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)
Solution:
Let's draw the given triangle ABC.
i) PC is the median of ΔABC.
∴ ar (ΔBPC) = ar (ΔAPC)..........(i)
RC is the median of ΔAPC.
∴ ar (ΔARC) = 1/2 ar (ΔAPC)......(ii)
[Median divides the triangle into two triangles of equal area]
PQ is the median of ΔBPC.
∴ ar (ΔPQC) = 1/2 ar (ΔBPC).................. (iii)
From equation (i) and (iii), we get,
ar (ΔPQC) = 1/2 ar (ΔAPC)................. (iv)
From equation (ii) and (iv), we get,
ar (ΔPQC) = ar (ΔARC)..........(v)
We are given that P and Q are the mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = 1/2 AC
⇒ ar (ΔAPQ) = ar (ΔPQC)..........(vi) [triangles between same parallel are equal in area ]
From equation (v) and (vi), we get
ar (ΔAPQ) = ar (ΔARC)..........(vii)
R is the mid-point of AP. Therefore, RQ is the median of ΔAPQ.
∴ ar (ΔPRQ) = 1/2 ar (ΔAPQ)......................... (viii)
From (vii) and (viii), we get,
∴ ar (ΔPRQ) = 1/2 ar (ΔARC)
ii) PQ is the median of ΔBPC.
∴ ar(ΔPQC) = 1/2 ar (ΔBPC) = 1/2 × 1/2 ar (ΔABC)............. (ix)
Also,
ar (ΔPRC) = 1/2 ar (ΔAPC) [Using (iv)]
ar (ΔPRC) = 1/2 × 1/2 ar (ΔABC) = 1/4 ar (ΔABC)............................. (x)
Adding equation (ix) and (x), we get,
ar (ΔPQC) + ar (ΔPRC) = ( 1/4 + 1/4 ) ar (ΔABC)
⇒ ar (quadrilateral PQCR) = 1/2 ar (ΔABC).....................(xi)
Subtracting ar (ΔPRQ) from the both sides,
ar (quadrilateral PQCR) - ar (ΔPRQ) = 1/2 ar (ΔABC) - ar (ΔPRQ)
⇒ ar (ΔRQC) = 1/2 ar (ΔABC) - 1/2 ar (ΔARC) [Using result (i)]
⇒ ar (ΔRQC) = 1/2 ar (ΔABC) - 1/2 × 1/2 ar (ΔAPC)
⇒ ar (ΔRQC) = 1/2 ar (ΔABC) - 1/4 ar (ΔAPC)
⇒ ar (ΔRQC) = 1/2 ar (ΔABC) - 1/4 × 1/2 ar (ΔABC) [PC is median of DABC]
⇒ ar (ΔRQC) = 1/2 ar (ΔABC) - 1/8 ar (ΔABC)
⇒ ar (ΔRQC) = (1/2 - 1/8) x ar (ΔABC)
⇒ ar (ΔRQC) = 3/8 × ar (ΔABC)
iii) ar (ΔPRQ) = 1/2 ar (ΔARC) [Using result (i)]
⇒ 2ar (ΔPRQ) = ar (ΔARC) ............ (xii)
ar (ΔPRQ) = 1/2 ar (ΔAPQ) [RQ is the medium of ΔAPQ]............(xiii)
But ar (ΔAPQ) = ar(ΔPQC) [Using reason of eq. (vi)] ..........(xiv)
From eq. (xiii) and (xiv), we get,
ar (ΔPRQ) = 1/2 ar (ΔPRQ)............(xv)
But,
ar (ΔBPQ) = ar (ΔPQC) [PQ is the median of DBPC] ..........(xvi)
From eq. (xv) and (xvi), we get,
ar (ΔPRQ) = 1/2 ar (ΔBPQ).......... (xvii)
Now from (xii) and (xvii), we get,
2 × 1/2 ar (ΔBPQ) = ar (ARC)
⇒ ar (ΔBPQ) = ar (ΔARC)
☛ Check: NCERT Solutions Class 9 Maths Chapter 9
Video Solution:
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)
Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.4 Question 7
Summary:
If 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, then ar (PRQ) = 1/2 ar(ARC), ar (RQC) = 3/8 ar (ABC), and ar (PBQ) = ar (ARC).
☛ Related Questions:
- In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
- In Fig. 9.32, 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 (BPC) = ar (DPQ).[Hint: Join AC.]
- In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show thati) 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.]
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