Advertisements
Advertisements
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) .
Advertisements
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)
APPEARS IN
संबंधित प्रश्न
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
In the following 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) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
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)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that
(1) ar (Δ BDE) = `1/2` ar (ΔABC)
(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)
(3) ar (BEF) = ar (ΔAFD)
(4) area (Δ ABC) = 2 area (ΔBEC)
(5) ar (ΔFED) `= 1/8` ar (ΔAFC)
(6) ar (Δ BFE) = 2 ar (ΔEFD)
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
In the following figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).
