In given triangle, $A B$ and $P Q$ are parallel to each other. $A P = c, P C = b, P Q = a$ and $A B = x$ . What is the value of $x$ ?

a + \frac{a b}{c}

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a + \frac{b c}{a}

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b + \frac{c a}{b}

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a + \frac{a c}{b}

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Solution

The correct option is D $a + \frac{a c}{b}$
$△ A B C \sim △ P Q C$
$\frac{A B}{P Q} = \frac{A C}{P C} \Rightarrow \frac{x}{a} = \frac{c + b}{b} \Rightarrow x = a (\frac{c + b}{b}) = \frac{a c}{b} + a$ .

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The steps to construct a triangle with given base angles $∠$ B and $∠$ C and BC + CA + AB, are:.
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A B C

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(iii) Bisect these angles. Let the bisectors of these angles intersect at a point A.

(iv) Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.

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In given triangle, AB and PQ are parallel to each other. AP=c,PC=b,PQ=a and AB=x. What is the value of x?

The correct option is D a+acb△ABC∼△PQCABPQ=ACPC⇒xa=c+bb⇒x=a(c+bb)=acb+a.

In given triangle, $A B$ and $P Q$ are parallel to each other. $A P = c, P C = b, P Q = a$ and $A B = x$ . What is the value of $x$ ?

The correct option is D $a + \frac{a c}{b}$
$△ A B C \sim △ P Q C$
$\frac{A B}{P Q} = \frac{A C}{P C} \Rightarrow \frac{x}{a} = \frac{c + b}{b} \Rightarrow x = a (\frac{c + b}{b}) = \frac{a c}{b} + a$ .