Question

In the given figure $\Delta$ABC is a right angled triangle with $\angle$B = ${90}^{\circ }$, where B is the centre of the circle. DE and FG are chords of length 24 cm and the radius of the circle is 13 cm. Find the length of the hypotenuse of the $\Delta$ABC (A and C are the foot of the perpendiculars from the centre to the respective chords).

• A. 5 cm
• B. 10 cm
• C. 5 cm
• D. 10 cm

Answer 1

The correct option is C

5 cm

Let PQ = `x' cm

In $\Delta$ABG -

$x^2$ + ${12}^2$ = ${13}^2$ (Perpendicular bisectors of two chords of a circle intersect at its centre)​

$\Rightarrow$ x = 5 cm

Equal chords of a circle are equidistant from the centre.

BC = 5 cm

In $\Delta$ABC,

${AC}^2$ = $5^2$ + $5^2$

AC = 5$\sqrt{2}$ cm