NCERT Solutions for Class 10 Maths Chapter 6 Triangles त्रिभुज
Table of Contents
Exercise 6.1
Question 1:
Fill in the blanks using the correct word given in brackets :
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________ . (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
कोष्ठकों में दिए शब्दों में से सही शब्दों का प्रयोग करते हुए, रिक्त स्थानों को भरिए :
(i) सभी वृत्त …………. होते है | (सर्वांगसम, समरूप)
(ii) सभी वर्ग …………. होते हैं| (समरूप, सर्वांगसम)
(iv) सभी …………. त्रिभुज समरूप होते है | (समद्विबाहु, समबाहु)
(v) भुजाओं की समान संख्या वाले दो बहुभुज समरूप होते हैं, यदि उनके संगत कोण (i) ………… हो तथा उनकी संगत (ii) ………… भुजाएँ हों | (बराबर, समानुपाती)
Solution
(i) similar समरूप
(ii) similar समरूप
(iii) equilateral समबाहु
(iv) (a) eqaul (b) proportional (a) बराबर, (b) समानुपाती
Question 2:
Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.
निम्नलिखित युग्मों के दो भिन्न -भिन्न उदाहरण दीजिए :
(i) समरूप आकृतियाँ
(ii) ऐसी आकृतियाँ जो समरूप नहीं हैं |
Solution
(i) A pair of equilateral triangles of different sides lengths, a pair of two rectangles of different lenghts and breadths, etc.
(ii) A pair of rectangle and a parallelogram with angles not equal to 900, a pair of isosceles triangle and a scalene triangle, etc.
(i) दो वृत्त और दो वर्ग
(ii) वृत्त और वर्ग तथा त्रिभुज और चतुर्भुज
Question 3:
State whether the following quadrilaterals are similar or not:
बताइए की निम्नलिखित चतुर्भुज समरूप है या नहीं :
Solution
PQRS and ABCD are not similar.
PQRS and ABCD चतुर्भुज समरूप नहीं है
Exercise 6.2
Question 1:
In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
आकृति 6.17 (i) और (ii) में, DE || BC में AD ज्ञात कीजिए :
Solution
(i) In △ ABC,
\(\quad\) DE∥BC
\(\quad\)∴ \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) [Using Basic proportionality theorem]
⇒\(\frac{1.5}{3}\) = \(\frac{1}{EC}\)
⇒EC = 2 cm
(ii) In △ ABC,
\(\quad\)DE∥BC
\(\quad\)∴ \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) [Using Basic proportionality theorem]
⇒ \(\frac{AD}{7.2}\)= \(\frac{1.8}{5.4}\)
⇒ AD = 2.4 cm.
Question 2:
E and F are points on the sides PQ and PR respectively of a \(\triangle\) PQR. For each of the following cases, state whether EF || QR :
किसी त्रिभुज PQR की भुजाओं PQऔर PR पर क्रमशः बिन्दु E और F स्थित हैं | निम्नलिखित में से प्रत्येक स्थिति के लिए, बताइए कि क्या EF|| QR है |
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution
(i) PE = 3.9 cm, EQ= 3cm, PF = 3.6 और FR= 2.4 cm
By Basic proportionality theorem,
we get, \(\frac{PE}{EQ}\) = \(\frac{3.9}{3}\) = 1.3 and \(\frac{PF}{FR}\)= \(\frac{3.6}{2.4}\) = 1.5
So, we get, \(\frac{PE}{EQ}\) ≠ \(\frac{PF}{FR}\)
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
By Basic proportionality theorem, we get,
\(\frac{PE}{QE}\) = \(\frac{4}{4.5}\) = \(\frac{8}{9}\) and, \(\frac{PF}{RF}\) = \(\frac{8}{9}\)
So, we get here, \(\frac{PE}{QE}\) = \(\frac{PF}{RF}\)
Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm and, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, \(\frac{PE}{EQ}\) = \(\frac{0.18}{1.10\) = \(\frac{9}{55}\)…………. (i)
and, \(\frac{PF}{FR}\) = \(\frac{0.36}{2.20}\) = \(\frac{9}{55}\) ………… (ii)
So, we get here, PE/EQ = PF/FR
Hence, EF is parallel to QR.
Question 3:
In Fig. 6.18, if LM || CB and LN || CD, prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).
Solution
In the given figure, LM || CB,
By Basic Proportionality Theorem,
we get, \(\frac{AM}{AB}\) = \(\frac{AL}{AC}\)……………………..(i)
Similarly, given, LN || CD and using basic proportionality theorem,
∴\(\frac{AN}{AD}\) = \(\frac{AL}{AC}\) ……………………………(ii)
From equation (i) and (ii) ,
we get, \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\)
Hence, proved.
