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# ABC is an equilateral triangle of side 4cm. If R, r and h are the circumradius, inradius, and altitude respectively, then $\dfrac{{R + r}}{h}$ is equal to?$\left( a \right)$ 4$\left( b \right)$ 2$\left( c \right)$ 1$\left( d \right)$ 3

Last updated date: 13th Jun 2024
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Hint: In this particular question use the concept that in an equilateral triangle the center of incircle and circumcircle both are same and use the concept that the radius of incircle is the ratio of the area of the equilateral triangle to the semi perimeter of the equilateral triangle, later on in the solution use the concept of Pythagoras theorem, so use these concepts too reach the solution of the question.

Consider the equilateral triangle ABC as shown above.
The side of the triangle is 4 cm (given).
Therefore, AB = BC = CD = 4cm
As we know that the area (A) of an equilateral triangle is given as,
$\Rightarrow A = \dfrac{{\sqrt 3 }}{4}{\left( {{\text{side}}} \right)^2}$ Square units.
So the area of an equilateral triangle is,
$\Rightarrow A = \dfrac{{\sqrt 3 }}{4}{\left( {\text{4}} \right)^2} = 4\sqrt 3 {\text{ c}}{{\text{m}}^2}$.
Now the incircle and the circumcircle of an equilateral triangle are also shown having center O, and radius r, and R cm respectively.
Therefore, OE = r cm and OB = R cm.
Now AD is the altitude of the equilateral triangle having height h cm.
Therefore, BD = DC = $\dfrac{{BC}}{2} = \dfrac{4}{2} = 2$ cm.
Now in triangle ABD apply Pythagoras theorem we have,
$\Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
Now substitute the variables we have,
$\Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}$
$\Rightarrow {\left( {\text{4}} \right)^2} = {\left( {\text{h}} \right)^2} + {\left( {\text{2}} \right)^2}$
$\Rightarrow {h^2} = 16 - 4 = 12$
$\Rightarrow h = 2\sqrt 3$cm.
Now as we know that in an equilateral triangle all angles are equal to 60 degrees.
BO is the bisector of the angle ABC, therefore, $\angle OBD = \dfrac{{\angle ABC}}{2} = \dfrac{{{{60}^o}}}{2} = {30^o}$
Now in triangle OBD,
$\Rightarrow \cos {30^o} = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} = \dfrac{{BD}}{{OB}}$
$\Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{2}{R}$, $\left[ {\because \cos {{30}^o} = \dfrac{{\sqrt 3 }}{2}} \right]$
$\Rightarrow R = \dfrac{4}{{\sqrt 3 }}$ cm.
Now as we know that the radius of incircle is the ratio of the area of the equilateral triangle to the semi perimeter of the equilateral triangle.
Now as we know that the perimeter of any shape is the sum of all the sides.
So the perimeter of an equilateral triangle is the sum of all the three sides,
Therefore, P = AB + BC + CA = 4 + 4 + 4 = 12 cm.
Now the semi perimeter of an equilateral triangle is, $s = \dfrac{P}{2} = \dfrac{{12}}{2} = 6$ cm.
So, the inradius of the equilateral triangle is, $r = \dfrac{A}{s} = \dfrac{{4\sqrt 3 }}{6}$ cm.
Now we have to find out the value of, $\dfrac{{R + r}}{h}$
$\Rightarrow \dfrac{{R + r}}{h} = \dfrac{{\dfrac{4}{{\sqrt 3 }} + \dfrac{{4\sqrt 3 }}{6}}}{{2\sqrt 3 }} = \dfrac{2}{3} + \dfrac{2}{6} = \dfrac{2}{3} + \dfrac{1}{3} = \dfrac{3}{3} = 1$

Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that that the perimeter of any shape is the sum of all the sides, and always recall the formula of an equilateral triangle which is stated above it saves lot of time during calculation, and always recall that in an equilateral triangle all the angles are equal to 60 degrees.