Question

point P is inside triangle ABC line segments APD BPE and CPF are drawn with the D on BC, E on CA,F on AB given that AP is equal to 6, BP is equal to 9 PD equal to 6,PE is equal to 3 and CF is equal to 20 then find the area of triangle ABC​

Answer 1

Given:

Point P is inside triangle ABC line segments APD BPE and CPF are drawn with the D on BC, E on CA,F on AB given that AP is equal to 6, BP is equal to 9 PD equal to 6,PE is equal to 3 and CF is equal to 20.

To find:

Find the area of triangle ABC​

Solution:

Construction: Extend AD to Q so that, PD = DQ = 6.

We get, BQ = 9, PQ = 12 and BP = 15

(∵ Δ CDP ≅ Δ BDQ)

Applying Pythagorean theorem on Δ APC we get,

AC = 3√13,

so AE = √13  and CE = 2√13

Now apply law of cosines on Δ CEP and Δ AEP.

Let PE = x.

Notice: ∠ CEB = 180° - ∠ AEB and

cos CEB = - cos AEB

so, we have two equations,

81 = 52 + y² - 2y√13 cos CEF ......(1)

36 = 13 + y² + y√13 cod CEF  ......(2)

solving equations (1) and (2), we get,

y = 5

Area of Δ ABC = Area of Δ AQB + Area of Δ APC

= 1/2 × 9 × 18 + 1/2 × 6 × 9

= 81 + 27

= 108

∴ Area of Δ ABC = 108 sq. units.