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ISEE Upper Level Math : Plane Geometry

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Example Question #9 : How To Find An Angle

A line intersects parallel lines and . $\angle 1$ and $\angle 2$ are corresponding angles; $\angle 1$ and $\angle 3$ are same side interior angles.

$m \angle 1 = \left (3x+2y \right )^{\circ }$

$m \angle 2 = \left ( 4x+21\right )^{\circ }$

$m \angle 3 = \left ( 2y-27\right )^{\circ}$

Evaluate x+y .

Possible Answers:

x+y = 120

x+y = 60

x+y = 30

x+y = 45

x+y = 90

Correct answer:

x+y = 60

Explanation:

When a transversal such as crosses two parallel lines, two corresponding angles - angles in the same relative position to their respective lines - are congruent. Therefore,

3x+2y = 4x+21

3x+2y- 3x - 21 = 4x+21 - 3x - 21

x = 2y - 21

Two same-side interior angles are supplementary - that is, their angle measures total 180 - so

3x+2y + 2y-27 = 180

3x+4y-27 = 180

3x+4y= 207

We can solve this system by the substitution method as follows:

3( 2y - 21)+4y= 207

6y-63+4y= 207

10y-63= 207

10y = 270

y = 27

Backsolve:

x = 2y - 21

x = 2 (27)- 21 = 54-21 = 33

x+y = 27+33 = 60 , which is the correct response.

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Example Question #1 : How To Find An Angle

Vertical_angles

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the measure of $\angle 1$ .

Possible Answers:

$88^{\circ }$

$128^{\circ }$

$144^{\circ }$

$96^{\circ }$

$120 ^{\circ }$

Correct answer:

$120 ^{\circ }$

Explanation:

The top and bottom angles, being vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure, so

2x+y = 2y ,

or, simplified,

2x+y - y= 2y - y

y = 2x

The right and bottom angles form a linear pair, so their degree measures total 180. That is,

2y+ 8x = 180

Substitute for :

2(2x)+ 8x = 180

4x+8x = 180

12x= 180

x = 15

The left and right angles, being vertical angles, have the same measure, so, since the right angle measures $\left (8x \right )^{\circ } =\left (8 \cdot 15 \right )^{\circ } = 120 ^{\circ }$ , this is also the measure of the left angle, $\angle 1$ .

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Example Question #11 : Lines

Thingy

Figure NOT drawn to scale

The above figure shows Trapezoid TRAP , with $\overline{RT}$ and $\overline{RA }$ tangent to the circle. $m \angle T = 106 ^{\circ }$ ; evaluate $m \angle A$ .

Possible Answers:

$138^{\circ }$

$128^{\circ }$

$158^{\circ }$

$148^{\circ }$

Correct answer:

$148^{\circ }$

Explanation:

By the Same-Side Interior Angle Theorem, since $\overline{TR} || \overline{AP}$ , $\angle T$ and $\angle P$ are supplementary - that is, their degree measures total $180 ^{\circ }$ . Therefore,

$m \angle P + m \angle T = 180 ^{\circ }$

$m \angle P = 180 ^{\circ } - m \angle T$

$m \angle P = 180 ^{\circ } - 106 ^{\circ } = 74^{\circ }$

$\angle P$ is an inscribed angle, so the arc it intercepts, $\overarc{TA}$ , has twice its degree measure;

$m \overarc{TA} = 2 \cdot m \angle P =2 \cdot 74^{\circ } = 148 ^{\circ }$ .

The corresponding major arc, $\overarc{TPA}$ , has as its measure

$m \overarc{TPA} = 360 ^{\circ }- m \overarc{TA} = 360 ^{\circ }- 148 ^{\circ }= 212 ^{\circ }$

The measure of an angle formed by two tangents to a circle is equal to half the difference of those of its intercepted arcs:

$m \angle R = \frac{1}{2} \left (m \overarc{TPA} - m \overarc{TA} \right )$

$m \angle R = \frac{1}{2} \left (212 ^{\circ } -148 ^{\circ } \right )= \frac{1}{2} \cdot 64 ^{\circ } = 32^{\circ }$

Again, by the Same-Side Interior Angles Theorem, $\angle R$ and $\angle A$ are supplementary, so

$m \angle A + m \angle R= 180 ^{\circ }$

$m \angle A = 180 ^{\circ } - m \angle R$

$m \angle A = 180 ^{\circ } - 32 ^{\circ } = 148^{\circ }$

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Example Question #1 : Triangles

Two sides of an isosceles triangle have lengths 3 feet and 4 feet. Which of the following could be the length of the third side?

Possible Answers:

$54 \textrm{ in}$

$32 \textrm{ in}$

$36 \textrm{ in}$

$42 \textrm{ in}$

$40 \textrm{ in}$

Correct answer:

$36 \textrm{ in}$

Explanation:

An isosceles triangle, by definition, has two sides of equal length. Having the third side measure either 3 feet or 4 feet would make the triangle meet this criterion.

3 feet is equal to $3 \times 12 = 36$ inches, and 4 feet is equal to $4 \times 12 = 48$ inches. We choose 36 inches, since that, but not 48 inches, is a choice.

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Example Question #41 : Isee Upper Level (Grades 9 12) Mathematics Achievement

The triangles are similar. Solve for .

Question_12

Possible Answers:

x=4

x=2

x=3

x=1

Correct answer:

x=2

Explanation:

Because the triangles are similar, proportions can be used to solve for the length of the side:

$\frac{9}{3}=\frac{6}{x}$

Cross-multiply:

9x=18

x=2

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Example Question #1 : Solve Simple Equations For An Unknown Angle In A Figure: Ccss.Math.Content.7.G.B.5

One of the base angles of an isosceles triangle is $42^{\circ}$ . Give the measure of the vertex angle.

Possible Answers:

$98 ^{\circ}$

$96 ^{\circ}$

$90^{\circ}$

$86 ^{\circ}$

$100^{\circ}$

Correct answer:

$96 ^{\circ}$

Explanation:

The base angles of an isosceles triangle are always equal. Therefore both base angles are $42^{\circ}$ .

Let the measure of the third angle. Since the sum of the angles of a triangle is $180^{\circ}$ , we can solve accordingly:

$42+42+x=180\Rightarrow x=180-84\Rightarrow x=96^{\circ}$

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Example Question #1 : Right Triangles

A right triangle has a hypotenuse of 10 and a side of 6. What is the missing side?

Possible Answers:

Correct answer:

Explanation:

To find the missing side, use the Pythagorean Theorem (a^2+b^2=c^2) . Plug in (remember c is always the hypotenuse!) so that 6^2+b^2=10^2 . Simplify and you get 36+b^2=100. Subtract 36 from both sides so that you get b^2=64. Take the square root of both sides. B is 8.

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Example Question #1 : How To Find The Length Of The Side Of A Right Triangle

Right_triangle

Refer to the above diagram. Which of the following quadratic equations would yield the value of as a solution?

Possible Answers:

$x^{2}+6x+409 = 0$

$x^{2}+6x-391 = 0$

$2x^{2}+6x-391 = 0$

$2x^{2}+6x+409 = 0$

$2x^{2}+6x-11 = 0$

Correct answer:

$2x^{2}+6x-391 = 0$

Explanation:

By the Pythagorean Theorem,

$x^{2}+ (x+3)^{2} = 20^{2}$

$x^{2}+ x^{2}+6x+9= 400$

$2x^{2}+6x-391 = 0$

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Example Question #1 : Triangles

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Which of the following quadratic equations would yield the value of as a solution?

Possible Answers:

$2x^{2} +4x-24= 0$

$x^{2} -4x+74= 0$

$x^{2} +4x+74= 0$

$x^{2} -4x-24= 0$

$2x^{2} -4x-24= 0$

Correct answer:

$x^{2} -4x-24= 0$

Explanation:

By the Pythagorean Theorem,

$x^{2}+ (x+5)^{2} = (x+7)^{2}$

$x^{2}+ x^{2} +10x+25 = x^{2} +14x +49$

$2x^{2} +10x+25 = x^{2} +14x +49$

$x^{2} -4x-24= 0$

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Example Question #41 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram.

AF = 6 , FD = 8

Find the length of $\overline{BC}$ .

Possible Answers:

$16\frac{2}{3}$

$22\frac{2}{9}$

$33\frac{1}{3}$

Correct answer:

$22\frac{2}{9}$

Explanation:

First, find .

Since $\overline{DF }$ is an altitude of right $\Delta ADB$ to its hypotenuse,

$\frac{FB}{FD}= \frac{FD}{AF}$

$\frac{FB}{8}= \frac{8}{6}$

$FB= \frac{8}{6} \cdot 8 = 10 \frac{2}{3}$

$AB = AF + FB = 6 + 10 \frac{2}{3} = 16 \frac{2}{3}$

$\Delta AFD \sim \Delta ABC$ by the Angle-Angle Postulate, so

$\frac{BC}{FD} = \frac{AB}{AF}$

$\frac{BC}{8} = \frac{16 \frac{2}{3}}{6}$

$BC = 16 \frac{2}{3}\cdot 8 \div 6 = 22\frac{2}{9}$

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ISEE Upper Level Math : Plane Geometry

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #9 : How To Find An Angle

Example Question #1 : How To Find An Angle

Example Question #11 : Lines

Example Question #1 : Triangles

Example Question #41 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #1 : Solve Simple Equations For An Unknown Angle In A Figure: Ccss.Math.Content.7.G.B.5

Example Question #1 : Right Triangles

Example Question #1 : How To Find The Length Of The Side Of A Right Triangle

Example Question #1 : Triangles

Example Question #41 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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