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# The base of an isosceles triangle is 6 cm and one of the equal sides is 12 cm. The radius of the circle through the vertices of the triangle isA) $\dfrac{{7\sqrt {15} }}{5}$B) $4\sqrt 3$C) $3\sqrt 5$D) $6\sqrt 3$E) None of these

Last updated date: 20th Jun 2024
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Hint: Here first we will let the radius of the circle be $r$ and the altitude of the triangle to be $h$ such that it is equal to $x+r$ illustrated in the figure. Then we will apply the Pythagoras theorem to get the value of $r$.
The Pythagoras theorem states that the square of the hypotenuse is equal to the sum of squares of base and height of a right-angled triangle.
${\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {altitude} \right)^2}$

Complete step by step solution:
Let the radius of the circle
$OA = OC = r$
Let the height/altitude of the isosceles $\Delta ABC$ be $h$ such that:
$h = r + x$……………………………..(1)
Now since it is given that the base of the triangle is 6cm
Hence,
$AB = 6cm$
which implies that:
$AM = MB = 3cm$
Also, it is given that one of the equal sides of the triangle is 12 cm
Therefore,
$AC = 12cm$
Now since $\Delta AOM$ is another right angled triangle and r is the hypotenuse of the triangle
Therefore, applying Pythagoras theorem in $\Delta AOM$
According to Pythagoras theorem:
${\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {altitude} \right)^2}$
Here,
$AC$ is the hypotenuse,
$AM$is the base,
$MC$ is the altitude.
Putting in the values we get:-
${\left( {AC} \right)^2} = {\left( {AM} \right)^2} + {\left( {MC} \right)^2}$
${\left( {12} \right)^2} = {\left( 3 \right)^2} + {\left( h \right)^2}$
$144 = 9 + {h^2}$
${h^2} = 144 - 9$
${h^2} = 135$
$h = \sqrt {135}$
Now, applying Pythagoras theorem in $\Delta AOM$
$OA$ is hypotenuse,
$AM$ is base,
$OM$ is the altitude.
Therefore putting in the values in the formula of Pythagoras theorem we get:-
${r^2} = {3^2} + {x^2}$
From equation 1 we get:-
$x = h - r$
Putting in this value in above equation we get:-
${r^2} = {3^2} + {\left( {h - r} \right)^2}$
Now applying the following identity:
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
We get:-
${r^2} = 9 + {h^2} + {r^2} - 2hr$
$2hr = 9 + {h^2}$
Solving for $r$,
$r = \dfrac{{9 + {h^2}}}{{2h}}$
Solving it further and putting the known values we get:-
$\Rightarrow r = \dfrac{{9 + {{\left( {\sqrt {135} } \right)}^2}}}{{2\left( {\sqrt {135} } \right)}}$
On simplifying the above values,
$\Rightarrow r = \dfrac{{9 + 135}}{{2\sqrt {135} }}$
On further simplifications, we get
$\Rightarrow r = \dfrac{{144}}{{2\sqrt {135} }}$
Now rationalizing the value of r by multiplying both the numerator and denominator by $\sqrt {135}$ in order to make the denominator a whole number we get:-
$\Rightarrow r = \dfrac{{144}}{{2\sqrt {135} }} \times \dfrac{{\sqrt {135} }}{{\sqrt {135} }}$
Simplifying it further we get:-
$\Rightarrow r = \dfrac{{144\sqrt {135} }}{{2 \times 135}}$
$\Rightarrow r = \dfrac{{144\sqrt {135} }}{{270}}$
On further simplification,
$\Rightarrow r = \dfrac{{8 \times 3\sqrt {15} }}{{15}}$
$r = \dfrac{{8\sqrt {15} }}{5}$

$\therefore$ The radius of the circle is $\dfrac{{8\sqrt {15} }}{5}cm$. Hence, option E is the correct option.

Note:
In an isosceles triangle, two sides are equal and the corresponding angles opposite to them are also equal.
Also, the altitude from a vertex to the opposite side bisects it into two halves.
The student can also consider $\Delta BOM$ instead of $\Delta AOM$ and then apply the Pythagoras theorem to get the value of $r$.