In - ABC- AC - 15 cm and DE- BC- If AB-AD - 3- find EC-

From basic proportionality theorem since DE∥BC, nbsp; AD/DB=AE/EC 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 —— (1) ...

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Standard IX

Mathematics

Properties of Similar Triangles

In Δ ABC, AC ...

Question

In $Δ$ ABC, AC = 15 cm and DE $∥$ BC. If $\frac{A B}{A D} = 3$ , find EC.

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Solution

From basic proportionality theorem since DE $∥$ BC,
$\frac{A D}{D B} = \frac{A E}{E C}$
$\frac{A B}{A D} = \frac{A D + D B}{A D} = 1 + \frac{D B}{A D} \Rightarrow 1 + \frac{D B}{A D} = 3 \Rightarrow \frac{D B}{A D} = 2 \Rightarrow \frac{A D}{D B} = \frac{A E}{E C} = \frac{1}{2} \Rightarrow 2 A E = E C \Rightarrow A C = A E + E C - - - - - - - (1)$
Substituting value of EC and AC in (1),
$\Rightarrow 15 = 3 A E \Rightarrow 5 = A E \Rightarrow E C = 10 c m$

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In Δ ABC, AC = 15 cm and DE∥ BC. If ABAD=3, find EC.

From basic proportionality theorem since DE∥BC, ADDB=AEEC ABAD=AD+DBAD=1+DBAD⇒1+DBAD=3⇒DBAD=2⇒ADDB=AEEC=12⇒2AE=EC⇒AC=AE+EC−−−−−−−(1) Substituting value of EC and AC in (1), ⇒15=3AE⇒5=AE⇒EC=10cm

In $Δ$ ABC, AC = 15 cm and DE $∥$ BC. If $\frac{A B}{A D} = 3$ , find EC.