Question

Class 10th , maths , exercise 6.2 question 1. In fig. 6.17 (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

Answer 1

Given :

• DE || BC.

In (i),

• AD = 1.5 cm.
• AE = 1 cm.
• BD = 3 cm.

In (ii),

• AE = 1.8 cm.
• EC = 5.4 cm.
• BD = 7.2 cm.

To Find :

The measure of EC in (i).

The measure of AD in (ii).

Solution :

(i) In ∆ABC,

DE || BC ---------[Given]

By using BPT (Basic proportionality theorem).

$\: \: \bf \therefore \: \: \dfrac{AD}{DB} = \dfrac{AE}{EC}$

Where,

• AD = 1.5 cm.
• AE = 1 cm.
• DB = 3 cm.

$\bf \implies \dfrac{1.5 \: \not cm}{3 \not cm} = \dfrac{1 \: cm}{EC} \\ \\ \\ \bf \implies EC = \dfrac{3}{1.5} \: cm \\ \\ \\ \bf \implies EC = \not 3 \times \dfrac{10}{ \not15} \: cm \\ \\ \\ \bf \implies EC = \dfrac{10}{5} \: cm \\ \\ \\ \bf \implies EC = 2 \: cm$

Hence,

The measure of EC is 2 cm.

________________________

(ii) In ∆ABC,

DE || BC ---------[Given]

By using BPT (Basic proportionality theorem).

$\: \: \bf \therefore \: \: \dfrac{AD}{DB} = \dfrac{AE}{EC}$

Where,

• AE = 1.8 cm.
• DB = 7.2 cm.
• EC = 5.4 cm.

$\bf\implies\dfrac{AD}{7.2\:cm}=\dfrac{1.8\: \not cm}{5.4\:\not cm}\\\\\\\bf\implies AD=\dfrac{1.8}{5.4}\times7.2\: cm \\\\\\\bf\implies AD=\dfrac{18}{10}\times\dfrac{\not10}{54}\times\dfrac{72}{\not10}\:cm\\\\\\\bf\implies AD=\dfrac{18}{10}\times\dfrac{\not72}{\not54}\:cm\\\\\\\bf\implies AD=\dfrac{18}{10}\times\dfrac{\not36}{\not27}\: cm\\\\\\\bf\implies AD=\dfrac{\not18}{10}\times\dfrac{4}{\not3}\: cm\\\\\\\bf\implies AD=\dfrac{6\times4}{10\times1}\: cm\\\\\\\bf \implies AD=\dfrac{24}{10}\: cm$

$\\\\\bf\implies AD=2.4\: cm$

Hence,

The measure of AD is 2.4 cm.

Answer 2

Given :

DE || BC.

In (i),

AD = 1.5 cm.

AE = 1 cm.

BD = 3 cm.

In (ii),

AE = 1.8 cm.

EC = 5.4 cm.

BD = 7.2 cm.

To Find :

The measure of EC in (i).

The measure of AD in (ii).

Solution :

(i) In ∆ABC,

DE || BC ---------[Given]

By using BPT (Basic proportionality theorem).

$\: \: \bf \therefore \: \: \dfrac{AD}{DB} = \dfrac{AE}{EC}$

Where,

AD = 1.5 cm.

AE = 1 cm.

DB = 3 cm.

$\bf \implies \dfrac{1.5 \: \not cm}{3 \not cm} = \dfrac{1 \: cm}{EC} \\ \\ \\ \bf \implies EC = \dfrac{3}{1.5} \: cm \\ \\ \\ \bf \implies EC = \not 3 \times \dfrac{10}{ \not15} \: cm \\ \\ \\ \bf \implies EC = \dfrac{10}{5} \: cm \\ \\ \\ \bf \implies EC = 2 \: cm$

Hence,

The measure of EC is 2 cm.

________________________

(ii) In ∆ABC,

DE || BC ---------[Given]

By using BPT (Basic proportionality theorem).

$\: \: \bf \therefore \: \: \dfrac{AD}{DB} = \dfrac{AE}{EC}$

Where,

AE = 1.8 cm.

DB = 7.2 cm.

EC = 5.4 cm.

$\bf\implies\dfrac{AD}{7.2\:cm}=\dfrac{1.8\: \not cm}{5.4\:\not cm}\\\\\\\bf\implies AD=\dfrac{1.8}{5.4}\times7.2\: cm \\\\\\\bf\implies AD=\dfrac{18}{10}\times\dfrac{\not10}{54}\times\dfrac{72}{\not10}\:cm\\\\\\\bf\implies AD=\dfrac{18}{10}\times\dfrac{\not72}{\not54}\:cm\\\\\\\bf\implies AD=\dfrac{18}{10}\times\dfrac{\not36}{\not27}\: cm\\\\\\\bf\implies AD=\dfrac{\not18}{10}\times\dfrac{4}{\not3}\: cm\\\\\\\bf\implies AD=\dfrac{6\times4}{10\times1}\: cm\\\\\\\bf \implies AD=\dfrac{24}{10}\: cm$

$\\\\\bf\implies AD=2.4\: cm$

Hence,

The measure of AD is 2.4 cm.