Question

In the angle ABC, D and E are points on side AB and AC respectively such that DE||BC. If AE= 2cm, AD= 3cm and BD=4.5cm, then find CE​

Answer 1

Given: In the ∆ABC, D and E are points on the sides AB and AC respectively. And,

• DE || BC
• AE = 2 cm
• AD = 3 cm
• BD = 4.5 cm

To find: What's the value of CE.

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Here, in the attached figure we can see that DE || BC.

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$\underline{\boldsymbol{By\; using\; Basic \; Proportionality \; theorem\; :}}$

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• If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

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Therefore,

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$:\implies\sf \dfrac{AD}{BD} = \dfrac{AE}{CE} \\\\\\:\implies\sf \dfrac{3}{4.5} = \dfrac{2}{CE} \\\\\\:\implies\sf CE = \dfrac{2 \times 4.5}{3}\\\\\\:\implies\sf CE = \cancel\dfrac{9}{3}\\\\\\:\implies{\underline{\boxed{\frak{\pink{ CE = 3\; cm}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence, \; required\; value \; of \; CE\; is \;\bf{ 3\;cm}.}}}$

Answer 2

Answer:

Given :-

• In the angle ABC, D and E are points on side AB and AC respectively such that DE || BC. If AE = 2 cm, AD = 3 cm and BD = 4.5 cm.

To Find :-

• What is the value of CE.

Solution :-

Given :

• DE || BC
• AE = 2 cm
• AD = 3 cm
• BD = 4.5 cm

$\sf\bold{\red{By\: Using\: The\: Basic\: Proportionality\: Theorem\: :-}}$

As we know that,

$\leadsto$ $\sf\bold{\dfrac{AD}{BD} =\: \dfrac{AE}{CE}}$

According to the question

$\implies$ $\dfrac{3}{4.5} =\: \dfrac{2}{CE}$

By doing cross multiplication we get,

$\implies$ $\sf 3CE =\: 4.5 \times 2$

$\implies$ $\sf 3CE =\: \dfrac{45 \times \cancel{2}}{\cancel{10}}$

$\implies$ $\sf 3CE =\: \dfrac{\cancel{45}}{\cancel{5}}$

$\implies$ $\sf 3CE =\: 9$

$\implies$ $\sf CE =\: \dfrac{\cancel{9}}{\cancel{3}}$

$\implies$ $\sf\bold{\purple{CE =\: 3\: cm}}$

$\therefore$ The value of CE is 3 cm .