NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Ans : (i) Similar
(ii) Similar 
(iii) Equilateral
(iv) (a) Equal
(b) Proportional 

Ans : (i) Two equilateral triangles with sides 1 cm and 2 cm.

Two squares with sides 1 cm and 2 cm.

(ii) Trapezium and square
 

Ans : Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e 1:2, but their corresponding angles are not equal. 

Ans : (i) 

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\( \begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{1.5}{3}=\frac{1}{x}} \\ {x=\frac{3 \times 1}{1.5}} \\ {x=2} \\ {\therefore \mathrm{EC}=2 \mathrm{cm}}\end{array} \) 
(ii) 

Let AD = x cm 
It is given that DE || BC
By using basic proportionality theorem, we obtain
\( \begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{x}{7.2}=\frac{1.8}{5.4}} \\ {x=\frac{1.8 \times 7.2}{5.4}} \\ {x=2.4} \\ {\therefore \mathrm{AD}=2.4 \mathrm{cm}}\end{array} \) 

Ans : (i) 

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \) 
\( \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=1.5 \) 
Hence, \( \frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}} \) 
Therefore, EF is not parallel to QR.
(ii)  

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
\( \frac{P E}{E Q}=\frac{4}{4.5}=\frac{8}{9} \) 
\( \frac{P F}{F R}=\frac{8}{9} \) 
Hence \( \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}} \) 
\( Therefore, EF is parallel to QR. \) 
(iii)  

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
\( \frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64} \) 
\( \frac{\mathrm{PF}}{\mathrm{PR}}=\frac{0.36}{2.56}=\frac{9}{64} \) 
\( \begin{array}{l}{\text { Hence, } \frac{\mathrm{PE}}{\mathrm{PQ}}=\frac{\mathrm{PF}}{\mathrm{PR}}} \\ {\text { Therefore, EF is parallel to QR. }}\end{array} \)