Question

ABC is an equilateral triangle P and Q are points of lines AB and AC respectively such that PQ II BC. If PQ = 5cm, then the area of triangle APQ is?​

Answer 1

Answer:

25 cm²

Step-by-step explanation:

Here PQ | | BC , So

In ∆ ABC and ∆ APQ

∠ ABC = ∠APQ ( As We know PQ | | BC and take AB as transversal line so these angles are corresponding angles )

∠ ACB = ∠AQP ( As We know PQ | | BC and take AC as transversal line so these angles are corresponding angles )

So,

∆ ABC ~ ∆ APQ ( By AA rule )

So, we can say that

Area of ∆ ABCArea of ∆ APQ = (Corresponding side )2(Corresponding side )2So,⇒Area of ∆ APQ = (PQ )2⇒Area of ∆ APQ = (5 )2⇒Area of ∆ APQ =25 cm2 (Ans)

Answer 2

GIVEN :-

• ABC is an equilateral triangle .
• P and Q are points of lines AB and AC respectively such that PQ II BC. If PQ = 5cm.

TO FIND :-

• The area of Δ APQ.

SOLUTION :-

Since, the Δ is an equilateral so AB = BC = AC.

Now from mid point theoren we have,

➳ 2PQ = BC

➳ 5 * 2 = BC

➳ 10 cm = BC

Since, the Δ is an equilateral so AB = BC = AC = 10 cm.

Also,

➳ AP = AQ = 10/2 cm.

➳ AP = AQ = 5 cm.

Now in Δ APQ we have

• AP = AQ = PQ = 5 cm [ Hence Δ APQ ia also an equilateral traingle ]

Now as we know that,

➳ Area of triangle = √3/4 * ( side )²

➳ Area of triangle = √3/4 * ( 5 )²

➳ Area of triangle = 25 * √3/4

➳ Area of triangle = ( 25√3 )/4 cm².

Hence The area of Δ APQ is ( 25√3 )/4 cm².