# 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.]
- Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC)[Hint: From A and C, draw perpendiculars to BD.]

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