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Nootan solutions for Mathematics [English] Class 9 ICSE chapter 10 - Pythagoras Theorem [Latest edition]

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Nootan solutions for Mathematics [English] Class 9 ICSE chapter 10 - Pythagoras Theorem - Shaalaa.com
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Solutions for Chapter 10: Pythagoras Theorem

Below listed, you can find solutions for Chapter 10 of CISCE Nootan for Mathematics [English] Class 9 ICSE.


Exercise 10AExercise 10B
Exercise 10A [Pages 210 - 212]

Nootan solutions for Mathematics [English] Class 9 ICSE 10 Pythagoras Theorem Exercise 10A [Pages 210 - 212]

1. (i)Page 210

Find whether the given sides of the triangle form a right-angled triangle or not:

3 cm, 4 cm and 5 cm

1. (ii)Page 210

Find whether the given sides of the triangle form a right-angled triangle or not:

5 cm, 13 cm and 12 cm

1. (iii)Page 210

Find whether the given sides of the triangle form a right-angled triangle or not:

8 cm, 9 cm and 12 cm

2.Page 210

A rectangular garden is 30 m broad and 40 m long. Find the length of its diagonal.

3.Page 210

A ladder 13 m long rests against a vertical wall. Its top reaches to a window on the wall at 12 m high. Find the distance of the foot of the ladder from the wall.

4.Page 210

Two poles of height 12 m and 24 m stand vertically on a plane ground. If the distance between their tops is 13 m. Find the distance between their feet.

5.Page 210

An aeroplane leaves an airport and flies due North at a speed of 200 km\hr. At the same time another aeroplane leaves the same airport and flies due West at a speed of 150 km\hr. How far apart will be the two aeroplanes after 4 hours?.

6.Page 210

A ladder 25 m long reaches a window which is 15 m above the ground on one side of the street. When it turned to the other side keeping its foot at the same point, it touches a wall at a height of 20 m from the ground. Find the width of the street.

7.Page 210

The side of a rhombus is 10 cm. Its one diagonal is 12 cm. Find the length of other diagonal.

8.Page 210

Find the length of the altitude of an equilateral triangle of side 2a cm. 

9.Page 210

Find the altitude of an equilateral triangle of side `6sqrt(3)` cm.

10.Page 210

In ΔАВC, ∠B = 90°. If AC = (x + 4) cm, BC = (x + 2) cm and AB = (3x + 1) cm, find the sides of triangle.

11.Page 210

In the adjoining figure, ∠PQR = 90°, PR = 10 cm, QR = 6 cm and SR = 9 cm. Find PS.

12.Page 210

In the adjoining figure, ∠RQS = 90°, ∠QPS = 90°, RS = 25 cm, QR = 20 cm, PQ = 9 cm. Find PS.

13.Page 210

In the adjoining figure, DC || AB. BC = 8 cm, AD = 17 cm, CD = 24 cm.

Find

  1. AB 
  2. Area of trapezium ABCD

14.Page 211

ΔАВС is an isosceles triangle in which AB = AC = 17 cm and BC = 16 cm. Find the length of perpendicular drawn from A to BC.

15.Page 211

In the adjoining figure, SR || PQ and ∠PQR = 90°. If PQ = 7 cm, PR = 25 cm and SR = 17 cm, find the length of PS.

16.Page 211

The sides of a right-angled triangle are 2x, x + 5 and 3x + 1. If the hypotenuse is 3x + 1, find the sides of triangle.

17.Page 211

In ΔАВC, ∠ABC = 90° and D is any point on BC. Prove that : AD2 + BC2 = AC2 + BD2.

18.Page 211

In ΔPQR, ∠QPR = 90° and PM ⊥ QR. Prove that : PM2 = QM.RM.

19.Page 211

In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2, prove that ∠ACD = 90°.

20.Page 211

ΔАВС is a isosceles triangle in which AB = AC and ∠A = 90°. Prove that : BC2 = 2AC2.

21.Page 211

In ΔABC, ∠A = 90° and BC2 = 2AC2, prove that : ΔABC is isosceles.

22.Page 211

In ΔABC, ∠ABC is an acute angle. Prove that : AC2 = AB2 + BC2 – 2BC.BD.

23.Page 211

ΔABC is an equilateral triangle. Side BC is trisected at D. Prove that : 9AD2 = 7AB2.

24.Page 211

In ΔАBC, ∠ABC = 90°. X and Y are mid-points of the sides AB and BC respectively.

Prove that:

  1. CX2 + AY2 = 5XY2
  2. 4(CX2 + AY2) = 5AC2
25.Page 211

In the adjoining figure, AB > AC, BE = EC and ∠ADC = 90°.

Prove that:

  1. AB2 – AC2 = 2BC.ED
  2. AB2 + AC2 = 2(AE2 + BE2)

26.Page 211

In ΔАВС, AB = AC and D is any point on side BC produced. Prove that: AD2 = AB2 + BD · CD.

27.Page 211

Prove that the sum of the squares on the sides of a rhombus is equal to the sum of square on its diagonals.

28.Page 211

The diagonals of a rhombus ABCD intersect each other at O. Prove that: `OA^2 + OC^2 = 2AB^2 - 1/2 BD^2`.

29.Page 211

In ◻ABCD, ∠B = 90° and ∠D = 90°. Prove that : 2AC2 = AB2 + BC2 + CD2 + DA2.

30.Page 211

In the adjoining figure, PQ = QR and ∠PSQ = 90°.

Prove that : PR2 = 2PQ.RS.

31.Page 212

In the adjoining figure, ∠PQR = 90° and XY || QR. If PX : QX = 1 : 2, PQ = 6 cm, PY = 4 cm, find PR and QR.

32.Page 212

If O is any point in the interior of a rectangle, ABCD, prove that : OA2 + OC2 = OB2 + OD2.

Exercise 10B [Page 212]

Nootan solutions for Mathematics [English] Class 9 ICSE 10 Pythagoras Theorem Exercise 10B [Page 212]

Multiple Choice Questions Choose the correct answer from the given four options in each of the following questions :

1.Page 212

Each side of an equilateral triangle is 10 cm. The length of the each its altitude is ______.

  • 5 cm

  • `5sqrt(2)  cm`

  • `5sqrt(3)  cm`

  • `3sqrt(5)  cm`

2.Page 212

A ladder 17 m long reaches a window above the ground. If the distance of the foot of ladder from wall is 8 m, the height of the window is ______.

  • 12 m

  • 15 m

  • 12 cm

  • 14 m

3.Page 212

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of rhombus is ______.

  • 8 cm

  • 9 cm

  • 11 cm

  • 10 cm

4.Page 212

In ΔАВC, BC = CA and ∠ACB = 90°. Then correct relation is ______.

  • AB2 = 2AC2

  • 2AB2 = AC2

  • BC2 = 2AC2

  • 2BC2 = AC2

5.Page 212

In a rhombus ABCD, AC2 + BD2 is equal to ______.

  • AB2

  • 2BC2

  • 3CD2

  • 4DA2

6.Page 212

If the sides of a rectangle are 6 cm and 8 cm then the length of its diagonal is ______.

  • 9 cm

  • 10 cm

  • 12 cm

  • 14 cm

7.Page 212

In ΔABC, ∠ACB > 90°. The correct relation is ______.

  • AB2 > BC2 + AC2

  • BC2 > AC2 + AB2

  • AC2 > AB2 + BC2

  • AB2 = BC2 + AC2

8.Page 212

In ΔАВС, ∠ACB < 90°. The correct relation is ______.

  • AB2 = BC2 + AC2

  • BC2 < AC2 + AB2

  • AB2 < BC2 + AC2

  • AB2 > BC2 + AC2

9.Page 212

In ΔABC, AB = AC and BD ⊥ AC then BD2 – CD2 is equal to ______.

  • 2CD·AD

  • 2CD·AC

  • 2AC·BC

  • 2BC·AC

10.Page 212

In ΔАВС, ∠ACB = 90°. P and Q are the points on CA and CB respectively which divides these sides in the ratio 2 : 1. Then 9(AQ2 + BP2) is equal to ______.

  • 5AB2

  • 8AB2

  • 10AB2

  • 13AB2

Solutions for 10: Pythagoras Theorem

Exercise 10AExercise 10B
Nootan solutions for Mathematics [English] Class 9 ICSE chapter 10 - Pythagoras Theorem - Shaalaa.com

Nootan solutions for Mathematics [English] Class 9 ICSE chapter 10 - Pythagoras Theorem

Shaalaa.com has the CISCE Mathematics Mathematics [English] Class 9 ICSE CISCE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Nootan solutions for Mathematics Mathematics [English] Class 9 ICSE CISCE 10 (Pythagoras Theorem) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

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Concepts covered in Mathematics [English] Class 9 ICSE chapter 10 Pythagoras Theorem are .

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