Advertisements
Chapters
2: Compound Interest
3: Expansions
4: Factorisation
5: Simultaneous Linear Equations
6: Indices/Exponents
7: Logarithms
8: Triangles
9: Mid-point Theorem
▶ 10: Pythagoras Theorem
11: Rectilinear Figures
12: Constructions of Polygons
13: Theorems on Area
14: Circles
15: Statistics
16: Mensuration
17: Trigonometric Ratios
18: Trigonometric Ratios of Some Standard Angles and Complementary Angles
Chapter 19: Co-ordinate Geometry: An Introduction
![Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 10 - Pythagoras Theorem Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 10 - Pythagoras Theorem - Shaalaa.com](/images/mathematics-english-class-9-icse_6:f26eb985e8254aa987299226050d7c71.jpg)
Advertisements
Solutions for Chapter 10: Pythagoras Theorem
Below listed, you can find solutions for Chapter 10 of CISCE Nootan for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई.
Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 10 Pythagoras Theorem Exercise 10A [Pages 210 - 212]
Find whether the given sides of the triangle form a right-angled triangle or not:
3 cm, 4 cm and 5 cm
Find whether the given sides of the triangle form a right-angled triangle or not:
5 cm, 13 cm and 12 cm
Find whether the given sides of the triangle form a right-angled triangle or not:
8 cm, 9 cm and 12 cm
A rectangular garden is 30 m broad and 40 m long. Find the length of its diagonal.
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.
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.
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?.
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.
The side of a rhombus is 10 cm. Its one diagonal is 12 cm. Find the length of other diagonal.
Find the length of the altitude of an equilateral triangle of side 2a cm.
Find the altitude of an equilateral triangle of side `6sqrt(3)` cm.
In ΔАВC, ∠B = 90°. If AC = (x + 4) cm, BC = (x + 2) cm and AB = (3x + 1) cm, find the sides of triangle.
In the adjoining figure, ∠PQR = 90°, PR = 10 cm, QR = 6 cm and SR = 9 cm. Find PS.

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

In the adjoining figure, DC || AB. BC = 8 cm, AD = 17 cm, CD = 24 cm.
Find
- AB
- Area of trapezium ABCD

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.
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.

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.
In ΔАВC, ∠ABC = 90° and D is any point on BC. Prove that : AD2 + BC2 = AC2 + BD2.
In ΔPQR, ∠QPR = 90° and PM ⊥ QR. Prove that : PM2 = QM.RM.

In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2, prove that ∠ACD = 90°.
ΔАВС is a isosceles triangle in which AB = AC and ∠A = 90°. Prove that : BC2 = 2AC2.
In ΔABC, ∠A = 90° and BC2 = 2AC2, prove that : ΔABC is isosceles.
In ΔABC, ∠ABC is an acute angle. Prove that : AC2 = AB2 + BC2 – 2BC.BD.

ΔABC is an equilateral triangle. Side BC is trisected at D. Prove that : 9AD2 = 7AB2.
In ΔАBC, ∠ABC = 90°. X and Y are mid-points of the sides AB and BC respectively.
Prove that:
- CX2 + AY2 = 5XY2
- 4(CX2 + AY2) = 5AC2
In the adjoining figure, AB > AC, BE = EC and ∠ADC = 90°.
Prove that:
- AB2 – AC2 = 2BC.ED
- AB2 + AC2 = 2(AE2 + BE2)

In ΔАВС, AB = AC and D is any point on side BC produced. Prove that: AD2 = AB2 + BD · CD.
Prove that the sum of the squares on the sides of a rhombus is equal to the sum of square on its diagonals.
The diagonals of a rhombus ABCD intersect each other at O. Prove that: `OA^2 + OC^2 = 2AB^2 - 1/2 BD^2`.
In ◻ABCD, ∠B = 90° and ∠D = 90°. Prove that : 2AC2 = AB2 + BC2 + CD2 + DA2.
In the adjoining figure, PQ = QR and ∠PSQ = 90°.
Prove that : PR2 = 2PQ.RS.

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

If O is any point in the interior of a rectangle, ABCD, prove that : OA2 + OC2 = OB2 + OD2.
Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 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 :
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`
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
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
In ΔАВC, BC = CA and ∠ACB = 90°. Then correct relation is ______.
AB2 = 2AC2
2AB2 = AC2
BC2 = 2AC2
2BC2 = AC2
In a rhombus ABCD, AC2 + BD2 is equal to ______.
AB2
2BC2
3CD2
4DA2
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
In ΔABC, ∠ACB > 90°. The correct relation is ______.
AB2 > BC2 + AC2
BC2 > AC2 + AB2
AC2 > AB2 + BC2
AB2 = BC2 + AC2
In ΔАВС, ∠ACB < 90°. The correct relation is ______.
AB2 = BC2 + AC2
BC2 < AC2 + AB2
AB2 < BC2 + AC2
AB2 > BC2 + AC2
In ΔABC, AB = AC and BD ⊥ AC then BD2 – CD2 is equal to ______.
2CD·AD
2CD·AC
2AC·BC
2BC·AC
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
![Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 10 - Pythagoras Theorem Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 10 - Pythagoras Theorem - Shaalaa.com](/images/mathematics-english-class-9-icse_6:f26eb985e8254aa987299226050d7c71.jpg)
Nootan solutions for माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 10 - Pythagoras Theorem
Shaalaa.com has the CISCE Mathematics माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 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 माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई 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.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. Nootan textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई chapter 10 Pythagoras Theorem are .
Using Nootan माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई solutions Pythagoras Theorem exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Nootan Solutions are essential questions that can be asked in the final exam. Maximum CISCE माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई students prefer Nootan Textbook Solutions to score more in exams.
Get the free view of Chapter 10, Pythagoras Theorem माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई additional questions for Mathematics माठेमटिक्स [इंग्रजी] इयत्ता ९ आयसीएसई CISCE, and you can use Shaalaa.com to keep it handy for your exam preparation.
