148 MATHEMATICS
Solution : Let AB be the ladder and CA be the wall
with the window at A (see Fig. 6.49).
Also, BC = 2.5 m and CA = 6 m
From Pythagoras Theorem, we have:
AB^2 =BC^2 + CA^2= (2.5)^2 + (6)^2= 42.25So, AB = 6.5
Thus, length of the ladder is 6.5 m.
Example 12 : In Fig. 6.50, if AD BC, prove that
AB^2 + CD^2 = BD^2 + AC^2.
Solution : From ✁ ADC, we have
AC^2 =AD^2 + CD^2
(Pythagoras Theorem) (1)From ✁ ADB, we have
AB^2 =AD^2 + BD^2
(Pythagoras Theorem) (2)Subtracting (1) from (2), we have
AB^2 – AC^2 =BD^2 – CD^2or, AB^2 + CD^2 =BD^2 + AC^2
Example 13 : BL and CM are medians of a
triangle ABC right angled at A. Prove that
4 (BL^2 + CM^2 ) = 5 BC^2.
Solution : BL and CM are medians of the
✁ ABC in which ✂ A = 90° (see Fig. 6.51).
From ✁ABC,
BC^2 =AB^2 + AC^2 (Pythagoras Theorem) (1)From ✁ ABL,
BL^2 =AL^2 + AB^2Fig. 6.49Fig. 6.50Fig. 6.51