Name: Areas of Similar Triangles
Uploaded: 2019-07-19
Duration: 6 min 54 s
Description: The corresponding sides of similar triangles are proportional. Are the ratios of areas of similar triangles the same as that of their sides? Watch this video...

Question

Let

P

be an interior point of a triangle

A B C

and let

B P

and

C P

meet

A C

and

A B

in

E

and

F

respectively. If

[B P F] = 4, [B P C] = 8

and

[C P E] = 13

, find

[A F P E]

. (Here,

[]

denotes the area of a triangle or a quadrilateral as the case may be)

Accepted Answer

More generally, let $[B P F] = u, [B P C] = v$ and $[C P E] = w$ . Join $A P$ .

Let $[A F P] = x$ and $[A E P] = y$ .

We know that a line joining the vertex of a triangle to the opposite side divides the triangle in two parts whose ratio of areas equals the ratio in which the side is divided.

Using the triangles $A F C$ and $B F C$ , we get

$\frac{F P}{P C} = \frac{[A F P]}{[A E P] + [C P E]} = \frac{[B P F]}{[B P C]}$

$\Rightarrow \frac{x}{y + w} = \frac{u}{v}$

$\Rightarrow v x - u y = u w . . . . . . . (i)$

Again, using the triangles $A E B$ and $C E B$ , we get

$\frac{E P}{P B} = \frac{[A E P]}{[A F P] + [B P F]} = \frac{[C P E]}{[B P C]}$

$\Rightarrow \frac{y}{x + u} = \frac{w}{v}$

$\Rightarrow w x - v y = - u w . . . . . . . . (ii)$

Subtracting equation $(i) \times v$ from equation $(ii) \times u$ , we get

$(u w x - u v y) - (v^{2} x - u v y) = - u^{2} w - u v w$

$\Rightarrow (u w - v^{2}) x = - u w (u + v)$

$\Rightarrow x = \frac{u w (u + v)}{v^{2} - u w}$

From equation $(i)$ , we have

$y = \frac{v x - u w}{u} = \frac{v}{u} x - w$

Substituting the value of $x$ obtained, we get

$y = \frac{v}{u} (\frac{u w (u + v)}{v^{2} - u w}) - w$

$\Rightarrow y = \frac{v w (u + v)}{v^{2} - u w} - w$

$\Rightarrow y = \frac{u v w + v^{2} w - v^{2} w + u w^{2}}{v^{2} - u w}$

$\Rightarrow y = \frac{u w (v + w)}{v^{2} - u w}$

So, $[A F P E] = x + y = \frac{u w (u + v)}{v^{2} - u w} + \frac{u w (v + w)}{v^{2} - u w} = \frac{u w (u + 2 v + w)}{v^{2} - u w}$ .

Putting the values $u = 4, v = 8, w = 13$ , we get

$[A F P E] = \frac{4 \cdot 13 (4 + 2 \cdot 8 + 13)}{8^{2} - 4 \cdot 13} = \frac{52 \cdot 33}{12} = 13 \cdot 11 = 143$

Let $P$ be an interior point of a triangle $A B C$ and let $B P$ and $C P$ meet $A C$ and $A B$ in $E$ and $F$ respectively. If $[B P F] = 4, [B P C] = 8$ and $[C P E] = 13$ , find $[A F P E]$ . (Here, $[]$ denotes the area of a triangle or a quadrilateral as the case may be)

Important Questions on Geometry

Learn and Prepare for any exam you want !

Let P be an interior point of a triangle ABC and let BP and CP meet AC and AB in E and F respectively. If BPF=4,BPC=8 and CPE=13, find AFPE. (Here, denotes the area of a triangle or a quadrilateral as the case may be)

Important Questions on Geometry

Learn and Prepare for any exam you want !

Let $P$ be an interior point of a triangle $A B C$ and let $B P$ and $C P$ meet $A C$ and $A B$ in $E$ and $F$ respectively. If $[B P F] = 4, [B P C] = 8$ and $[C P E] = 13$ , find $[A F P E]$ . (Here, $[]$ denotes the area of a triangle or a quadrilateral as the case may be)