In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP. Prove that:
(i) ar(Δ PBQ) = ar(Δ ARC)
(ii) ar(Δ PQR) = 1/2 ar(Δ ARC)
(iii) ar(Δ RQC) = 3/8 ar(Δ ABC)
Answers
Given : In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP.
To Prove :
(i) ar(Δ PBQ) = ar(Δ ARC)
(ii) ar(Δ PQR) = 1/2 ar(Δ ARC)
(iii) ar(Δ RQC) = 3/8 ar(Δ ABC)
Proof :
(i) We know that each median of a triangle divides it into two triangles of equal area.
Since, CR is a median of ΔCAP
∴ ar (ΔCRA) = ar (ΔCAP) ………..(i)
Also, CP is a median of ΔCAB .
∴ar (ΔCAP) = ar (ΔCPB) ……..(ii)
From eq (i) and (ii), we get
∴ ar (ΔARC) = ½ ar (ΔCPB) ……... (iii)
PQ is a median of ΔPBC.
∴ ar (ΔCPB) = 2 ar (ΔPQB) ……….. (iv)
From eq (iii) and (iv), we get,
ar (ΔARC) = ar (ΔPBQ) …………...(v)
(ii) Since QP and QR medians of Δ's QAB and QAP .
∴ ar (ΔQAP) = ar (ΔQBP) ………….(vi)
and, ar (ΔQAP) = 2 ar (ΔQRP) ……...(vii)
From eq (vi) and (vii), we get,
ar (ΔPRQ) = 1/2ar (ΔPBQ) ………..(viii)
From eq (v) and (viii), we get,
ar (ΔPRQ) = ½ ar (ΔARC) ………..(ix)
(iii) Since CR is a median of ΔCAP.
∴ ar (ΔARC) = ½ ar (ΔCAP)
= ½ × {½ ar (ΔABC)}
(∵ CP is a median of Δ ABC)
ar (ΔARC) = ¼ ar (ΔABC) ………... (x)
Since, RQ is a median of Δ RBC.
∴ ar (ΔRQC) = 1/2 ar (ΔRBC)
= ½ [ar (ΔABC) – ar (ΔARC)]
= ½ [ar (ΔABC) - ¼ ar (∆ABC)]
[From eq (x) ]
= ½ [ar (ΔABC) (1- ¼)]
= ½ [ar (ΔABC) (¾ )]
= ½ × ¾ ar (ΔABC)
ar (ΔRQC) = ⅜ ar (ΔABC)
Hence proved….
HOPE THIS ANSWER WILL HELP YOU…..
Similar questions :
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).
https://brainly.in/question/15909759
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) = 1/8 ar(Δ ABC)
https://brainly.in/question/15909771
Answer:-
Given :
In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP.
To Prove :
(i) ar(Δ PBQ) = ar(Δ ARC)
(ii) ar(Δ PQR) = 1/2 ar(Δ ARC)
(iii) ar(Δ RQC) = 3/8 ar(Δ ABC)
Proof :
(i) We know that each median of a triangle divides it into two triangles of equal area.
Since, CR is a median of ΔCAP
∴ ar (ΔCRA) = ar (ΔCAP) ………..(i)
Also, CP is a median of ΔCAB .
∴ar (ΔCAP) = ar (ΔCPB) ……..(ii)
From eq (i) and (ii), we get
∴ ar (ΔARC) = ½ ar (ΔCPB) ……... (iii)
PQ is a median of ΔPBC.
∴ ar (ΔCPB) = 2 ar (ΔPQB) ……….. (iv)
From eq (iii) and (iv), we get,
ar (ΔARC) = ar (ΔPBQ) …………...(v)
(ii) Since QP and QR medians of Δ's QAB and QAP .
∴ ar (ΔQAP) = ar (ΔQBP) ………….(vi)
and, ar (ΔQAP) = 2 ar (ΔQRP) ……...(vii)
From eq (vi) and (vii), we get,
ar (ΔPRQ) = 1/2ar (ΔPBQ) ………..(viii)
From eq (v) and (viii), we get,
ar (ΔPRQ) = ½ ar (ΔARC) ………..(ix)
(iii) Since CR is a median of ΔCAP.
∴ ar (ΔARC) = ½ ar (ΔCAP)
= ½ × {½ ar (ΔABC)}
(∵ CP is a median of Δ ABC)
ar (ΔARC) = ¼ ar (ΔABC) ………... (x)
Since, RQ is a median of Δ RBC.
∴ ar (ΔRQC) = 1/2 ar (ΔRBC)
= ½ [ar (ΔABC) – ar (ΔARC)]
= ½ [ar (ΔABC) - ¼ ar (∆ABC)]
[From eq (x) ]
= ½ [ar (ΔABC) (1- ¼)]
= ½ [ar (ΔABC) (¾ )]
= ½ × ¾ ar (ΔABC)
ar (ΔRQC) = ⅜ ar (ΔABC)
Hence proved….
Hope it's help You❤️