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प्रश्न
In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
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उत्तर

(1) We know that each median of a Δle divides it into two triangles of equal area
Since, OR is a median of ΔCAP
∴ ar (ΔCRA) = `1/2` ar (ΔCAP) ....... (1)
Also, CPis a median of ΔCAB
∴ ar ( ΔCAP) ar (ΔCPB) ....... (2)
From (1) and (2) we get
∴ area (Δ ARC ) = `1/2 ar (CPB)` ....... (3)
PQ is the median of ΔPBC
∴ area( Δ CPB) = 2area (Δ PBQ) ......... (4)
From (3) and (4) we get
∴ area (Δ ARC) = area (PBQ) ....... (5)
(2) Since QP and QR medians of s QAB and QAP respectively.
∴ ar (ΔQAP) = area (ΔPBQ) ............ (6)
And area (ΔQAP) = 2ar (QRP) ......... (7)
From (6) and (7) we have
Area (ΔPRQ) = `1/2` ar (ΔPBQ) ......... (8)
From (5) and (8) we get
Area (ΔPRQ) = `1/2` area (ΔARC)
(3) Since, ∠R is a median of ΔCAP
∴ area (ΔARC) = `1/2` ar (ΔCAP)
`= 1/2 xx1/2[ ar (ABC)]`
= `1/4` area (ABC)
Since RQ is a median of ΔRBC
∴ ar (ΔRQC) =`1/2` ar (Δ RBC)
= `1/2`[ ar (ΔABC)- ar (ARC) ]
= `1/2`[ar (ΔABC) - `1/4`(Δ ABC )]
= `3/8`(Δ ABC)
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संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)

Show that the diagonals of a parallelogram divide it into four triangles of equal area.
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the following 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 ⊥ DE meets BC at Y. Show that:-

(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
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