In $△ A B C$ , $¯ ¯¯¯¯¯¯ ¯ A C$ $= 24 c m$ , $¯ ¯¯¯¯¯¯ ¯ B C$ $= 10 c m$ , and $¯ ¯¯¯¯¯¯ ¯ A B$ $= 26 c m$ . The radius of the inscribed circle is:

26 c m

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4 c m

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13 c m

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8 c m

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Solution

The correct option is B $4 c m$
Given
sides are,24,10 and26cm
Semiperimeter s= $\frac{a + b + c}{2} = \frac{24 + 10 + 26}{2} = 30$

Area of triangle ABC= $\sqrt{s (s - a) (s - b) (s - c)} = \sqrt{30 \times 6 \times 20 \times 4} = 30 \times 4 = 120 c m^{2}$

Raius of inscribed circle= $\frac{A R E A}{s e m i p e r i m e t e r} = \frac{120}{30}$ =4cm

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In △ABC, ¯¯¯¯¯¯¯¯AC =24cm, ¯¯¯¯¯¯¯¯BC =10cm, and ¯¯¯¯¯¯¯¯AB =26cm. The radius of the inscribed circle is:

The correct option is B 4cmGiven sides are,24,10 and26cmSemiperimeter s=a+b+c2=24+10+262=30Area of triangle ABC=√s(s−a)(s−b)(s−c)=√30×6×20×4=30×4=120cm2Raius of inscribed circle=AREAsemiperimeter=12030=4cm

In $△ A B C$ , $¯ ¯¯¯¯¯¯ ¯ A C$ $= 24 c m$ , $¯ ¯¯¯¯¯¯ ¯ B C$ $= 10 c m$ , and $¯ ¯¯¯¯¯¯ ¯ A B$ $= 26 c m$ . The radius of the inscribed circle is: