Let (5, \fraca4) be the circumcenter of a triangle with vertices A(a,-2), B(a, 6) and C(\fraca4,-2) . Let \alpha denote the circumradius, \beta denote the area and \gamma denote the perimeter of the triangle. Then \alpha+\beta+\gamma is

Properties of Triangle PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q33 with Solution

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Correct answer: C

Use the fact that the circumcenter is equidistant from all three vertices.

Let the circumcenter be $O\left(5,\frac a4\right).$

Given vertices are $A(a,-2),\quad B(a,6),\quad C\left(\frac a4,-2\right).$

Since $O$ is the circumcenter, $OA=OB=OC.$

First observe the triangle shape.

Points $A(a,-2)$ and $B(a,6)$ have the same $x$ -coordinate, so $AB$ is vertical.

Points $A(a,-2)$ and $C\left(\frac a4,-2\right)$ have the same $y$ -coordinate, so $AC$ is horizontal.

Hence, $AB \perp AC,$ so triangle $ABC$ is right-angled at $A$ .

Find $a$ using circumcenter property.

For a right triangle, the circumcenter is the midpoint of the hypotenuse $BC$ .

Midpoint of $BC$ is $\left(\frac{a+\frac a4}{2},\frac{6+(-2)}{2}\right)=\left(\frac{5a}{8},2\right).$

This is given as $\left(5,\frac a4\right).$

So we compare coordinates:

From $x$ -coordinates: $\frac{5a}{8}=5 \implies a=8.$
From $y$ -coordinates: $\frac a4=2 \implies a=8,$ which matches.

Thus, $a=8.$

Write the actual coordinates.

Substituting $a=8$ : $A(8,-2),\quad B(8,6),\quad C(2,-2).$

Find side lengths.

$AB$ is vertical: $AB=|6-(-2)|=8.$
$AC$ is horizontal: $AC=|8-2|=6.$
$BC$ by distance formula:

\sqrt{6^2+8^2}=\sqrt{100}=10.$$ 6. **Find circumradius $\alpha$.** For a right triangle, circumradius is half the hypotenuse: $$\alpha=\frac{BC}{2}=\frac{10}{2}=5.$$ 7. **Find area $\beta$.** Since the triangle is right-angled at $A$, $$\beta=\frac12 \cdot AB \cdot AC = \frac12 \cdot 8 \cdot 6 = 24.$$ 8. **Find perimeter $\gamma$.** $$\gamma=AB+BC+CA=8+10+6=24.$$ 9. **Compute the required sum.** $$\alpha+\beta+\gamma=5+24+24=53.$$ Therefore, the correct option is $$\boxed{53}.$$

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View written solutionFree

Correct answer: C

Use the fact that the circumcenter is equidistant from all three vertices.

Let the circumcenter be $O\left(5,\frac a4\right).$

Given vertices are $A(a,-2),\quad B(a,6),\quad C\left(\frac a4,-2\right).$

Since $O$ is the circumcenter, $OA=OB=OC.$

First observe the triangle shape.

Points $A(a,-2)$ and $B(a,6)$ have the same $x$ -coordinate, so $AB$ is vertical.

Points $A(a,-2)$ and $C\left(\frac a4,-2\right)$ have the same $y$ -coordinate, so $AC$ is horizontal.

Hence, $AB \perp AC,$ so triangle $ABC$ is right-angled at $A$ .

Find $a$ using circumcenter property.

For a right triangle, the circumcenter is the midpoint of the hypotenuse $BC$ .

Midpoint of $BC$ is $\left(\frac{a+\frac a4}{2},\frac{6+(-2)}{2}\right)=\left(\frac{5a}{8},2\right).$

This is given as $\left(5,\frac a4\right).$

So we compare coordinates:

From $x$ -coordinates: $\frac{5a}{8}=5 \implies a=8.$
From $y$ -coordinates: $\frac a4=2 \implies a=8,$ which matches.

Thus, $a=8.$

Write the actual coordinates.

Substituting $a=8$ : $A(8,-2),\quad B(8,6),\quad C(2,-2).$

Find side lengths.

$AB$ is vertical: $AB=|6-(-2)|=8.$
$AC$ is horizontal: $AC=|8-2|=6.$
$BC$ by distance formula:

\sqrt{6^2+8^2}=\sqrt{100}=10.$$ 6. **Find circumradius $\alpha$.** For a right triangle, circumradius is half the hypotenuse: $$\alpha=\frac{BC}{2}=\frac{10}{2}=5.$$ 7. **Find area $\beta$.** Since the triangle is right-angled at $A$, $$\beta=\frac12 \cdot AB \cdot AC = \frac12 \cdot 8 \cdot 6 = 24.$$ 8. **Find perimeter $\gamma$.** $$\gamma=AB+BC+CA=8+10+6=24.$$ 9. **Compute the required sum.** $$\alpha+\beta+\gamma=5+24+24=53.$$ Therefore, the correct option is $$\boxed{53}.$$

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