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Question: An isosceles triangle with base 24 and legs of 15 is inscribed in a circle. Find the radius. A.7 ...

An isosceles triangle with base 24 and legs of 15 is inscribed in a circle. Find the radius.
A.7
B.121212\dfrac{1}{2}
C.25
D.251225\dfrac{1}{2}
E.Cannot be determined

Explanation

Solution

We will calculate the radius of the circumcircle by drawing a perpendicular to the base of the triangle, say x, and then from the figure, in triangle ADC, we will use Pythagoras theorem: (hypotenuse)2=(base)2+(perpendicular)2{\left( {{\text{hypotenuse}}} \right)^2} = {\left( {{\text{base}}} \right)^2} + {\left( {{\text{perpendicular}}} \right)^2}to calculate the value of r.

Complete step-by-step answer:
We are given that the base of an isosceles triangle is 24 and its sides are of 15 and it is inscribed in a circle.
Let us draw a perpendicular from the centre of the circle to the base of the triangle.
The figure of this setup:

From the figure, we can see that OA is the radius of the circumcircle. Since D is the midpoint of the base BC (ABC being an isosceles triangle), then DC = BD =242\dfrac{{24}}{2}= 12
Now, using the Pythagoras theorem in triangle ADC, we get
\Rightarrow (hypotenuse)2=(base)2+(perpendicular)2{\left( {{\text{hypotenuse}}} \right)^2} = {\left( {{\text{base}}} \right)^2} + {\left( {{\text{perpendicular}}} \right)^2}= (AC)2=(DC)2+(AD)2{\left( {AC} \right)^2} = {\left( {DC} \right)^2} + {\left( {AD} \right)^2}
Putting their values, we get
\Rightarrow (15)2=(12)2+(r+x)2{\left( {15} \right)^2} = {\left( {12} \right)^2} + {\left( {r + x} \right)^2}
225=144+(r+x)2 225144=(r+x)2 81=(r+x)2  \Rightarrow 225 = 144 + {\left( {r + x} \right)^2} \\\ \Rightarrow 225 - 144 = {\left( {r + x} \right)^2} \\\ \Rightarrow 81 = {\left( {r + x} \right)^2} \\\
r+x=9\Rightarrow r + x = 9 equation (1)
Now, in right triangle ODB, using the Pythagoras theorem, we get
OB2=BD2+OD2\Rightarrow O{B^2} = B{D^2} + O{D^2}
r2=122+x2 r2x2=144 (r+x)(rx)=144  \Rightarrow {r^2} = {12^2} + {x^2} \\\ \Rightarrow {r^2} - {x^2} = 144 \\\ \Rightarrow \left( {r + x} \right)\left( {r - x} \right) = 144 \\\
Substituting the value of r + x, we get
9(rx)=144\Rightarrow 9\left( {r - x} \right) = 144
(rx)=16\Rightarrow \left( {r - x} \right) = 16 equation (2)
Now, adding both the equations (1) and (2), we get
(r+x)+(rx)=16+9 2r=25 r=252=1212  \Rightarrow \left( {r + x} \right) + \left( {r - x} \right) = 16 + 9 \\\ \Rightarrow 2r = 25 \\\ \Rightarrow r = \dfrac{{25}}{2} = 12\dfrac{1}{2} \\\
We obtained the value of the radius of the circumcircle as 121212\dfrac{1}{2} .
Hence, option(B) is correct.

Note: In this question, you may get confused with the method used because we have to construct a perpendicular in order to obtain the value of the radius of the circle using the Pythagoras theorem twice. You can also solve this question by putting the value of x in terms of r from (1) in equation (2).