Question

In a triangle ABC, Angle A=25 , Angle B=35 and AB=16 units . In another triangle PQR, Angle P= 35 , Angle Q=120 and PR 4 units .
Which of the following is true?
(a) ar(ABC) =2ar(PQR)
(b) ar(ABC) =4ar(PQR)
(c) ar(ABC) =8ar(PQR)
(d) ar(ABC) =16ar(PQR)​

Answer 1

Answer:

(d)

Step-by-step explanation:

∆ABC is similar to ∆RPQ by AA similarity criterion.

Hence, corresponding sides will be proportional.

AB/RP = BC/PQ = CA/QR = 16/4 = 4/1

Hence, area will be in ratio of 16/1.

REASON:

Ar of ∆ = base×height/2

If we multiply all dimensions by 4,

ar of new ∆ = base×4×height×4/2 = base×height×8 which is 16 times of previous triangle.

Hope you got it.

Answer 2

Ar (ΔABC) = 16Ar (ΔPQR) if in ΔABC , ∠A=25° , ∠B = 35° , AB = 16 units and ΔPQR , ∠P=35° , ∠Q = 120° , PR = 4 units

Given:

• ΔABC , ∠A=25° , ∠B = 35° , AB = 16 units
• ΔPQR , ∠P=35° , ∠Q = 120° , PR = 4 units

To Find: (Choose correct option:)

• (a) ar(ABC) =2ar(PQR)
• (b) ar(ABC) =4ar(PQR)
• (c) ar(ABC) =8ar(PQR)
• (d) ar(ABC) =16ar(PQR)​

Solution:

"Sum of angles of a triangle is 180°"

Step 1 :

Find angle C using sum of angles of triangles and substituting ∠A=25° , ∠B = 35°

∠A + ∠B + ∠C = 180°

=> 25° + 35° + ∠C = 180°

=> ∠C = 120°

Step 2 :

Find angle R using sum of angles of triangles and substituting ∠Q=120° , ∠P = 35°

∠P + ∠Q + ∠R = 180°

=> 35° + 120° + ∠R = 180°

=> ∠R = 25°

Step 3 :

Compare ΔABC and ΔPQR  

∠A = ∠R = 25°

∠B = ∠P = 35°

∠C = ∠Q = 120°

=> ΔABC ~ ΔRPQ   ( Using AAA Similarity)

Step 4 :

Ratio of area of similar triangle is equal to ratio of corresponding sides hence,

Ar (ΔABC) / Ar (ΔPQR)  =  (AB/ RP)²

Step 5 :

Substitute B = 16 and RP = 4   and simplify

Ar (ΔABC) / Ar (ΔPQR)  =  (16/ 4)²

=> Ar (ΔABC) / Ar (ΔPQR)  =16

=> Ar (ΔABC) = 16Ar (ΔPQR)

Correct option is (d) ar(ABC) =16ar(PQR)​

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