S Chand ICSE Mathematics Class 9 Solution Tenth Chapter Pythagoras Theorem Exercise 10B
(1) ABCD is a square, prove that AC2 = 2AB2
(2) In Fig. 10.26, AB = BC = CA = 2a and segment AD ⊥ side BC, Show that
(i) AD = a √3, (ii) area of ΔABC = a2√3
(3) In Fig 10.27, prove that AB2 – AD2 = CD2 – CB2,
(4) In a ΔABC, AD ⊥ BC, Prove that:
AB2 + CD2 = AC2 + BD2.
(5) In a quadrilateral ABCD, the diagonals AC, BD intersect at right angles. Prove that:
AB2 + CD2 = BC2 + DA2.
(6) In ΔABC, ∠B = 90o and D is the mid point of BC, Prove that
(i) AC2 = AD2 + 3CD2 (ii) BC2 = 4(AD2 – AB2)