Case Study

Ankita took part in a Rangoli Competition. She made a beautiful Rangoli in the shape of a triangle ABC as shown. In the triangle, P, Q and R are mid-points of the sides AB, BC and CA respectively. She decorated it by putting a garland along the sides of △PQR. The lengths of the sides of the triangle are AB = 20 cm, BC = 26 cm and AC = 24 cm.

Based on this information, answer the following questions:

1. The length of AP is :
   (a) QB
   (b) QR
   (c) AR
   (d) RP

2. The length of PQ is :
   (a) 12 cm
   (b) 13 cm
   (c) 10 cm
   (d) 15 cm

3. The length of the garland is :
   (a) 30 cm
   (b) 32 cm
   (c) 35 cm
   (d) 40 cm

4. Area of △PQR is :
   (a) area of △ABC
   
   (b) $\dfrac{1}{2}$ area of △ABC
   
   (c) $\dfrac{1}{3}$ area of △ABC
   
   (d) $\dfrac{1}{4}$ area of △ABC
   
   

5. APQR is a :
   (a) Rectangle
   (b) Parallelogram
   (c) Square
   (d) Rhombus

1. Given,

P is the mid-point of AB.

⇒ AP = $\dfrac{1}{2}$ AB ……….(1)

By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Since, Q and R are the mid-points of BC and AC respectively.

⇒ QR = $\dfrac{1}{2}$ AB ……..(2)

From equation (1) and (2), we get :

⇒ AP = QR

Hence, option (b) is the correct option.

2. By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Since, P and Q are the mid-points of AB and BC respectively.

PQ || AC

⇒ PQ = $\dfrac{1}{2} AC = \dfrac{1}{2}$ × 24 = 12 cm.

Hence, option (a) is the correct option.

3. By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Since, P and Q are the mid-points of AB and BC respectively.

PQ || AC

⇒ PQ = $\dfrac{1}{2} AC = \dfrac{1}{2}$ × 24 = 12 cm.

Since, P and R are the mid-points of AB and AC respectively.

PR || BC

⇒ PR = $\dfrac{1}{2}BC = \dfrac{1}{2}$ × 26 = 13 cm.

Since, R and Q are the mid-points of AC and BC respectively.

QR || AB

⇒ QR = $\dfrac{1}{2} AB = \dfrac{1}{2}$ × 20 = 10 cm.

Length of the garland = PQ + QR + PR = 12 + 10 + 13 = 35 cm.

Hence, option (c) is the correct option.

4. By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

In △ARP and △PQR,

Since, R and Q are the mid-points of AC and BC respectively.

QR || AB

⇒ QR || AP

Since, P and Q are the mid-points of AB and BC respectively.

PQ || AC

⇒ PQ || AR

⇒ ∠ARP = ∠QPR (Alternate angles are equal)

⇒ ∠APR = ∠QRP (Alternate angles are equal)

⇒ PR = PR (Common side)

∴ △ARP ≅ △PQR (By A.S.A axiom) …(1)

In △QRC and △PQR,

⇒ ∠CRQ = ∠PQR (Alternate angles)

⇒ ∠CQR = ∠PRQ (Alternate angles)

⇒ QR = QR (Common side)

∴ △QRC ≅ △PQR (By A.S.A axiom) …(2)

In △PBQ and △PQR,

Since, R and Q are the mid-points of AC and BC respectively.

QR || AB

⇒ QR || AP

Since, P and Q are the mid-points of AB and BC respectively.

PQ || AC

⇒ PQ || AR

⇒ ∠BPQ = ∠PQR (Alternate angles)

⇒ ∠BQP = ∠QPR (Alternate angles)

⇒ PQ = PQ (Common side)

∴ △PBQ ≅ △PQR (By A.S.A axiom) …(3)

From eq.(1), (2) and (3), we have:

Area of △PBQ = Area of △QRC = Area of △ARP = Area of △PQR

From figure,

⇒ Area of △ABC = Area of △PBQ + Area of △QRC + Area of △ARP + Area of △PQR

⇒ Area of △ABC = Area of △PQR + Area of △PQR + Area of △PQR + Area of △PQR

⇒ Area of △ABC = 4 Area of △PQR

⇒ Area of △PQR = $\dfrac{1}{4}$ Area of △ABC

Hence, option (d) is the correct option.

5. By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

In △ARP and △PQR,

Since, R and Q are the mid-points of AC and BC respectively.

QR || AB

⇒ QR || AP

⇒ QR = $\dfrac{1}{2}AB$

⇒ QR = AP (∵ P is the midpoint of AB)

Since, P and Q are the mid-points of AB and BC respectively.

PQ || AC

⇒ PQ || AR

⇒ PQ = $\dfrac{1}{2}AC$

⇒ PQ = AR (∵ R is the midpoint of AC)

Since, in quadrilateral APQR opposite sides are parallel and equal.

∴ APQR is a parallelogram.

Hence, option (b) is the correct option.