Question

Q19) In AABC, line PQ | side
BC , AP = 3, BP= 6, AQ = 5 then
the value of CQ is *​

Answer 1

Answer:

10

Step-by-step explanation:

hope u will understand the ans

Answer 2

Answer:

The value of $CQ$ is $5$.

Step-by-step explanation:

Given: $P Q \| B C$,

$\begin{aligned}&A P=3 \\&B P=6 \\&A Q=5\end{aligned}$

To Find: The value of $CQ$.

Solution: Let the value of $CQ$ is $x$ as shown in the figure.

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{APQ}$,

$\begin{array}{ll}\angle \mathrm{A}=\angle \mathrm{A} & \text { (common in both) } \\ \angle \mathrm{ABC}=\angle \mathrm{APQ} & \text { (corresponding angles) } \\ \angle \mathrm{ACB}=\angle \mathrm{AQP} \quad & \text { (corresponding angles) } \\ \text { Therefore, } \triangle \mathrm{ABC}=\triangle \mathrm{APQ} \quad & \text { (by } \mathrm{AAA} \text { ) } \\ \end{array}$

$\frac{A P}{A B}=\frac{A Q}{A C}=\frac{P Q}{B C}$

On substitution in $\frac{A P}{A B}=\frac{A Q}{A C}$,

  $\begin{aligned}&\frac{3}{6}=\frac{5}{A C} \end{aligned}$

$\text{AC}=\frac{6\times5}{3}$

$\text{AC}=\frac{30}{3}$

$\begin{aligned}&\mathrm{AC}=10 \end{aligned}$

As we know that $\begin{aligned}& \mathrm{AC}=\mathrm{AQ}+\mathrm{QC}\end{aligned}$,

$\begin{aligned}&10=5+\mathrm{x} \end{aligned}$

 $\begin{aligned}&\mathrm{x}=10-5 \end{aligned}$

    $\begin{aligned}&=5\end{aligned}$

Hence, the value of $\text{CQ}$ is $5$.