You visited us 1 times! Enjoying our articles? Unlock Full Access!

Section Formula

In ABC, D a...

Question

In $△$ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = $4 x - 3$ , BD = $3 x - 1$ , AE = $8 x - 7$ and CE = $5 x - 3$ find the value $x$ .

Open in App

Solution

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

It is given that $A D = 4 x - 38$ , $B D = 3 x - 1$ , $A E = 8 x - 7$ and $C E = 5 x - 3$ . Let $A C = x$

Using the basic proportionality theorem, we have

$\frac{A D}{B D} = \frac{A E}{A C} \Rightarrow \frac{4 x - 3}{3 x - 1} = \frac{8 x - 5}{x} \Rightarrow x (4 x - 3) = (3 x - 1) (8 x - 5) \Rightarrow 4 x^{2} - 3 x = 3 x (8 x - 5) - 1 (8 x - 5) \Rightarrow 4 x^{2} - 3 x = 24 x^{2} - 15 x - 8 x + 5 \Rightarrow 4 x^{2} - 3 x = 24 x^{2} - 23 x + 5 \Rightarrow 24 x^{2} - 23 x + 5 - 4 x^{2} + 3 x = 0 \Rightarrow 20 x^{2} - 20 x + 5 = 0 \Rightarrow 5 (4 x^{2} - 4 x + 1) = 0 \Rightarrow 4 x^{2} - 4 x + 1 = 0 \Rightarrow (2 x)^{2} - (2 \times 2 x \times 1) x + 1^{2} = 0 (∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b) \Rightarrow (2 x - 1)^{2} = 0 \Rightarrow (2 x - 1) = 0 \Rightarrow 2 x = 1 \Rightarrow x = \frac{1}{2}$

Hence, $x = \frac{1}{2}$ .

Suggest Corrections

Similar questions

Δ A B C

D

E

A B

A C

D E | | B C

A D = 4 x - 3, A E = 8 x - 7, B D = 3 x - 1

C E = 5 x - 3

x

Δ A B C, D a n d E

A B

A C

D E ∥ B C

A D = 4 x - 3, A E = 8 x - 7, B D = 3 x - 1 a n d C E = 5 x - 3

x

Δ A B C

D

E

A B

A C

D E | | B C

A D = 8 x - 7, D B = 5 x - 3, A E = 4 x - 3

E C = (3 x - 1)

x

△ A B C, D

E

A B

A C

D E ∥ B C

A D = 2.4 c m, A E = 3.2 c m, D E ∥ B C

A D = 2.4 c m, A E = 3.2 c m, D E = 2 c m

B C = 5 c m

B D

C E

In triangle ABC, D and E are points on AB and AC such that DE || BC. If AD = 4x-3, AE = 8x-7,
BD = 3x-1 and CE = 5x-3, find the value of x

Join BYJU'S Learning Program