In an acute-angled triangle XYZ, an altitude XP is drawn

Solution

Since, XP ⊥ YZ, both ΔXYP and ΔOYP are right triangles.Similarly, both ΔXZP and ΔOZP are also right triangles.Using Pythagoras theorem, we have;In ΔXYP, XY² = YP² + XP²Or, XP² = XY² - YP² ....... (I)And ΔXZP, we have;XZ² = XP² + PZ²So, XP² = XZ² - PZ² ...... (II)So, XY² - YP² = XZ² - PZ²Now, using equation (I) and (II) , we have;Or, XZ² - XY² = PZ² - YP² ...... (III)Now, in ΔOYP, we have;And OP² = OY² - YP² ....... (IV)And in ΔOZP, we have;OP² = OZ² - PZ² ...... (V)Now, using equation (IV) and (V) , we have;OY² - YP² = OZ² - PZ²Or, PZ² - PY² = OZ² - OY² ...... (VI)Now, using equation (III) and (VI) , we have;OZ² - OY² = XZ² - XY²Or, XY² + OZ² = XZ² + OY²So, 12² + 9² = XZ² + 5²Or, 144 + 81 = XZ² + 25Or, XZ² = 225 - 25Or, XZ² = 200And XZ = ± 10√2Since, side of a triangle cannot be negative, XZ = 10√2 cm