(3)

Since, $∆$ABC ~ $∆$EDF

Therefore, the ratio of their corresponding sides is proportional.

$\Rightarrow \frac{BC}{EF}=\frac{AB}{DE}\Rightarrow BC.DE\ne AB.EF$

(3)

Given that, AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3

By Basic Proportionality theorem, we have $\frac{AD}{BD}=\frac{AE}{EC}$

$\Rightarrow \frac{3x-2}{7x-5}=\frac{5x-4}{5x-3}\Rightarrow (3x-2)(5x-3)=(5x-4)(7x-5)$

$\Rightarrow 15x^2-19x+6=35x^2-53x+20$

$\Rightarrow 20x^2-34x+14=0\Rightarrow 10x^2-17x+7=0$

$\Rightarrow (x-1)(10x-7)=0\Rightarrow x=1,x=\frac{7}{10}$

(3)

As $DE\parallel BC$

$\therefore$ $\frac{AE}{AC}=\frac{DE}{BC}$ $\text{[From BPT]}$

$\Rightarrow \frac{a}{a+b}=\frac{x}{y}$

$\Rightarrow x(a+b)=ay\Rightarrow x=\frac{ay}{a+b}$

(3)

$AD\parallel BC$

and $\frac{AO}{OC}=\frac{DO}{OB}=\frac{1}{2}$

$\therefore$  $\frac{AD}{BC}=\frac{AO}{OC}\Rightarrow \frac{4}{BC}=\frac{1}{2}\Rightarrow BC=8\text{ cm}$

(4)

We have, $∆ABC~∆PQR$

         $\Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}=\frac{AM}{PN} \text{(corresponding sides of similar triangles)}$

But $\frac{A{B}^2}{P{Q}^2}=\frac{4}{9}\Rightarrow \frac{AB}{PQ}=\frac{2}{3}$

i.e., $\frac{AB}{PQ}=\frac{AM}{PN}=\frac{2}{3}$

Hence, $AM:PN=2:3$

(2)

The ratio of the corresponding sides of similar triangles is same as the ratio of their perimeter

$\therefore$  $∆ABC~∆PQR\text{ or }∆PQR~∆ABC$

$\Rightarrow \frac{PQ}{AB}=\frac{QR}{BC}=\frac{PR}{AC}=\frac{48}{56}\Rightarrow \frac{PQ}{AB}=\frac{6}{7}$

(3)     15.3 cm

(3)

$AD=5\text{ cm}$

$DB=2.5\text{ cm}$

$BC=12\text{ cm}$

$DE\parallel BC$

$∆ABC~∆ADE$

$\Rightarrow \frac{AD}{AB}=\frac{DE}{BC}\Rightarrow \frac{5}{7.5}=\frac{DE}{12}\Rightarrow DE=\frac{60}{7.5}=8\text{ cm}$

(4)

Trapezium is a quadrilateral in which diagonal divide each other proportionally.

(2)

$$\begin{gathered} In △ABC, we have PQ \parallel BC \\ AP/PB = AQ/QC = 4/6 = 8/QC \\ \Rightarrow 2/3 = 8/QC \Rightarrow QC = 12 cm \\ \therefore AC = AQ + QC = 8 + 12 = 20 cm \end{gathered}$$

(2)

Two triangles are similar if their corresponding sides are in proportion

(3)

Since ABCD is a trapezium with AB ? CD and diagonals intersect at O

AO/OC = OB/OD ⇒ 2/5 = (x - 2)/(2x + 5)

=  2(2x + 5) = 5(x - 2)

= 4x + 10 = 5x - 10 $\Rightarrow$ x = 20

Option (c) is correct.

(1)

Given right angled triangle ABC, using Pythagoras theorem we have

$$\begin{gathered} AB² = AC² + BC² \\ \Rightarrow 5² = AC² + 9 \Rightarrow AC² = 25 – 9 = 16 \Rightarrow \\ AC = 4 cm, Hence \left(iii\right) is correct. \\ Given DE \parallel BC, u\sin g BPT theorem, \\ AD/AB = AE/AC \Rightarrow 2/5 = AE/4 \\ \Rightarrow AE = 8/5 cm, Hence \left(i\right) is correct. \end{gathered}$$

(4)

Checking for right-angled triangle:

$$\begin{gathered} Given, AB = 12 cm, BC = 5 cm and AC = 13 cm \\ \left(AB\right)² + \left(BC\right)² = \left(12\right)² + \left(5\right)² = 144 + 25 = 169 = \left(13\right)² \\ So, \left(AB\right)² + \left(BC\right)² = \left(AC\right)², by Pythagoras Theorem, \\ \Delta ABC is right angled at B. \\ Hence, \left(i\right) is correct, and \left(iii\right) is incorrect \\ Now, AD/AB = 3/12 = 1/4 and AE/AC = (13/4 \times 1/13) = 1/4 \\ By BPT if AD/AB = AE/AC, then DE \parallel BC, Hence \left(ii\right) is correct. \\ As B is a right angle and DE \parallel BC, angle D will also be 90°, so DE ⟂ AB, \\ Hence \left(iv\right) is correct. \end{gathered}$$

(4)

$$\begin{gathered} Given PQ \parallel BC \Rightarrow AP/AB = AQ/AC (By BPT) \therefore \left(iv\right) is correct. \\ 4/\left(4+6\right)=8/AC\Rightarrow AC=20cm,\therefore \left(i\right)iscorrect. \end{gathered}$$

QC = 20 − 8 = 12 cm

4, 8, 12 are in Arithmetic Progression.

Hence, (iii) is correct.

(3)

$In \Delta AGF and \Delta DBG:$

$\angle GAF = \angle BDG = 90° and$

$\angle AGF = \angle DBG (Corresponding angles)$

$$\begin{gathered} \therefore \Delta AGF ~ \Delta DBG (By AA similarity) \dots \left(A\right) \\ \therefore \left(iii\right) is correct. \end{gathered}$$

$Similarly, in \Delta AGF and \Delta EFC:$ 

$\angle GAF = \angle CEF = 90° and$

$\angle AFG = \angle FCE (Corresponding angles)$

$$\begin{gathered} \therefore \Delta AGF ~ \Delta EFC (By AA similarity) \dots \left(B\right) \\ \therefore \left(iv\right) is correct. \end{gathered}$$

$$\begin{gathered} From \left(A\right) and \left(B\right): \\ \Delta DBG ~ \Delta EFC \end{gathered}$$

$\frac{DB}{EF}=\frac{DG}{EC}$

$But EF = DG = DE (side of square)$

$$\begin{gathered} \frac{DB}{DE}=\frac{DE}{EC} \\ \Rightarrow D{E}^2=DB\times EC \\ \therefore \left(ii\right) is correct. \end{gathered}$$

(1)

$In \Delta AGF and \Delta ADC$

$\angle AGF=\angle ADC (Each90°)$

$\angle GAF=\angle DAC (Each45°)$

$\therefore \Delta AGF~\Delta ADC(By AA similaritycriterion)$

$\Rightarrow \frac{AG}{AD}=\frac{AF}{AC}\Rightarrow \frac{AC}{AD}=\frac{AF}{AG}\dots \left(i\right)$

$also,\angle DAC=\angle GAF(Each45°)$

$\Rightarrow \angle DAC-\angle GAC=\angle GAF-\angle GAC$

$$\begin{gathered} \Rightarrow \angle DAG=\angle CAF \\ and\frac{AC}{AD}=\frac{AF}{AG}\left(From\right(i\left)\right) \end{gathered}$$

$\therefore \Delta ACF~\Delta ADG (By SAS similaritycriterion)$