Question : The sides AB, BC, and AC of a $\triangle {ABC}$ are 12 cm, 8 cm, and 10 cm respectively. A circle is inscribed in the triangle touching AB, BC, and AC at D, E, and F respectively. The difference between the lengths of AD and CE is:
Option 1: 4 cm
Option 2: 5 cm
Option 3: 3 cm
Option 4: 2 cm
Correct Answer: 4 cm
Solution : Given: The sides AB, BC, and AC of a $\triangle {ABC}$ are 12 cm, 8 cm, and 10 cm respectively. Tangents from a fixed point outside the circle always have the same length. ⇒ AD = AF, FC = CE, and BE = BD Let the AD and CE be $x$ and $z$ respectively. ⇒ BE = BD = 12 - $x$ Also, CE = BC – BE ⇒ $z=8-12+x$ ⇒ $x-z=4$ ⇒ AD – CE = 4 cm So, the difference between the lengths of AD and CE is 4 cm. Hence, the correct answer is 4 cm.
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Question : A circle is inscribed in a triangle ABC. It touches sides AB, BC, and AC at points R, P, and Q, respectively. If AQ = 2.6 cm, PC = 2.7 cm, and BR = 3 cm, then the perimeter (in cm) of the triangle $\triangle \mathrm{ABC}$ is:
Question : A circle is inscribed in a triangle ABC. It touches sides AB, BC and AC at points R, P and Q, respectively. If AQ = 3.5 cm, PC = 4.5 cm and BR = 7 cm, then the perimeter (in cm) of the triangle $\triangle \mathrm{ABC}$ is:
Question : ABC is a right-angled triangle with AB = 6 cm and BC = 8 cm. A circle with centre O has been inscribed inside $\triangle ABC$. The radius of the circle is:
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