Question

In the figure (not drawn to scale) given
below. P is a point on AB such that AP : PB
= 4:3. PQ is parallel to AC and QD is
parallel to CP. In triangle ARC, angle ARC = 90° and
in triangle POS, PSQ = 90°. The length of QS is
6 cm. What is ratio AP: PD?
(A) 10:3
(B) 2:1
C) 3
(D) 8:3​

Answer 1

Answer:

7:3 IND.

Step-by-step explanation:

Given,AP/PB =4/3

AP/PB = QC/QB = 4/3

BD/PD = QB/QC =3/ 4

PB = BD + PD = 3+4 = 7

PB/PD = 7/4

AP/PB * PB/PD = 4/3 * 7/4

AP/PD =7/3

thus AP : PD = 7 : 3

Answer 2

Answer:

The ratio of AP:PD = 7:3

Step-by-step explanation:

Given,

PQ is parallel to AC and QD is parallel to CP

AP:PB = 4:3

∠ARC = 90°

∠PSQ = 90°

To find,

The ratio AP:PD

Solution:

Recall the theorem

Basic proportionality theorem

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

Since PQ is parallel to AC, By Basic proportionality theorem we have,

$\frac{AP}{PB} = \frac{CQ}{QB}$

Since $\frac{AP}{PB} = \frac{4}{3}$, we have

$\frac{CQ}{QB} = \frac{4}{3}$ ------------------(1)

Again, since QD is parallel to CP, by Basic proportionality theorem we have,

$\frac{PD}{DB}= \frac{CQ}{QB}$

$\frac{PD}{DB}= \frac{CQ}{QB} = \frac{4}{3}$ ---------------(2)

From (1) and (2) we have

$\frac{AP}{PB} = \frac{PD}{DB} = \frac{CQ}{QB}$

$\frac{AP}{PB} = \frac{PD}{DB}$ -------------(3)

$\frac{AP}{PD} = \frac{PB}{DB}$

$= \frac{PD+DB}{DB}$

$= \frac{PD}{DB}+ \frac{DB}{DB}$

$= \frac{PD}{DB}+ 1$

$= \frac{4}{3} + 1$

$= \frac{7}{3}$

$\frac{AP}{PD} = \frac{7}{3}$

The ratio of AP:PD = 7:3

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