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Byju's Answer

Standard IX

Mathematics

SSS Criteria for Congruency

State true or...

Question

State true or false:
In the following diagram, lines $l, m$ and $n$ are parallel to each other. Two transversals $p$ and $q$ intersect the parallel lines at points $A, B, C$ and $P, Q, R$ respectively as shown, then:
$\frac{A B}{B C} = \frac{P Q}{Q R}$

True

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False

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Solution

The correct option is A True
Given $l | | m | | n$ and have two transversal lines $p$ and $q$ . They cut at $A, B, C$ and $P, Q, R$ respectively.
$P X$ perpendicular to $O A$ and $A Y$ perpendicular to $O Q$ .
Since $P X$ perpendicular to $O A$ , $P X$ is height of $△ O P A$ and $B A P$
$∴ a r (△ O P A) = \frac{1}{2} b h = \frac{1}{2} \times O A \times P X$
$∴ a r (△ B A P) = \frac{1}{2} b h = \frac{1}{2} \times A B \times P X$
$\frac{a r (△ B A P)}{a r (△ O P A)} = \frac{\frac{1}{2} A B . P X}{\frac{1}{2} O A . P X} \Rightarrow \frac{a r (B A P)}{a r c (O P A)} = \frac{A B}{O A}$ ...(i)
Since $A Y$ perpendicular to $O P$ . So, $A Y$ is the height of $△ O A P$ and $△ Q A P$
$\frac{a r (△ O A P)}{a r (△ Q A P)} = \frac{\frac{1}{2} \times P Q \times A Y}{\frac{1}{2} \times O P \times A Y} = \frac{P Q}{O P}$ ...(ii)
But $△ B A P$ and $△ Q A P$ are on the same base $A P$ and between same parallel straight lines $B Q$ and $A P$
$∴ △ B A P = △ Q A P$ ...(iii)
$∴ \frac{A B}{O A} = \frac{P Q}{O P}$
Since $P X$ is perpendicular to $O A$ ,
$P X$ is the height of $△ O A P$ and $△ C A P$
$∴ a r (△ O A P) = \frac{1}{2} b h = \frac{1}{2} \times O A \times P X$
$∴ a r (△ C A P) = \frac{1}{2} b h = \frac{1}{2} \times A C \times P X$
$\Rightarrow \frac{a r (C A P)}{a r (△ O A P)} = \frac{A C}{O A} = \frac{A B + A C}{O A} = \frac{A B}{O A} + \frac{B C}{O A}$ ....(iv)
Similarly, $A Y$ is perpendicular to $O P$ .
So, $A Y$ is height of $△ O A P$ and $△ R A P$
$∴ \frac{a r (R A P)}{a r (O A P)} = \frac{\frac{1}{2} \times R P \times A Y}{\frac{1}{2} \times O P \times A Y} = \frac{R P}{O P}$
$= \frac{P Q + Q R}{O P} = \frac{P Q}{O P} + Q R O P$ ...(v)
But $△ C A P$ and $△ R A P$ are on the same base $A P$ and between same parallel $C R$ and $A P$ . ( $C R | | A P$ as $l | | n$ )
$∴ △ C A P = △ R A P$ ...(vi)
From equations (iv), (v) and (vi), we get
$\frac{A B}{O A} + \frac{B C}{O A} = \frac{P Q}{O P} + \frac{Q R}{O P}$
$\Rightarrow \frac{A B}{O A} + \frac{B C}{O A} = \frac{P Q}{O P} + \frac{B C}{O A}$
$\Rightarrow \frac{A B}{O A} = \frac{P Q}{O P}$
Dividing the equation by equation A,
$\frac{\frac{A B}{O A}}{\frac{B C}{O A}} = \frac{\frac{P Q}{O P}}{\frac{Q R}{O P}}$
$\Rightarrow \frac{A B}{B C} = \frac{P Q}{Q R}$
$∴$ The statement is true.

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Similar questions

In the follow diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines l, m and n at points A,B,C and P,Q,R respectively as shown below. Prove that $\frac{A B}{B C} = \frac{P Q}{Q R}$ .

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State true or false:In the following diagram, lines l,m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A,B,C and P,Q,R respectively as shown, then:ABBC=PQQR

State true or false:
In the following diagram, lines $l, m$ and $n$ are parallel to each other. Two transversals $p$ and $q$ intersect the parallel lines at points $A, B, C$ and $P, Q, R$ respectively as shown, then:
$\frac{A B}{B C} = \frac{P Q}{Q R}$