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

If $R$ is the radius of the circumcircle of the $\Delta ABC$ and $\Delta$ is its area,then

In $\triangle ABC$,suppose the exradii opposite to angles $A, B$ and $C$ are denoted by $r_1, r_2$ and $r_3$ respectively. If $r_1=2, r_2=3, r_3=6$ and $R$ is the radius of the circumcircle,then the value of $r_1+r_2+r_3-r$ is:

If the lengths of the sides of a triangle are $15, 20, 25$ units,find the circumradius of the triangle. (in $units$)

In an equilateral triangle of side $2\sqrt{3} \text{ cm}$,the circum-radius is ..... $\text{cm}$.

The radius of the incircle of a triangle with sides $18 \, cm$,$24 \, cm$,and $30 \, cm$ is ..... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module