AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.

Given,

AM = 18 cm and MB = 8 cm

From figure,

AB = AM + MB = 18 + 8 = 26 cm

Radius of the circle = $\dfrac{26}{2}$ = 13 cm.

Let CD is the shortest chord drawn through M.

∴ CD ⊥ AB

From figure,

OM = AM - AO = 18 - 13 = 5 cm

OC (radius) = OA = 13 cm

Now in right ∆OMC,

OC² = OM² + MC² (By pythagoras theorem)

13² = 5² + MC²

MC² = 13² - 5²

MC² = 169 - 25 = 144

MC = $\sqrt{144}$ = 12 cm.

M is Mid-Point of CD (As perpendicular from center to the chord bisects it.)

CD = 2 × MC = 2 × 12 = 24 cm.

Hence, length of shortest chord = 24 cm.