1. $$\triangle ABC$$ is a right angled triangle with $$AB=6 cm$$, $$BC = 8 cm$$. O is the in-centre of the triangle. The radius of the in-circle is:
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By: anil on 05 May 2019 03.31 am
Let the inradius of the triangle be $$r$$ cm In right $$ riangle$$ ABC, => $$(AC)^=(AB)^2+(BC)^2$$ => $$(AC)^=(6)^2+(8)^2$$
=> $$(AC)^2=36+64=100$$ => $$AC=sqrt{100}=10$$ cm Area of triangle = $$ riangle=r imes s$$, where $$r$$ is inradius and $$s$$ is semi-perimeter. => Area = $$ riangle=frac{1}{2} imes8 imes6=24$$ $$cm^2$$ Semi-perimeter = $$s=frac{(10+8+6)}{2}=frac{24}{2}=12$$ cm $$ herefore$$ Inradius of triangle = $$r=frac{ riangle}{s}=frac{24}{12}=2$$ cm => Ans - (C)
By: anil on 05 May 2019 03.31 am
Let the inradius of the triangle be $$r$$ cm In right $$ riangle$$ ABC, => $$(AC)^=(AB)^2+(BC)^2$$ => $$(AC)^=(6)^2+(8)^2$$
=> $$(AC)^2=36+64=100$$ => $$AC=sqrt{100}=10$$ cm Area of triangle = $$ riangle=r imes s$$, where $$r$$ is inradius and $$s$$ is semi-perimeter. => Area = $$ riangle=frac{1}{2} imes8 imes6=24$$ $$cm^2$$ Semi-perimeter = $$s=frac{(10+8+6)}{2}=frac{24}{2}=12$$ cm $$ herefore$$ Inradius of triangle = $$r=frac{ riangle}{s}=frac{24}{12}=2$$ cm => Ans - (C)
By: anil on 05 May 2019 03.31 am
Let the inradius of the triangle be $$r$$ cm In right $$ riangle$$ ABC, => $$(AC)^=(AB)^2+(BC)^2$$ => $$(AC)^=(6)^2+(8)^2$$
=> $$(AC)^2=36+64=100$$ => $$AC=sqrt{100}=10$$ cm Area of triangle = $$ riangle=r imes s$$, where $$r$$ is inradius and $$s$$ is semi-perimeter. => Area = $$ riangle=frac{1}{2} imes8 imes6=24$$ $$cm^2$$ Semi-perimeter = $$s=frac{(10+8+6)}{2}=frac{24}{2}=12$$ cm $$ herefore$$ Inradius of triangle = $$r=frac{ riangle}{s}=frac{24}{12}=2$$ cm => Ans - (C)
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=> $$(AC)^2=36+64=100$$ => $$AC=sqrt{100}=10$$ cm Area of triangle = $$ riangle=r imes s$$, where $$r$$ is inradius and $$s$$ is semi-perimeter. => Area = $$ riangle=frac{1}{2} imes8 imes6=24$$ $$cm^2$$ Semi-perimeter = $$s=frac{(10+8+6)}{2}=frac{24}{2}=12$$ cm $$ herefore$$ Inradius of triangle = $$r=frac{ riangle}{s}=frac{24}{12}=2$$ cm => Ans - (C)