Hong Kong
Given a right-angled ^@\Delta ABC^@. The lengths of the sides containing the right angle are ^@7 \space cm^@ and ^@24 \space cm^@. A circle is inscribed in ^@\Delta ABC^@. Find the radius of the circle.
Answer:
^@ 3 \space cm^@
- In ^@\Delta ABC^@, we have @^ \begin{aligned} \angle B = 90^\circ, AB = 7 \space cm \space \text{ and } BC = 24 \space cm \end{aligned} @^
- A circle is inscribed in ^@\Delta ABC^@. Let ^@O^@ be its centre and ^@M^@, ^@N^@ and ^@P^@ be the points where it touches the sides ^@AB^@, ^@BC^@ and ^@CA^@ respectively. @^ \begin{aligned} \text { Then, } OM \perp AB, ON \perp BC, OP \perp CA. \end{aligned} @^
- Let ^@ r \space cm^@ be the radius of the circle.
Then, ^@OM = ON = OP = r\space cm^@.
^@ \text{ Now, } AB^2 + BC^2 = CA^2 \space\space [\text{ By pythagoras' theorem }]^@
^@ \implies (7)^2 cm^2 + (24)^2 cm^2 = CA^2 \\ \implies CA = 25 \space cm ^@. - Now, @^ \begin{aligned} &ar(\Delta ABC) = ar(\Delta AOB) + ar(\Delta BOC) + ar(\Delta COA) \\ \implies& \dfrac { 1 } { 2 } \times AB \times BC = \bigg( \dfrac { 1 } { 2 } \times AB \times OM \bigg) + \bigg( \dfrac { 1 } { 2 } \times BC \times ON \bigg) + \bigg( \dfrac { 1 } { 2 } \times CA \times OP \bigg) \\ \implies& \dfrac { 1 } { 2 } \times 7 \times 24 = \bigg( \dfrac { 1 } { 2 } \times 7 \times r \bigg) + \bigg( \dfrac { 1 } { 2 } \times 24 \times r \bigg) + \bigg( \dfrac { 1 } { 2 } \times 25 \times r \bigg) \\ \implies& 168 = 7 r + 24 r + 25 r = 56 r \\ \implies& r = \dfrac { 168 } { 56 } = 3 \space cm. \end{aligned} @^ Hence, the radius of the circle is ^@ 3 \space cm^@.
