Advertisements
Online Mock Tests
Chapters
2: Profit , Loss and Discount
3: Compound Interest
4: Expansions
5: Factorisation
6: Changing the subject of a formula
7: Linear Equations
8: Simultaneous Linear Equations
9: Indices
10: Logarithms
11: Triangles and their congruency
12: Isosceles Triangle
13: Inequalities in Triangles
14: Constructions of Triangles
15: Mid-point and Intercept Theorems
16: Similarity
▶ 17: Pythagoras Theorem
18: Rectilinear Figures
19: Quadrilaterals
20: Constructions of Quadrilaterals
21: Areas Theorems on Parallelograms
22: Statistics
23: Graphical Representation of Statistical Data
24: Perimeter and Area
25: Surface Areas and Volume of Solids
26: Trigonometrical Ratios
27: Trigonometrical Ratios of Standard Angles
28: Coordinate Geometry
![Frank solutions for Mathematics [English] Class 9 ICSE chapter 17 - Pythagoras Theorem Frank solutions for Mathematics [English] Class 9 ICSE chapter 17 - Pythagoras Theorem - Shaalaa.com](/images/mathematics-english-class-9-icse_6:c41cc344f5174c64a036c55d113af73f.jpg)
Advertisements
Solutions for Chapter 17: Pythagoras Theorem
Below listed, you can find solutions for Chapter 17 of CISCE Frank for Mathematics [English] Class 9 ICSE.
Frank solutions for Mathematics [English] Class 9 ICSE 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: AB2 + CD2 = AC2 + BD2
In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD2 = 7 AB2.
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: AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2
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: AF2 + BD2 + CE2 = AE2 + CD2 + BF2
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2`
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 = AD2 - BC x CE + `(1)/(4)"BC"^2`
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2AD2 + `(1)/(2)"BC"^2`
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 - AB2 = 2BC x ED
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2(AD2 + CD2)
A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB2 + OD2 = OC2 + OA2
ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2= AC2 + BD2
AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD2 = 3AB2.
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: 9AQ2 = 9AC2 + 4BC2
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: 9BP2 = 9BC2 + 4AC2
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(AQ2 + BP2) = 13AB2
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.
Solutions for 17: Pythagoras Theorem
![Frank solutions for Mathematics [English] Class 9 ICSE chapter 17 - Pythagoras Theorem Frank solutions for Mathematics [English] Class 9 ICSE chapter 17 - Pythagoras Theorem - Shaalaa.com](/images/mathematics-english-class-9-icse_6:c41cc344f5174c64a036c55d113af73f.jpg)
Frank solutions for Mathematics [English] Class 9 ICSE chapter 17 - 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. Frank solutions for Mathematics Mathematics [English] Class 9 ICSE CISCE 17 (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. Frank textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in Mathematics [English] Class 9 ICSE chapter 17 Pythagoras Theorem are Regular Polygon, Right-angled Triangles and Pythagoras Property, Right-angled Triangles and Pythagoras Property.
Using Frank Mathematics [English] Class 9 ICSE 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 Frank Solutions are essential questions that can be asked in the final exam. Maximum CISCE Mathematics [English] Class 9 ICSE students prefer Frank Textbook Solutions to score more in exams.
Get the free view of Chapter 17, Pythagoras Theorem Mathematics [English] Class 9 ICSE additional questions for Mathematics Mathematics [English] Class 9 ICSE CISCE, and you can use Shaalaa.com to keep it handy for your exam preparation.