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Difficulty: Very hardPythagorean Theorem and Special Right Triangles

A regular hexagon ABCDEFABCDEF has a side length of 88 centimeters. A point PP lies on the side CDCD such that the ratio of the length of CPCP to the length of PDPD is 1:31:3. What is the length, in centimeters, of the segment APAP?

Answer: 14 cm

Answer

The length of the segment APAP is 1414 centimeters.
The correct answer is 1414. Dropping a perpendicular from PP to the main diagonal ADAD creates a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle △PHD\triangle PHD with hypotenuse PD=6PD = 6. The legs are DH=3DH = 3 and PH=33PH = 3\sqrt{3}. This leaves AH=13AH = 13. Applying the Pythagorean Theorem to the right triangle △AHP\triangle AHP with legs 1313 and 333\sqrt{3} yields AP=132+(33)2=14AP = \sqrt{13^2 + (3\sqrt{3})^2} = 14.

Step-by-Step Solution

1
Determine the length of the main diagonal ADAD of the regular hexagon.
AD=16AD = 16 cm
In a regular hexagon with side length ss, the main diagonal connecting opposite vertices has a length of 2s2s. Given s=8s = 8, we find AD=2×8=16AD = 2 \times 8 = 16.
2
Calculate the length of the segment PDPD on the side CDCD.
PD=6PD = 6 cm
The point PP divides the side CDCD of length 88 in the ratio CP:PD=1:3CP:PD = 1:3. Thus, PD=31+3×8=6PD = \frac{3}{1+3} \times 8 = 6.
3
Identify the angles and type of triangle formed by dropping a perpendicular from PP to diagonal ADAD.
△PHD\triangle PHD is a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle.
The diagonal ADAD bisects the interior angle ∠CDE=120∘\angle CDE = 120^\circ of the regular hexagon, making ∠ADC=60∘\angle ADC = 60^\circ. Since PH⊥ADPH \perp AD, the triangle △PHD\triangle PHD has angles 90∘90^\circ, 60∘60^\circ, and 30∘30^\circ.
4
Find the lengths of the legs DHDH and PHPH of the special right triangle △PHD\triangle PHD.
DH=3DH = 3 cm and PH=33PH = 3\sqrt{3} cm
Using the ratios of a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle with hypotenuse PD=6PD = 6, the leg adjacent to the 60∘60^\circ angle is DH=6cos⁡(60∘)=3DH = 6 \cos(60^\circ) = 3, and the leg opposite to the 60∘60^\circ angle is PH=6sin⁡(60∘)=33PH = 6 \sin(60^\circ) = 3\sqrt{3}.
5
Calculate the length of the segment AHAH.
AH=13AH = 13 cm
Since HH lies on the diagonal ADAD, we subtract the length of DHDH from the total length of the diagonal: AH=AD−DH=16−3=13AH = AD - DH = 16 - 3 = 13.
6
Apply the Pythagorean Theorem to the right triangle △AHP\triangle AHP to find the length of APAP.
AP=14AP = 14 cm
In the right triangle △AHP\triangle AHP with legs AH=13AH = 13 and PH=33PH = 3\sqrt{3}, the hypotenuse is AP=AH2+PH2=132+(33)2=169+27=196=14AP = \sqrt{AH^2 + PH^2} = \sqrt{13^2 + (3\sqrt{3})^2} = \sqrt{169 + 27} = \sqrt{196} = 14.

Key Concept

Applying special right triangle ratios and the Pythagorean Theorem in multi-step geometric figures.
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Topics

Pythagorean Theorem and Special Right Triangles

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