NCERT Solutions for Class 10 Maths Chapter 6

NCERT Solutions for Class 10 Maths Chapter 6 exercise 6.6 (optional), 6.5, 6.4, 6.3, 6.2 and 6.1 Triangles in PDF form for High School UP Board and CBSE Board exams 2019-20 in Hindi Medium and English medium. Download Class 10 Solutions Apps based on latest NCERT Solutions for offline use.


Class 10:Maths – गणित
Chapter 6:Triangles

NCERT Solutions for class 10 Maths chapter 6

NCERT Solutions for class 10 Maths chapter 6

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Class 10 Maths Solutions – Triangles

Complete Exercises solutions and a brief description about triangles, similarity of triangles, theorems and the facts related to this chapter are given below. It will help the students to enhance their knowledge about the chapter triangles and the mathematician involved. Click here to see the NCERT Textbooks of all subjects of class 10.

English Medium & Hindi Medium




What is meant by Similarity or Similar Triangle?

Similarity of geometric figures is an important concept of Euclidean geometry. Similarity in a geometric transformation of one figure into the other figure such that the measure of all linear elements of one figure are in proportion to the corresponding linear elements of the other figure. Two triangles (or any polygons of the same number of sides) are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). All congruent figures are similar but the similar figures need not be congruent. Download Class 10 Maths App for offline use as well as कक्षा 10 गणित App for offline use.




Objectives of the chapter – Similar Triangles
  • To identify similar figures, distinguish between congruent and similar triangles, prove that if a line is drawn parallel to one side of a triangle then the other two sides are divided in the same ratio, state and use the criteria (Criteria means a standard which is established so that judgement or decision, especially a scientific one can be made) for similarity of triangles viz. AAA, SSS and SAS.
  • To verify and use results given in the curriculum based on similarity theorems. To prove the Baudhayan/Pythagoras Theorem and apply these results in verifying experimentally (or proving logically) problems based on similar triangles.
  • Download Class 10 Maths App for offline use as well as कक्षा 10 गणित App for offline use.



Do You Know?

A Greek mathematician Thales gave an important relation relating to two equiangular triangles that ‘The ratio of any two corresponding sides in two similar triangles is always the same. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio’. Which is known as the Basic Proportionality Theorem or the Thales Theorem.
There are so many other important theorems based on similar triangles like If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. Or if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. Check CBSE board papers, for the questions based on BPT in board exams.



  • Area Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
  • Pythagoras theorem and its converse: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Converse of the theorem is ‘In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle’.



Historical Facts!
  • Pythagoras theorem is famous because of its wide range of applications. In ancient Indian civilization, ‘Sulb Sutras’ written by Bodhayan (800 BC) depict Pythagoras theorem. भास्कराचार्य and ब्रह्मगुप्त gave different proofs of Pythagoras theorem.
  • Leonardo De Vinchi, the great artist, sculpturist, and architect, famous for his painting ‘Monalisa’ also gave a beautiful proof for this theorem.
  • Thales of Miletus (624 – 546 BC, Greece) was the first known philosopher and mathematician. He is credited with the first use of deductive reasoning in geometry. He discovered many propositions in geometry. He is believed to have found the heights of the pyramids in Egypt, using shadows and principle of similar triangles. Height of pyramids can also be find using applications of trigonometry.
  • Brahma Gupta’s theorem (628 A.D.): The rectangle contained by any two sides of a triangle, is equal to rectangle contained by altitude drawn to the third side and the circum diameter.
  • According to Galileo Galilei, the universe cannot be read until we have learnt the language in which it is written. It is written in mathematical language and the letters are triangles, circles and other geometrical figures, without which it is humanly impossible to comprehend a single word.
  • Download Class 10 Maths App for offline use as well as कक्षा 10 गणित App for offline use.

Sides of triangles are given below. Determine this is a right triangles. In case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm.

Sides of triangle: 7 cm, 24 cm and 25 cm.
Squaring these sides, we get 49, 576 and 625.
49 + 576 = 625
⇒ 7^2+〖24〗^2=〖25〗^2
These sides satisfy the Pythagoras triplet,
hence these are sides of right angled triangle.
We know that the hypotenuses is the longest side in right angled triangle.
Hence, its length is 25 cm.

यदि दो समरूप त्रिभुजों के क्षेत्रफल बराबर हों तो सिद्ध कीजिए कि वे त्रिभुज सर्वांगसम होते हैं।

माना, ∆ABC ~ ∆DEF, इसलिए
ar(∆ABC)/ar(∆DEF) =〖AB〗^2/〖DE〗^2 =〖BC〗^2/〖EF〗^2 =〖AC〗^2/〖DF〗^2
…(1)
दिया है, ar(∆ABC) = ar(∆DEF)
इसलिए, ar(∆ABC)/ar(∆DEF) = 1
समीकरण (1) से,
〖AB〗^2/〖DE〗^2 =〖BC〗^2/〖EF〗^2 =〖AC〗^2/〖DF〗^2 =1
⇒ AB = DE, BC = EF और AC = DF
∴ ∆ABC ≅ ∆DEF [SSS सर्वांगसम प्रमेय से]

ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC^2.

Given that the triangle ABC is an isosceles triangle
such that AC = BC and ∠C = 90°,
In ∆ABC, by Pythagoras theorem
AB^2 = AC^2 + BC^2
⇒ AB^2 = AC^2 + AC^2
[Because AC = BC] ⇒ AB^2 = 2AC^2

ABC is an isosceles triangle with AC = BC. If AB^2 = 2AC^2, prove that ABC is a right triangle.

Given that: AB^2 = 2AC^2
⇒ AB^2 = AC^2 + AC^2
⇒ AB^2 = AC^2 + BC^2
[Because AC = BC] These sides satisfy the Pythagoras theorem.
Hence, the triangle ABC is a right angled triangle.

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Let OA is wall and AB is ladder.
In ∆AOB, by Pythagoras theorem
AB^2 = OA^2 + OB^2
⇒〖10〗^2 =8^2 + BO^2
⇒ 100 = 64 + BO^2
⇒ BO^2 = 36
⇒ BO = 6 m
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.

दो समरूप त्रिभुजों की भुजाएँ 4:9 के अनुपात में हैं। इन त्रिभुजों के क्षेत्रफलों का अनुपात है: (A) 2:3 (B) 4:9 (C) 81:16 (D) 16:81

हम जानते हैं कि समरूप त्रिभुजों के क्षेत्रफलों का अनुपात उनकी संगत भुजाओं के वर्गों के अनुपात के बराबर होता है।

इसलिए,
त्रिभुजों के क्षेत्रफलों का अनुपात =(4/9)^2 = 16/81
अतः, विकल्प (D) सही है।

S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS. Show that Δ RPQ ~ Δ RTS.

In ∆RPQ and ∆RST,
∠RTS = ∠QPS [Given] ∠R = ∠R [Common] ∴ ∆RPQ ∼ ∆RTS [AA similarity]

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