Question

Look at the statements given below:
I. ∆ABC ∼ ∆DEF and the altitudes of these triangles are in the ratio 1 : 2, then ar(∆ABC) : ar(∆DEF) = 1 : 4.
II. In ∆ABC, DE ∥ BC and AD : DB = 1 : 2, then $\frac{\mathrm{ar}(∆ADE)}{\mathrm{ar}(∆ABC)}=\frac{1}{4}.$
III. In a ∆ABC, P and Q are points on AB and AC respectively such that AB = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC = 3PQ.

Which is true?
(a) I only
(b) II only
(c) I and II
(d) I and III

Answer 1

(d) I and III are true.

I .
$\because$ ∆ABC ∼ ∆DEF

$\therefore \frac{ar(△ABC)}{ar(△DEF)}={{\left(\frac{1}{2}\right)}^2}^{}=\frac{1}{4}=1:4$

II.
We have:
AD : DB = 1 : 2
$$\begin{gathered} DE\parallel BC \\ Therefore,byB.P.T., \\ \frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{2} \end{gathered}$$

$\Rightarrow \frac{DB}{AD}=\frac{2}{1}$

$\Rightarrow \frac{DB}{AD}+1=3$

$\Rightarrow \frac{AB}{AD}=3\Rightarrow \frac{AD}{AB}=\frac{1}{3}$

$\therefore \frac{ar(△ADE)}{ar(△ABC)}=\frac{A{D}^2}{A{B}^2}=\frac{1}{9}$

III.
We have:
AP = 3 cm
PB = 6 cm
AQ = 5 cm
QC = 10 cm
BC = 3PQ

$$\begin{gathered} \frac{BC}{PQ}=\frac{AB}{AP}=\frac{AC}{AQ}=\frac{3}{1} \\ \Rightarrow BC=3PQ \end{gathered}$$
Thus, I and III are true, II is false.