Question

An equilateral triangle has its two vertices P(5, 4) and Q(13, –2), then the radius of its circumcircle is (in units)

Answer 1

SOLUTION

GIVEN

An equilateral triangle has its two vertices P(5, 4) and Q(13, –2)

TO DETERMINE

The radius of its circumcircle (in units)

FORMULA TO BE IMPLEMENTED

For an equilateral triangle with each sides of length a unit, the radius of circumcircle is

$\displaystyle \sf{r = \frac{a}{ \sqrt{3} } \: } \: unit$

EVALUATION

Here it is given that An equilateral triangle has its two vertices P(5, 4) and Q(13, –2)

Then the length of the side PQ

$= \sf{ \sqrt{ {(13 - 5)}^{2} + {( - 2 - 4)}^{2} } \: } \: \: unit$

$= \sf{ \sqrt{ {( 8)}^{2} + {( - 6)}^{2} } \: } \: \: unit$

$= \sf{ \sqrt{64 + 36 }} \: \: unit$

$= \sf{ \sqrt{100 } \: } \: \: unit$

$= \sf{ 10} \: \: unit$

Since the triangle is equilateral

So the length of each side is equal

Therefore the length of each side = a = 10 unit

Hence the radius of the circumcircle

$= \displaystyle \sf{r = \frac{a}{ \sqrt{3} } \: } \: unit$

$\displaystyle \sf{= \frac{10}{ \sqrt{3} } \: } \: unit$

$\displaystyle \sf{= \frac{10 \sqrt{3} }{3 } \: } \: unit$

FINAL ANSWER

$\displaystyle \sf{The \: radius \: of \: the \: circumcircle = \frac{10 \sqrt{3} }{3 } \: } \: unit$

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Answer 2

Answer:

aise question Google per poochhne chahie gadhedi tu to pahle long answer deti hai aur hamen hi bolati hai