#### Chapters

Chapter 2: Profit , Loss and Discount

Chapter 3: Compound Interest

Chapter 4: Expansions

Chapter 5: Factorisation

Chapter 6: Changing the subject of a formula

Chapter 7: Linear Equations

Chapter 8: Simultaneous Linear Equations

Chapter 9: Indices

Chapter 10: Logarithms

Chapter 11: Triangles and their congruency

Chapter 12: Isosceles Triangle

Chapter 13: Inequalities in Triangles

Chapter 14: Constructions of Triangles

Chapter 15: Mid-point and Intercept Theorems

Chapter 16: Similarity

Chapter 17: Pythagoras Theorem

Chapter 18: Rectilinear Figures

Chapter 19: Quadrilaterals

Chapter 20: Constructions of Quadrilaterals

Chapter 21: Areas Theorems on Parallelograms

Chapter 22: Statistics

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Perimeter and Area

Chapter 25: Surface Areas and Volume of Solids

Chapter 26: Trigonometrical Ratios

Chapter 27: Trigonometrical Ratios of Standard Angles

Chapter 28: Coordinate Geometry

## Chapter 17: Pythagoras Theorem

### Frank solutions for Class 9 Maths ICSE Chapter 17 Pythagoras Theorem Exercise 17.1

Find the length of the perpendicular of a triangle whose base is 5cm and the hypotenuse is 13cm. Also, find its area.

Find the length of the hypotenuse of a triangle whose other two sides are 24cm and 7cm.

Calculate the area of a right-angled triangle whose hypotenuse is 65cm and one side is 16cm.

A man goes 10 m due east and then 24 m due north. Find the distance from the straight point.

A ladder 25m long reaches a window of a building 20m above the ground. Determine the distance of the foot of the ladder from the building.

A right triangle has hypotenuse p cm and one side q cm. If p - q = 1, find the length of third side of the triangle.

A ladder 15m long reaches a window which is 9m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12m high. Find the width of the street.

The foot of a ladder is 6m away from a wall and its top reaches a window 8m above the ground. If the ladder is shifted in such a way that its foot is 8m away from the wall to what height does its tip reach?

Two poles of height 9m and 14m stand on a plane ground. If the distance between their 12m, find the distance between their tops.

The length of the diagonals of rhombus are 24cm and 10cm. Find each side of the rhombus.

Each side of rhombus is 10cm. If one of its diagonals is 16cm, find the length of the other diagonals.

In ΔABC, AD is perpendicular to BC. Prove that: AB^{2} + CD^{2} = AC^{2} + BD^{2}

In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD^{2 }= 7 AB^{2}.

From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF^{2} + BD^{2} + CE^{2 }= OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2}

From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC^{2} = AD^{2 }+ BC x DE + `(1)/(4)"BC"^2`

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB^{2} = AD^{2} - BC x CE + `(1)/(4)"BC"^2`

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB^{2} + AC^{2} = 2AD^{2 }+ `(1)/(2)"BC"^2`

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC^{2} - AB^{2 }= 2BC x ED

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB^{2} + AC^{2} = 2(AD^{2} + CD^{2})

A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB^{2} + OD^{2 }= OC^{2} + OA^{2}

ABCD is a rhombus. Prove that AB^{2} + BC^{2} + CD^{2} + DA^{2}= AC^{2} + BD^{2}

AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD^{2} = 3AB^{2}.

The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that `2"AB"^2 = 2"AC"^2 + "BC"^2`

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.

Prove that: 9AQ^{2 }= 9AC^{2} + 4BC^{2}

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.

Prove that: 9BP^{2} = 9BC^{2} + 4AC^{2}

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.

Prove that : 9(AQ^{2} + BP^{2}) = 13AB^{2}

In the given figure, PQ = `"RS"/(3)` = 8cm, 3ST = 4QT = 48cm.

SHow that ∠RTP = 90°.

In a right-angled triangle ABC,ABC = 90°, AC = 10 cm, BC = 6 cm and BC produced to D such CD = 9 cm. Find the length of AD.

In the given figure. PQ = PS, P =R = 90°. RS = 20 cm and QR = 21 cm. Find the length of PQ correct to two decimal places.

In a right-angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.

PQR is an isosceles triangle with PQ = PR = 10 cm and QR = 12 cm. Find the length of the perpendicular from P to QR.

In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle.

## Chapter 17: Pythagoras Theorem

## Frank solutions for Class 9 Maths ICSE chapter 17 - Pythagoras Theorem

Frank solutions for Class 9 Maths ICSE chapter 17 (Pythagoras Theorem) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Class 9 Maths ICSE solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Class 9 Maths ICSE chapter 17 Pythagoras Theorem are Pythagoras Theorem, Regular Polygon, Pythagoras Theorem.

Using Frank Class 9 solutions Pythagoras Theorem exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Frank Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 9 prefer Frank Textbook Solutions to score more in exam.

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