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ICSE Class 9 Maths Selina Solutions Chapter 13 Pythagoras Theorem

In this article we have provided ICSE Class 9 Maths Selina Solutions Chapter 13 prepared by our experts to help students to prepare better for their examinations.

Neha Tanna11 Jul, 2024

ICSE Class 9 Maths Selina Solutions Chapter 13: Selina Solutions are thought to be very helpful for ICSE Class 9 Maths exam preparation. We have provided you with thorough answers to the ICSE Class 9 Maths Selina Solutions Chapter 13 Pythagoras Theorem tasks below. The subject area experts have developed these answers under the CISCE-mandated syllabus for the ICSE.

The PDF answers for ICSE Class 9 Maths Selina Solutions Chapter 13 Pythagoras Theorem are available here. These Selina solutions are available for free download, allowing students to practise them offline as well.

ICSE Class 9 Maths Selina Solutions Chapter 13 Overview

ICSE Class 9 Maths Selina Solutions Chapter 13, focusing on the Pythagoras Theorem, serve as an essential resource for students aiming to grasp fundamental geometric principles. This chapter elucidates the theorem's applications in solving problems involving right-angled triangles, emphasizing its utility in both theoretical and practical scenarios. The solutions offered are meticulously crafted to provide clarity and step-by-step guidance, aiding students in understanding the underlying concepts effectively. Through a variety of exercises and examples, the chapter reinforces the theorem's significance in geometry, equipping students with the skills necessary for tackling related problems confidently. Ultimately, Selina Solutions Chapter 13 not only facilitates comprehensive exam preparation but also promotes a deeper appreciation and mastery of geometric principles, thereby enhancing overall mathematical proficiency.

ICSE Class 9 Maths Selina Solutions Chapter 13 PDF

Below we have provided ICSE Class 9 Maths Selina Solutions Chapter 13 in detail. This chapter will help you to clear all your doubts regarding the chapter Inequalities. Students are advised to prepare from these ICSE Class 9 Maths Selina Solutions Chapter 13 before the examinations to perform better.

ICSE Class 9 Maths Selina Solutions Chapter 13 PDF

ICSE Class 9 Maths Selina Solutions Chapter 13 Pythagoras Theorem

Below we have provided ICSE Class 9 Maths Selina Solutions Chapter 13 -

1. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

Solution:

The pictorial representation of the given problem is given below, Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. (i)Here, AB is the hypotenuse. Therefore, applying the Pythagoras theorem, we get, AB ² = BC ² + CA ² 13 ² = 5 ² + CA ² CA ² = 13 ² – 5 ² CA ² = 144 CA = 12m Therefore, the distance of the other end of the ladder from the ground is 12m

2. A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

Solution:

Here, we need to measure the distance AB as shown in the figure below, Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Therefore, in this case AB ² = BC ² +CA ² AB ² = 50 ² + 40 ² AB ² = 2500 + 1600 AB ² = 4100 AB = 64.03 Therefore, the required distance is 64.03 m.

3. In the figure: ∠PSQ = 90 ^o , PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Solution:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the triangle PQS and applying Pythagoras theorem we get, PQ ² = PS ² + QS ² 10 ² = PS ² + 6 ² PS ² = 100 – 36 PS ² = 64 PS = 8 Now, we consider the triangle PRS and applying Pythagoras theorem we get, PR ² = RS ² + PS ² PR ² = 15 ² + 8 ² PR = 17 Therefore, the length of PR = 17cm

4. The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ABD = BCD = 90 ^o . Calculate the length of AB.

Solution:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the triangle BDC and applying Pythagoras theorem we get, DB ² = DC ² + BC ² DB ² = 12 ² + 3 ² = 144 + 9 = 153 Now, we consider the triangle ABD and applying Pythagoras theorem we get, DA ² = DB ² + BA ² 13 ² = 153 + BA ² BA ² = 169 – 153 BA = 4 The length of AB is 4 cm.

5. AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Solution:

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of same measure and the perpendicular AD will divide BC in two equal parts. Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Here, we consider the triangle ABD and applying Pythagoras theorem we get, AB ² = AD ² + BD ² AD ² = 100 ² – 5 ² AD ² = 100 – 25 AD ² = 75 = 8.7 Therefore, the length of AD is 8.7 cm

6. In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC= 3 cm. Calculate the length of OC.

Solution:

We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the triangle ABO and applying Pythagoras theorem we get, AB ² = AO ² + OB ² AO ² = AB ² – OB ² AO ² = AB ² – (BC + OC) ² Let OC = x AO ² = AB ² – (BC + x) ² ……… (1) First, we consider the triangle ACO and applying Pythagoras theorem we get AC ² = AO ² + x ² AO ² = AC ² – x ² ……… (2) From 1 and 2 AB ² – (BC + x) ² = AC ² – x ² 8 ² – (6 + x) ² = 3 ² – x ² X = 1 7/12 cm Therefore, the length of OC will be 19/12 cm

7. In triangle ABC, AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm ² . Find x.

Solution:

Here, the diagram will be, We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Since, ABC is an isosceles triangle, therefore perpendicular from vertex will cut the base in two equal segments. First, we consider the triangle ABD and applying Pythagoras theorem we get, AB ² = AD ² + BD ² AD ² = x ² – 5 ² AD ² = x ² – 25 AD = √(x ² – 25) ………. (1) Now, Area = 60 ½ (10) AD = 60 ½ (10) [√(x ² – 25)] = 60 x = 13 Therefore x = 13 cm

8. If the sides of triangle are in the ratio 1 :2: 1, show that is a right-angled triangle.

Solution:

Let, the sides of the triangle be, x ² + x ² = 2x ² = (√2x) ² ……… (1) Here, in (i) it is shown that, square of one side of the given triangle is equal to the addition of square of other two sides. This is nothing but Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Therefore, the given triangle is right angled triangle.

9. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.

Solution:

The diagram of the given problem is given below, We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Here 11 – 6 = 5m Base = 12 m Applying Pythagoras theorem, we get h ² = 5 ² + 12 ² = 25 + 144 = 169 h = 13 therefore, the distance between the tips will be 13m

10. In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.

Take M be the point on CD such that AB = DM. So, DM = 7cm and MC = 10 cm Join points B and M to form the line segment BM. So BM || AD also BM = AD. In triangle BAD BD ² = AD ² + BA ² 25 ² = AD ² + 7 ² AD ² = 576 AD = 24 In triangle CMB CB ² = CM ² + MB ² CB ² = 10 ² + 24 ² CB ² = 676 CB = 26 cm

ICSE Class 9 Maths Selina Solutions Chapter 13 Exercise 13B

1. In the figure, given below, AD parallel to BC. Prove that: c ² = a ² + b ² – 2ax

Solution:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the triangle ABD and applying Pythagoras theorem we get, AB ² = AD ² + BD ² c ² = h ² + (a – x) ² h ² = c ² – (a – x) ² ……… (1) First, we consider the triangle ACD and applying Pythagoras theorem we get AC ² = AD ² + CD ² b ² = h ² + x ² h ² = b ² – x ² ………. (2) from 1 and 2 c ² – (a – x) ² = b ² – x ² c ² – a ² – x ² + 2ax = b ² – 2ax c ² = a ² + b ² – 2ax hence the proof.

2. In equilateral Δ ABC, AD parallel to BC and BC = x cm. Find, in terms of x, the length of AD .

Solution:

In equilateral Δ ABC, AD parallel to BC. Therefore, BD = DC = x/2 cm. Applying Pythagoras theorem, we get In right angled triangle ADC AC ² = AD ² + DC ² x ² = AD ² + (x/2) ² AD ² = (x) ² – (x/2) ² AD ² = (x/2) ² AD = (x/2) cm

3. ABC is a triangle, right-angled at B. M is a point on BC. Prove that:

AM ² + BC ² = AC ² + BM ² .

Solution:

The pictorial form of the given problem is as follows, Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the triangle ABM and applying Pythagoras theorem we get, AM ² = AB ² + BM ² AB ² = AM ² – BM ² ……….. (1) Now we consider the triangle ABC and applying Pythagoras theorem we get AC ² = AB ² + BC ² AB ² = AC ² – BC ² …… (2) From 1 and 2 we get AM ² – BM ² = AC ² + BM ² AM ² + BC ² = AC ² + BM ² Hence the proof.

4. M and N are the mid-points of the sides QR and PQ respectively of a triangle PQR, right-angled at Q. Prove that:

(i) PM ² + RN ² = 5 MN ²

(ii) 4 PM ² = 4 PQ ² + QR ²

(iii) 4 RN ² = PQ ² + 4 QR ²

(iv) 4 (PM ² + RN ² ) = 5 PR ²

Solution:

Draw, PM, MN, NR Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Since, M and N are the mid-points of the sides QR and PQ respectively, therefore, PN = NQ, QM = RM (i) First, we consider the triangle PQM and applying Pythagoras theorem we get, PM ² = PQ ² + MQ ² = (PN + NQ) ² + MQ ² = PN ² + NQ ² + 2 PN. NQ + MQ ² = MN ² + PN ² + 2 PN. NQ [ we know MN ² = NQ ² + MQ ² ] ………….. (1) Now we consider the triangle RNQ and applying Pythagoras theorem, RN ² = NQ ² + RQ ² = NQ ² + (QM + RM) ² = NQ ² + QM ² + 2 QM. RM + RM ² …………. (2) Adding 1 and 2 we get PM ² + RN ² = MN ² + PN ² + 2PN. NQ + MN ² + RM ² + 2QM. RM PM ² + RN ² = 2MN ² + PN ² + RM ² + 2PN. NQ + 2QM. RM PM ² + RN ² = 2MN ² + NQ ² + QM ² + 2(QN) ² + 2 (QM) ² PM ² + RN ² = 2MN ² + MN ² + 2MN ² PM ² + RN ² = 5MN ² Hence the proof. (ii) Now consider the triangle PQM and apply Pythagoras theorem we get PM ² = PQ ² + MQ ² 4PM ² = 4PQ ² + 4 MQ ² [multiplying both sides by 4] 4PM ² = 4PQ ² + 4 (½ QR ² ) [MQ = ½ QR] 4PM ² = 4PQ ² + QR ² Hence the proof. (iii) now consider triangle RQN and apply Pythagoras theorem we get RN ² = NQ ² + RQ ² 4RN ² = 4NQ ² + 4 QR ² [multiplying both sides by 4] 4RN ² = 4QR ² + 4 (½ PQ ² ) [NQ = ½ PQ] 4RN ² = PQ ² + 4QR ² Hence the proof. (iv) now consider the triangle PQM and apply Pythagoras theorem, PM ² = PQ ² + MQ ² = (PN + NQ) ² + MQ ² = PN ² + NQ ² + 2 PN. NQ + MQ ² = MN ² + PN ² + 2 PN. NQ [ we know MN ² = NQ ² + MQ ² ] ………….. (1) Now we consider the triangle RNQ and applying Pythagoras theorem, RN ² = NQ ² + RQ ² = NQ ² + (QM + RM) ² = NQ ² + QM ² + 2 QM. RM + RM ² = MN ² + RM ² + 2 QM. RM …………. (2) Adding 1 and 2 we get PM ² + RN ² = MN ² + PN ² + 2PN. NQ + MN ² + RM ² + 2QM. RM PM ² + RN ² = 2MN ² + PN ² + RM ² + 2PN. NQ + 2QM. RM PM ² + RN ² = 2MN ² + NQ ² + QM ² + 2(QN) ² + 2 (QM) ² PM ² + RN ² = 2MN ² + MN ² + 2MN ² PM ² + RN ² = 5MN ² 4 (PM ² + RN ² ) = 4. 5 (NQ ² + MQ ² ] 4 (PM ² + RN ² ) = 4. 5 [(½ PQ) ² + (½ QR) ² ] 4 (PM ² + RN ² ) = 5PR ² Hence the proof.

5. In triangle ABC, ∠B = 90 ^o and D is the mid-point of BC. Prove that: AC ² = AD ² + 3CD ² .

Solution:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. In triangle ABC, ∠ B = 90 ^o and D is the mid-point of BC. Join AD. Therefore, BD=DC First, we consider the triangle ADB and applying Pythagoras theorem we get, AD ² = AB ² + BD ² AB ² = AD ² – BD ² …. (1) Similarly, we get from rt. angle triangles ABC we get, AC ² = AB ² + BC ² AB ² = AC ² – BC ² …. (1) From 1 and 2 we get AC ² – BC ² = AD ² – BD ² AC ² = AD ² – BD ² + BC ² AC ² = AD ² – CD ² + 4CD ² [BD = CD = ½ BC] AC ² = AD ² + 3CD ² Hence the proof.

6. In a rectangle ABCD, prove that: AC ² + BD ² = AB ² + BC ² + CD ² + DA ² .

Solution:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Since, ABCD is a rectangle angles A, B, C and D are rt. angles. First, we consider the triangle ACD and applying Pythagoras theorem we get, AC ² = DA ² + CD ² ……. (1) Similarly, we get from rt. angle triangle BDC we get, BD ² = BC ² + CD ² = BC ² + AB ² [In a rectangle opposite sides are equal CD = AB] Adding (i) and (ii), AC ² + BD ² = AB ² + BC ² + CD ² + DA ² Hence the proof.

7. In a quadrilateral ABCD, ∠B = 90 ⁰ and ∠D = 90 ⁰ . Prove that: 2AC ² – AB ² = BC ² + CD ² + DA ²

Solution:

In quadrilateral ABCD ∠ B = 90 ⁰ and ∠ D = 90 ⁰ So triangle ABC and triangle ADC are right angles. For triangle ABC, apply Pythagoras theorem, AC ² = AB ² + BC ² AB ² = AC ² – BC ² ……….. (i) For triangle ADC, apply Pythagoras theorem, AC ² = AD ² + DC ² ……….. (ii) LHS = 2AC ² – AB ² = 2AC ² – (AC ² – BC ² ) from 1 = 2AC ² – AC ² + BC ² = AC ² + BC ² = AD ² + DC ² + BC ² from 2 = RHS

8. O is any point inside a rectangle ABCD. Prove that: OB ² + OD ² = OC ² + OA ² .

Solution:

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH. Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. Using Pythagorean theorem, we have from the above diagram: OA ² = AH ² + OH ² = AH ² + AE ² OC ² = CG ² + OG ² = EB ² + HD ² OB ² = EO ² + BE ² = AH ² + BE ² OD ² = HD ² + OH ² = HD ² + AE ² Adding these equalities, we get: OA ² + OC ² = AH ² + HD ² + AE ² + EB ² OB ² + OD ² = AH ² + HD ² + AE ² + EB ² From which we prove that for any point within the rectangle there is the relation OA ² + OC ² = OB ² + OD ² Hence Proved.

9. In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that: AR ² + BP ² + CQ ² = AQ ² + CP ² + BR ²

Solution:

Here, we first need to join OA, OB, and OC after which the figure becomes as follows, Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the and applying Pythagoras theorem we get, AO ² = AR ² + OR ² AR ² = AO ² – OR ² ……. (1) Similarly, from triangles, BPO, COQ, AOQ, CPO and BRO we get the following results, BP ² = BO ² – OP ² ……. (2) CQ ² = OC ² – OQ ² ……. (3) AQ ² = AO ² – OQ ² ……. (4) CP ² = OC ² – OP ² ……. (5) BR ² = OB ² – OR ² ……. (6) Adding 1, 2 and 3 we get AR ² + BP ² + CQ ² = AO ² – OR ² + BO ² – OP ² + OC ² – OQ ² ……. (7) Adding 4, 5 and 6 we get AQ ² + CP ² + BR ² = AO ² – OQ ² + OC ² – OP ² + OB ² – OR ² ………… (8) From 7 and 8, we get, AR ² + BP ² + CQ ² = AQ ² + CP ² + BR ² Hence proved.

10. Diagonals of rhombus ABCD intersect each other at point O. Prove that: OA ² + OC ² = 2AD ² – BD ² /2

Solution:

We know diagonals of the rhombus are perpendicular to each other. In quadrilateral ABCD, ∠AOD = ∠COD = 90 ^o We know triangle AOD and COD are right angle triangle. In triangle AOD, apply Pythagoras theorem, AD ² = OA ² + OD ² OA ² = AD ² – OD ² ………… (1) In triangle COD, apply Pythagoras theorem, CD ² = OC ² + OD ² OC ² = CD ² – OD ² ………… (2) LHS = OA ² + OC ² = AD ² – OD ² + CD ² – OD ² from 1 and 2 = AD ² – AD ² – 2(BD/2) ² [AD = CD and OD = BD/2] = 2AD ² – BD ² /2 = RHS

Benefits of ICSE Class 9 Maths Selina Solutions Chapter 13

ICSE Class 9 Maths Selina Solutions Chapter 13 on the Pythagoras Theorem offer several benefits for students:

Clear Explanation: The ICSE Class 9 Maths Selina Solutions Chapter 13 provided are known for their clarity and step-by-step explanations. They help students understand the concepts underlying the Pythagoras Theorem thoroughly.

Comprehensive Coverage: The ICSE Class 9 Maths Selina Solutions Chapter 13 cover all the exercises and problems given in the textbook. This ensures that students have sufficient practice and exposure to different types of problems related to the Pythagoras Theorem.

Practice Material: The ICSE Class 9 Maths Selina Solutions Chapter 13 include worked-out examples and practice exercises that help students apply the theorem in various geometrical and real-life contexts. This enhances their problem-solving skills.

Concept Reinforcement: By using the Selina Solutions, students can reinforce their understanding of the Pythagoras Theorem through practice. They can see how the theorem applies to different types of triangles and geometric configurations.

Exam Preparation: Since ICSE exams often include questions directly related to the Pythagoras Theorem, using Selina Solutions ensures that students are well-prepared for their exams. They can practice questions of varying difficulty levels, which helps in building confidence.

ICSE Class 9 Maths Selina Solutions Chapter wise
Chapter 1 Rational and Irrational Numbers	Chapter 11 Inequalities
Chapter 2 Compound Interest (Without Using Formula)	Chapter 12 Mid-Point and Its Converse (Including Intercept Theorem)
Chapter 3 Compound Interest (Using Formula)	Chapter 13 Pythagoras Theorem
Chapter 4 Expansions	Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram, Rectangles, Rhombus, Square, and Trapezium)
Chapter 5 Factorisation	Chapter 15 Construction of Polygons
Chapter 6 Simultaneous (Linear) Equations (Including Problems)	Chapter 16 Area Theorems
Chapter 7 Indices (Exponents)	Chapter 17 Circle
Chapter 8 Logarithms	Chapter 18 Statistics
Chapter 9 Triangles (Congruency in Triangles)	Chapter 19 Mean and Median
Chapter 10 Isosceles Triangles	Chapter 20 Area and Perimeter of Plane Figures

ICSE Class 9 Maths Selina Solutions Chapter 13 FAQs

What is the Pythagorean theorem Class 9 Selina?

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

What is the Pythagorean theorem class 9th?

Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides “. The sides of this triangle have been named Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°.

What is the main Pythagorean theorem?

A² + B² = C²

What are the uses of Pythagorean theorem?

Some of the important real-life uses of the Pythagorean theorem are as follows: Used in construction and architecture. Used in two-dimensional navigation to find the shortest distance. Used to survey the steepness of the slopes of mountains or hills.

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Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.

What Makes Us Different

Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others