In the given figure, ^@ ABC ^@ is a right-angled triangle with ^@ AB = 7 \space cm ^@ and ^@ AC = 9 \space cm ^@. A circle with center ^@ O ^@ has been inscribed inside the triangle. Calculate the value of ^@r^@, the radius of the inscribed circle.
O A B C F D E 7 cm 9 cm


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Answer:

^@ 2.3 \space cm ^@

Step by Step Explanation:
  1. Let us join ^@O^@ to ^@A, B,^@ and ^@C^@ and draw ^@OD \perp AB^@, ^@OE \perp BC^@ and ^@OF \perp CA^@.
    O A B C F D r r r E 7 cm 9 cm

    We see that ^@OD, OE,^@ and ^@ OF ^@ are the radius of the circle with center ^@O^@.
    ^@ \implies OD = OE = OF = r \space cm ^@

    Also, ^@ \triangle ABC ^@ is a right-angled triangle. @^ \begin{aligned} \implies \text { Area of } \triangle ABC & = \dfrac { 1 }{ 2 } \times AB \times AC \\ & = \dfrac { 1 }{ 2 } \times 7 \space cm \times 9 \space cm \\ & = 31.5 \space cm^2 \end{aligned} @^
  2. Let us now find the area of ^@ \triangle ABC ^@ in terms of ^@ r ^@. @^ \begin{aligned} \text{Area of } \triangle ABC & = \text{Area of } \triangle OAB + \text{Area of } \triangle OBC + \text{Area of } \triangle OCA \\ & = \dfrac { 1 }{ 2 } \times AB \times OD + \dfrac { 1 }{ 2 } \times BC \times OE + \dfrac { 1 }{ 2 } \times CA \times OF \\ & = \dfrac { 1 }{ 2 } \times AB \times r + \dfrac { 1 }{ 2 } \times BC \times r + \dfrac { 1 }{ 2 } \times CA \times r \\ & = \dfrac { 1 }{ 2 } (AB + BC + CA) \times r \\ & = \dfrac { 1 }{ 2 } \text{(Perimeter of } \triangle ABC) \times r \end{aligned} @^
  3. Comparing the area of ^@ \triangle ABC ^@ obtained in step 1 and step 2, we have @^ \begin{aligned} & \text{ Area of } \triangle ABC = 31.5 \space cm^2 = \dfrac { 1 }{ 2 } \times (AB + BC + CA) \times r \\ \implies & 31.5 \space cm^2 = \dfrac { 1 }{ 2 } \times (7 + BC + 9 ) \times r && \ldots \text{(i)} \end{aligned} @^
  4. Applying Pythagoras theorem in ^@ \triangle ABC^@, we have @^ \begin{aligned} & BC^2 = AB^2 + AC^2 \\ \implies & BC = \sqrt{(7)^2 + (9)^2} = 11.4 \space cm \end{aligned} @^
  5. Now, substituting the value of ^@ BC ^@ in ^@ eq \space \text{(i)}^@, we have @^ \begin{aligned} & 31.5 = \dfrac { 1 } { 2 } \times (7 + 11.4 + 9) \times r \\ \implies & 31.5 \times 2 = 27.4 \times r \\ \implies & r = \dfrac { 31.5 × 2 } { 27.4 } = 2.3 \space cm \end{aligned} @^
  6. Hence, the radius of the inscribed circle is ^@ 2.3 \space cm ^@.

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