In Delta ABC,AB = AC. Let P_1 denote the inci | vedclass

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

Question

(D) Let $\alpha$ be half of the angle $\angle A$. Let $C_1$ and $C_2$ be the centers of circles $P_2$ and $P_1$ with radii $r_1 = 1$ and $r_2 = 2$ res

Vedclass · Accepted Answer

$16\sqrt{2}$

Vedclass · Answer

(D) Let $\alpha$ be half of the angle $\angle A$. Let $C_1$ and $C_2$ be the centers of circles $P_2$ and $P_1$ with radii $r_1 = 1$ and $r_2 = 2$ respectively.In the right triangle formed by the center $C_1$,the vertex $A$,and the point of tangency on $AB$,we have $\sin \alpha = \frac{r_1}{AC_1} = \frac{1}{AC_1}$.Since the circles are tangent,$AC_2 = AC_1 + r_1 + r_2 = AC_1 + 1 + 2 = AC_1 + 3$.Also,$\sin \alpha = \frac{r_2}{AC_2} = \frac{2}{AC_2}$.Equating the two expressions for $\sin \alpha$: $\frac{1}{AC_1} = \frac{2}{AC_1 + 3} \implies AC_1 + 3 = 2AC_1 \implies AC_1 = 3$.Thus,$\sin \alpha = \frac{1}{3}$.The height of the triangle $AD = AC_2 + r_2 = (AC_1 + 3) + 2 = 6 + 2 = 8$.In $\Delta ABD$,$	an \alpha = \frac{BD}{AD} = \frac{BD}{8}$.Since $\sin \alpha = \frac{1}{3}$,$\cos \alpha = \sqrt{1 - (\frac{1}{3})^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.Then $	an \alpha = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}}$.So,$BD = 8 	imes \frac{1}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The base $BC = 2 	imes BD = 4\sqrt{2}$.Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4\sqrt{2} 	imes 8 = 16\sqrt{2}$.

In $\Delta ABC$,$AB = AC$. Let $P_1$ denote the incircle of $\Delta ABC$. Circle $P_2$ is tangent to sides $AB$,$AC$ and to circle $P_1$. If the radii of circles $P_1$ and $P_2$ are $2$ and $1$ respectively,then the area of $\Delta ABC$ is:

Similar Questions

In a triangle $ABC$, $a=5$, $b=4$ and $\tan \frac{C}{2}=\sqrt{\frac{7}{9}}$, then its inradius $r=$

If the vertices of a triangle are $A(0, -6)$,$B(-6, 0)$,and $C(1, 1)$,find the coordinates of the excenter opposite to vertex $A$.

In a triangle $ABC$,if $r_1=3, r_2=4, r_3=6$,then $b=$

In a $\triangle ABC$,which of the following formulae are correct?
$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$II. r_1 = (s-a) \tan \frac{A}{2}$
$III. r_3 = \frac{\Delta}{s-c}$

For a triangle with side lengths $6$,$5$,and $9$,find the inradius of the triangle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module