# FRANK Solutions for Class 9 Maths Chapter 17 - Pythagoras Theorem

Revise Frank Solutions for ICSE Class 9 Mathematics Chapter 17 Pythagoras Theorem on TopperLearning. As you revise, you may come across problems asking you to find the area of a right-angled triangle. The chapter solutions by our experts will help you use the Pythagoras theorem step-by-step to find the right answer.

Frank textbook solutions for ICSE Class 9 Maths also includes model answers on methods to find the length of the side of a triangle. Practise similar problems by going through our Selina solutions as well. For revisiting the basics, check our e-learning concept videos available 24x7 on our study portal.

## Chapter 17 - Pythagoras Theorem Exercise Ex. 17.1

InABC, AD is perpendicular to BC. Prove that

AB^{2}
+ CD^{2} = AC^{2} + BD^{2}

From a point O in the interior of aABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that:

a. AF^{2} + BD^{2} + CE^{2 }= OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2}

b. AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

In a triangle ABC, AC > AB, D is the midpoint BC, and AEBC. Prove that:

Ina 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.

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