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PSAT Math : Other Polygons

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Example Question #101 : Plane Geometry

Heptagon

Note: Figure NOT drawn to scale.

The perimeter of the above polygon is 225. Also, B = A + C .

Evaluate .

Possible Answers:

$B = 34\frac{1}{6}$

$B = 37 \frac{1}{2}$

Insufficient information exists to answer the question.

$B = 51\frac{1}{4}$

$B = 56\frac{1}{4}$

Correct answer:

$B = 51\frac{1}{4}$

Explanation:

To get the expression equivalent to the perimeter, add the lengths of the sides:

P = A + A + B + B + C + C + 20

Since the perimeter is 225, we can simplify this to

225= 2A + 2B + 2C + 20

and, furthermore, since B = A + C ,

2A + 2B + 2C = 205

2B +2A + 2C = 205

$2B +2\left (A + C \right )= 205$

2B +2B= 205

4B = 205

$B = 51\frac{1}{4}$

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Example Question #1 : How To Find The Length Of A Diagonal Of A Polygon

Regular Octagon ABCDEFGH has sidelength 1.

Give the length of diagonal $\overline{AD}$ .

Possible Answers:

$\frac{2 + \sqrt{2}}{2}$

$\frac{2 + \sqrt{3}}{2}$

$1 + \sqrt{3}$

$1 + \sqrt{2}$

Correct answer:

$1 + \sqrt{2}$

Explanation:

The trick is to construct segments perpendicular to $\overline{AD}$ from and , calling the points of intersection and respectively.

Octagon_1

Each interior angle of a regular octagon measures

$\frac{\left ( 8 - 2\right ) 180^{\circ }}{8} = 135^{\circ }$ ,

and by symmetry, $m \angle HAD = m \angle ADE = 90^{\circ }$ ,

so $m \angle BAX = m \angle CDY= 45^{\circ }$ .

This makes $\Delta BAX$ and $\Delta CDY$ $45^{\circ } -45^{\circ } -90^{\circ }$ triangles.

Since their hypotenuses are sides of the octagon with length 1, then their legs - in particular, $\overline{AX}$ and $\overline{YD}$ - have length $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

Also, since a rectangle was formed when the perpendiculars were drawn, XY = BC = 1 .

The length of diagonal $\overline{AD}$ is

$AD = AX + XY + YD = \frac{\sqrt{2}}{2}+ 1 + \frac{\sqrt{2}}{2} = 1 + \sqrt{2}$ .

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Example Question #421 : Geometry

Regular Polygon ABCDEFGHIJKL (a twelve-sided polygon, or dodecagon) has sidelength 1.

Give the length of diagonal $\overline{AD}$ to the nearest tenth.

Possible Answers:

$1 + \sqrt{2}$

$\frac{2 + \sqrt{2}}{2}$

$1 + \sqrt{3}$

$\frac{2 + \sqrt{3}}{2}$

Correct answer:

$1 + \sqrt{3}$

Explanation:

The trick is to construct segments perpendicular to $\overline{AD}$ from and , calling the points of intersection and respectively.

Dodecagon

Each interior angle of a regular dodecagon measures

$\frac{\left ( 12 - 2\right ) 180^{\circ }}{12} = 150^{\circ }$ .

Since $\overline{BX}$ and $\overline{CY}$ are perpendicular to $\overline{AD}$ , it can be shown via symmetry that they are also perpendicular to $\overline{BC}$ . Therefore,

$\angle ABX$ and $\angle DCY$ both measure $60^{\circ }$

and $\Delta ABX$ and $\Delta DCY$ are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles with long legs $\overline{AX}$ and $\overline{YD}$ . Since their hypotenuses are sides of the dodecagon and therefore have length 1,

$AX = YD = \frac{\sqrt{3}}{2}$ .

Also, since Quadrilateral BCXY is a rectangle, XY = BC = 1 .

The length of diagonal $\overline{AD}$ is $AD = AX + XY + YD = \frac{\sqrt{3}}{2}+ 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}$ .

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Example Question #431 : Geometry

Rectangle_3

Note: Figure NOT drawn to scale.

Refer to the above figure. The white trapezoid is isosceles. Give the perimeter of the blue polygon.

Possible Answers:

$264 + 36 \sqrt{2}$

318

336

$264 + 18 \sqrt{2}$

$264 + 18 \sqrt{3}$

Correct answer:

$264 + 36 \sqrt{2}$

Explanation:

The lower base of this trapezoid is 100 - 64 = 36 units longer than the upper base, and being isosceles, it is symmetrical. As a result, the lower leg of the right triangle at bottom is half this difference, or 18, which is the same as the upper leg. That makes the right triangle isosceles, and, therefore, a 45-45-90 triangle. Subsequently, the hypotenuse, which is one leg of the trapezoid, has length $\sqrt{2}$ times a leg of the triangle, or $18 \sqrt{2}$ .

The blue polygon has two sides that are the legs of this isosceles trapezoid, both of which have length $18 \sqrt{2}$ ; its other three sides are of length 64, 50, 100, and 50. Add them:

$P = 100 + 64 + 50 + 50 + 18 \sqrt{2}+ 18 \sqrt{2} = 264 + 36 \sqrt{2}$

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PSAT Math : Other Polygons

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #101 : Plane Geometry

Example Question #1 : How To Find The Length Of A Diagonal Of A Polygon

Example Question #421 : Geometry

Example Question #431 : Geometry

All PSAT Math Resources

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