In -ABC- AC - 15 cm and DE - BC- If AB-AD - 3- then the length of EC is -

Given: DE ∥ BC From basic proportionality theorem; AD/DB=AE/EC Now, AB/AD=AD + DB/AD=1+DB/AD ⇒ 1+DB/AD=3 ⇒DB/AD=2 ⇒AD/DB=AE/EC=1/2 ⇒ 2AE=EC ⇒ AC = AE + EC —— ...

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Standard X

Mathematics

Basic Proportionality Theorem

In ΔABC, AC =...

Question

In $Δ A B C$ , AC = 15 cm and DE || BC. If $\frac{A B}{A D} = 3$ , then the length of EC is :

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Solution

Given: DE $∥$ BC
From basic proportionality theorem;

$\frac{A D}{D B} = \frac{A E}{E C}$

$Now, \frac{A B}{A D} = \frac{A D + D B}{A D} = 1 + \frac{D B}{A D} \Rightarrow 1 + \frac{D B}{A D} = 3 \Rightarrow \frac{D B}{A D} = 2 \Rightarrow \frac{A D}{D B} = \frac{A E}{E C} = \frac{1}{2} \Rightarrow 2 A E = E C \Rightarrow A C = A E + E C - - - - - - - (1)$
On substituting value of EC in (1), we get
$15 = 3 A E \Rightarrow 5 = A E \Rightarrow E C = 10 c m$

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In ΔABC, AC = 15 cm and DE || BC. If ABAD=3, then the length of EC is :

Given: DE ∥ BC From basic proportionality theorem; ADDB=AEEC Now, ABAD=AD + DBAD=1+DBAD⇒1+DBAD=3⇒DBAD=2⇒ADDB=AEEC=12⇒2AE=EC⇒AC=AE+EC−−−−−−−(1) On substituting value of EC in (1), we get 15=3AE⇒5=AE⇒EC=10 cm

In $Δ A B C$ , AC = 15 cm and DE || BC. If $\frac{A B}{A D} = 3$ , then the length of EC is :