Question

In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC.

(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC.
(ii) If $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{4}$ and AC = 15 cm, find AE.
(iii) If $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{3}$ and AC = 18 cm, find AE.
(iv) If AD = 4, AE = 8, DB = x − 4, and EC = 3x − 19, find x.
(v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.
(vi) if AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.
(vii) If AD = 2 cm, AB = 6 cm, and AC = 9 cm, find AE.
(viii) If $\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{4}{5}$ and EC = 2.5 cm, find AE.
(ix) If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, find the value of x.
(x) If AD = 8x − 7, DB = 5x − 3, AE = 4x − 3 and EC = (3x − 1), find the value of x.
(xi) If AD = 4x − 3, AE = 8x − 7, BD = 3x − 1 and CE = 5x − 3. find the volume x.
(xii) If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm find the length of AC.

Answer 1

(i) It is given that and DE || BC

We have to find the

Since

So (by Thales theorem)

Then

Hence

(ii)It is given that and

We have to find

Let

So (by Thales theorem)

Then

Hence

(iii)It is given that and

We have to find

Let and

So (by Thales theorem)

Then

Hence

(iv)It is given that ,, and .

We have to find

So (by Thales theorem)

Then

Hence

(v) It is given that , and .

We have to find CE.

So (by Thales theorem)

Then

Hence

(vi) It is given that , and .

We have to find .

So (by Thales theorem)

Then

Hence

(vii) It is given that , and .

We have to find .

Now

DB = 6 − 2 = 4 cm

So (by Thales theorem)

Then (Let)

Hence

(viii) It is given that and EC = 2.5 cm

We have to find .

So (by Thales theorem)

Then

$AE=\frac{4\times 2.5}{5}=2$ cm

Hence

(ix) It is given that ,, and .

We have to find the value of .

So (by Thales theorem)

Then

Hence

(x) It is given that AD = 8x − 7, DB = 5x − 3, AE = 4x − 3 and EC = 3x − 1.

We have to find the value of .

So (by Thales theorem)

Then,

$$\begin{gathered} \frac{8x-7}{5x-3}=\frac{4x-3}{3x-1} \\ \Rightarrow \left(8x-7\right)\left(3x-1\right)=\left(5x-3\right)\left(4x-3\right) \\ \Rightarrow 24x^2-29x+7=20x^2-27x+9 \\ \Rightarrow 4x^2-2x-2=0 \\ \Rightarrow 2\left[2x^2-x-1\right]=0 \\ \Rightarrow 2x^2-x-1=0 \\ \Rightarrow 2x^2-2x+x-1=0 \\ \Rightarrow 2x\left(x-1\right)+1\left(x-1\right)=0 \\ \Rightarrow \left(x-1\right)\left(2x+1\right)=0 \\ \Rightarrow x-1=0\mathrm{or}2x+1=0 \\ \Rightarrow x=1\mathrm{or}x=-\frac{1}{2}\left(\mathrm{rejected}\right) \end{gathered}$$

Hence,

(xi) It is given that AD = 4x − 3, BD = 3x − 1, AE = 8x − 7 and EC = 5x − 3.

We have to find the value of .

So (by Thales theorem)

Then

$\left(4x-3\right)\left(5x-3\right)=\left(3x-1\right)\left(8x-7\right)$

Then

Hence

(xii) It is given that , and .

So (by Thales theorem)

Then

Now