In Delta ABC, the lengths of sides AC and AB | vedclass

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(D) The area of $\Delta ABC$ is given by $\Delta = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A = 30 \, cm^{2}.$Substituting the values,we get $\frac{1}

Vedclass · Accepted Answer

$15$

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(D) The area of $\Delta ABC$ is given by $\Delta = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A = 30 \, cm^{2}.$Substituting the values,we get $\frac{1}{2} \cdot 5 \cdot 12 \cdot \sin A = 30$ $\Rightarrow 30 \sin A = 30$ $\Rightarrow \sin A = 1.$Thus,$A = 90^{\circ},$ which means $\Delta ABC$ is a right-angled triangle with the hypotenuse $BC = \sqrt{AB^{2} + AC^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm.$The circumradius $R$ of a right-angled triangle is half of the hypotenuse,so $R = \frac{BC}{2} = \frac{13}{2} = 6.5 \, cm.$The inradius $r$ is given by $r = \frac{\Delta}{s},$ where $s$ is the semi-perimeter.$s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \, cm.$So,$r = \frac{30}{15} = 2 \, cm.$The value of $2R + r = 2(6.5) + 2 = 13 + 2 = 15 \, cm.$

In $\Delta ABC,$ the lengths of sides $AC$ and $AB$ are $12 \, cm$ and $5 \, cm,$ respectively. If the area of $\Delta ABC$ is $30 \, cm^{2}$ and $R$ and $r$ are respectively the radii of the circumcircle and incircle of $\Delta ABC,$ then the value of $2R + r$ (in $cm$) is equal to ....... .

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