Question

In ABC, Line PQ||side BC. If AP =10, PB = 12, AQ = 15, then complete the following activity to find the value of QC?​

Answer 1

Answer:

QC = 18

Step-by-step explanation:

In ∆ ABC, line PQ II side BC

AP/PB = AQ/QC

10/12 = 15/QC

QC= 15x12 / 10

QC = 180/10

QC = 18

Answer 2

We need to recall the basic proportionality theorem.

• Basic proportionality theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

This problem is about the basic proportionality theorem.

Given:

In Δ$ABC$ , $PQ||BC$ .

$AP=10$ , $PB=12$ , $AQ=15$

From the basic proportionality theorem, we get

$\frac{AP}{PB}=\frac{AQ}{QC}$

$\frac{10}{12}=\frac{15}{QC}$

$QC=\frac{15*12}{10}$

$QC=\frac{180}{10}$

$QC=18$