Question

ABC is an equilateral triangle. P and Q are 2 points on AB and AC respectively such that PQ is parallel to BC. If PQ =5cm , then area of triangle APQ is: ?​

Answer 1

Answer:

Draw a perpendicular from A point to BC . Lets it cut BC at point O. Then

BO = OC. And it cuts PQ at point M. (PERPENDICULAR FROM ONE POINT TO OPPOSITE SIDE OF AN EQUILATERAL TRIANGLE BISECT THE SIDE.)

FOR TRIANGLE , ABO and APM.

<ABO =< APM AS. PQ||BC.

Cos (<ABO) = Cos(<APM).

BO/AB = PM/AP.

Here BO =AB/2 ( as it is a equilateral triangle)

So, PM = AP/2.

Same way for triangle ACO and AQM.

it can prove that QM = AQ/2.

IT CAN BE PROVED THAT AP=AQ. So AP= AQ =PQ.

MEANS TRIANGLE APQ is also a equilateral triangle.

Area of an equilateral triangle = (squrt(3/4) * (side)^2) = squrt(3/4) * 5^2 = 25 * squrt (3/4).

Hope it helps.........Pls mark this answer as brainiest..............

Answer 2

SoluTion :-

$\sf {As\ PQ\ ||\ BC}\\\\\\\\\sf {\Rightarrow \Delta APQ\ {\raise.17ex\hbox{ \scriptstyle\sim }}\ \Delta ABC}\\\\\\\\\sf {\Delta APQ\ is\ also\ an\ equilateral\ triangle}\\\\\\\\\sf {\Rightarrow \Delta APQ=\frac{\sqrt{3} }{4} (5)^2}\\\\\\\sf {\Rightarrow\ \ \frac{25 \sqrt{3} }{4} cm^{2}}$