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# Circumscribed circle of the hexagon | PRMO 2018 | Problem 7 Try this beautiful problem from Geometry based on Radius of the circumscribed circle of the hexagon

## Hexagon|PRMO| 2007

A point P in the interior of a regular hexagon is at distance 8,8,16 units from three consecutive vertices of the hexagon, respectively. If r is radius of the circumscribed circle of the hexagon, what is the integer closest to r ?

• $12$
• $14$
• $16$

### Key Concepts

Geometry

Triangle

Hexagon

Answer:$14$

PRMO (2018) Problem 7

Pre College Mathematics

## Try with Hints

Show $\triangle PAB$is similar to $\triangle PFC$.

Can you now finish the problem ..........

The circumradius of a regular hexagon = side of regular hexagon

can you finish the problem........

Note that CF = 2AB, PA = 2PC & PB = 2PF PFC, hence $\triangle PAB$is similar to $\triangle PFC$.Hence $A_1P_1C and B_1P_1F$  are collinear. Let each side of hexagon be equal to x . Let Q & R be foot of altitudes from P to base AB & CF respectively. So R is centre of hexagon

Now

$\frac{1}{3} \times \frac {\sqrt 3 x}{2}$ =$\sqrt{64-\frac{x^2}{4}}$

$\Rightarrow 64-\frac{x^2}{4}$

$\Rightarrow \frac{4 x^2}{12} =64$

$\Rightarrow x=8\sqrt 3$

We know that the circumradius of a regular hexagon = side of regular hexagon

Hence r=8$\sqrt 3$$\approx$ 13.856

Therefore r=14 ( nearest integer)

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Try this beautiful problem from Geometry based on Radius of the circumscribed circle of the hexagon

## Hexagon|PRMO| 2007

A point P in the interior of a regular hexagon is at distance 8,8,16 units from three consecutive vertices of the hexagon, respectively. If r is radius of the circumscribed circle of the hexagon, what is the integer closest to r ?

• $12$
• $14$
• $16$

### Key Concepts

Geometry

Triangle

Hexagon

Answer:$14$

PRMO (2018) Problem 7

Pre College Mathematics

## Try with Hints

Show $\triangle PAB$is similar to $\triangle PFC$.

Can you now finish the problem ..........

The circumradius of a regular hexagon = side of regular hexagon

can you finish the problem........

Note that CF = 2AB, PA = 2PC & PB = 2PF PFC, hence $\triangle PAB$is similar to $\triangle PFC$.Hence $A_1P_1C and B_1P_1F$  are collinear. Let each side of hexagon be equal to x . Let Q & R be foot of altitudes from P to base AB & CF respectively. So R is centre of hexagon

Now

$\frac{1}{3} \times \frac {\sqrt 3 x}{2}$ =$\sqrt{64-\frac{x^2}{4}}$

$\Rightarrow 64-\frac{x^2}{4}$

$\Rightarrow \frac{4 x^2}{12} =64$

$\Rightarrow x=8\sqrt 3$

We know that the circumradius of a regular hexagon = side of regular hexagon

Hence r=8$\sqrt 3$$\approx$ 13.856

Therefore r=14 ( nearest integer)

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