
The in-radius of a triangle whose sides are given by 3 , 5 , 6 is?
a) $\sqrt{\dfrac{8}{7}}$
b) $\sqrt{8}$
c) $\sqrt{7}$
d) $\sqrt{\dfrac{7}{8}}$
Answer
596.7k+ views
Hint: Find out the semiperimeter of the triangle by adding the three sides and diving the sum by 2. Find the area of the triangle by $\Delta= \sqrt{s(s-a)(s-b)(s-c)}$. Then find the in-radius by dividing the area by the semiperimeter .
Complete step-by-step answer:
Given , the three sides of the triangle are 3 , 5 , 6 .
Semiperimeter of the triangle = 7
Area of the triangle = $\Delta= \sqrt{s(s-a)(s-b)(s-c)}=\sqrt{ 7 \times 4 \times 2 \times 1}=2 \sqrt{14}$
Hence , the inradius of the triangle , by formula is r = $\dfrac{\Delta}{s}=\dfrac{2\sqrt{14}}{7}=\sqrt{\dfrac{8}{7}}$
Thus , the required option is a) $\sqrt{\dfrac{8}{7}}$
Note: Inradius is the radius of the circle which is inscribed inside the triangle. Circumradius (R) Circumradius is defined as the radius of that circle which circumscribes (surrounds) the triangle. Students often don’t understand the difference between the inradius and circumradius of a circle.
Complete step-by-step answer:
Given , the three sides of the triangle are 3 , 5 , 6 .
Semiperimeter of the triangle = 7
Area of the triangle = $\Delta= \sqrt{s(s-a)(s-b)(s-c)}=\sqrt{ 7 \times 4 \times 2 \times 1}=2 \sqrt{14}$
Hence , the inradius of the triangle , by formula is r = $\dfrac{\Delta}{s}=\dfrac{2\sqrt{14}}{7}=\sqrt{\dfrac{8}{7}}$
Thus , the required option is a) $\sqrt{\dfrac{8}{7}}$
Note: Inradius is the radius of the circle which is inscribed inside the triangle. Circumradius (R) Circumradius is defined as the radius of that circle which circumscribes (surrounds) the triangle. Students often don’t understand the difference between the inradius and circumradius of a circle.
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