The lengths of the sides of a triangle are 13 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the sides be $a = 13, b = 14, c = 15$. Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$. Area of the triangle $\Delta = \sqrt{s

Vedclass · Accepted Answer

$61$

Vedclass · Answer

(C) Let the sides be $a = 13, b = 14, c = 15$. Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$. Area of the triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$. Circumradius $R = \frac{abc}{4\Delta} = \frac{13 	imes 14 	imes 15}{4 	imes 84} = \frac{2730}{336} = \frac{65}{8}$. Inradius $r = \frac{\Delta}{s} = \frac{84}{21} = 4$. Now,$8R - r = 8 	imes \left(\frac{65}{8}ight) - 4 = 65 - 4 = 61$.

The lengths of the sides of a triangle are $13, 14, \text{ and } 15$. If $R$ and $r$ respectively denote the circumradius and inradius of this triangle,then $8R - r = $

Similar Questions

If in a triangle $\sin A : \sin C = \sin (A - B) : \sin (B - C)$,then $a^2, b^2, c^2$:

In a triangle,the lengths of the two larger sides are $10 \, cm$ and $9 \, cm$ respectively. If the angles of the triangle are in $A.P.$,then the length of the third side in $cm$ can be

If in a triangle $ABC$,side $a = (\sqrt{3} + 1) \text{ cm}$,$\angle B = 30^\circ$,and $\angle C = 45^\circ$,then the area of the triangle is:

In a triangle $ABC$,with usual notations,if $3b = a + c$,then $\cot \frac{A}{2} \cdot \cot \frac{C}{2} = $

In a $\triangle ABC$,if $\tan(A/2)$,$\tan(B/2)$,and $\tan(C/2)$ are in Arithmetic Progression,then which of the following options is always correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module