC , RQ is a Median

\(\therefore \) ar(RQC) = ar(RBQ)

\(\Rightarrow \) ar(RQC) = ar(PRQ) + ar(BPQ)

\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} \)× ar(ABC) + ar(BPQ)

\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} \) × ar(ABC) + ar(BPQ)

\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} × ar(ABC) + \frac{1}{2} \) ar(BPQ)

[\(\because \) PQ is the Median of BPC ]

\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} × ar(ABC) + \frac{1}{2} \frac{1}{2} \) ar(ABC)

[\(\because \) CP is the Median of ABC ]

\(\Rightarrow \) ar(RQC) = ( \(\frac{1}{4} + \frac{1}{8} \) ar(ABC)

Thus , ar(RQC) = ( \(\frac{3}{8} \) ar(ABC)

(ii) QP is the Median of \(\triangle \) ABQ

\(\therefore \) ar\(\triangle \) PBQ = \(\frac{1}{2} \) ar(ABQ)

= \(\frac{1}{2}× \frac{1}{2} \) × ar(ABC)

= \(\frac{1}{4} \) × ar(ABC) = ar(ARC )

From equation (4)

Thus ar(PBQ) = ar(ARC)

- Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
- In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).
- In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).
- In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(\(\triangle \)BPC) = ar(\(\triangle \)DPQ).[Hint Join AC.]
- In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that (i) ar (BDE) = \(\frac{1}{4} \) ar (ABC) (ii) ar (BDE) = \(\frac{1}{2} \) ar (BAE) (iii) ar (ABC) = 2 ar (BEC) (iv) ar (BFE) = ar (AFD) (v) ar (BFE) = 2 ar (FED) (vi) ar (FED) = \(\frac{1}{8} \) ar (AFC) [Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
- Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) x ar(CPD) = ar(APD) x ar(BPC). [Hint From A and C, draw perpendiculars to BD.]
- In 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 AX \(\perp \) DE meets BC at Y. Show that (i) \(\triangle \)MBC \(\cong \) \(\triangle \)ABD (ii) ar(BYXD) = 2 ar(MBC) (iii) ar(BYXD) = ax(ABMN) (iv) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE (v) ar(CYXE) = 2 ar(FCB) (vi) ar(CYXE) = ar(ACFG) (vii) ar(BCED) = ar(ABMN) + ar(ACFG)

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