Plane Geometry - ISEE Upper Level Quantitative Reasoning

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Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

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Answer

The two marked angles are same-side exterior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

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Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

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Answer

The two marked angles are same-side interior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

$2y \div 2 = \left ( 184 -x \right )\div 2$

$y= 92 - $\frac{1}{2}$x$

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Question

Examine the above diagram. Which of the following statements must be true whether or not and are parallel?

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Answer

Four statements can be eliminated by the various parallel theorems and postulates. Congruence of alternate interior angles or corresponding angles forces the lines to be parallel, so

$\angle 5 \cong \angle 4 \Rightarrow l \parallel m$ and

$\angle 2 \cong \angle 6 \Rightarrow l \parallel m$ .

Also, if same-side interior angles or same-side exterior angles are supplementary, the lines are parallel, so

$m \angle 4 + m \angle 7 = 180 ^{\circ } \Rightarrow l \parallel m$ and

$m \angle 1 + m \angle 6 = 180 ^{\circ } \Rightarrow l \parallel m$ .

However, $\angle 5 \cong \angle 8$ whether or not $l \parallel m$ since they are vertical angles, which are always congruent.

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Question

Examine the above diagram. What is ?

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Answer

By angle addition,

$\left ( x - 13\right )+ 100 + (y + 25) = 180$

$x - 13\right+ 100 + y + 25 = 180$

$x + y - 13\right+ 100+ 25 = 180$

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Question

$\angle 1$ and $\angle 2$ are supplementary; $\angle 1$ and $\angle 3$ are complementary.

$m \angle 2 = 100 ^{\circ }$ .

What is $m \angle 3$ ?

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Answer

Supplementary angles and complementary angles have measures totaling and , respectively.

$m \angle 2 = 100 ^{\circ }$ , so its supplement $\angle 1$ has measure

$m \angle 1 = $180^{\circ }$ - m \angle 2 = $180^{\circ }$ - 100 ^{\circ } = $80^{\circ }$$

$\angle 3$ , the complement of $\angle 1$ , has measure

$m \angle 3 = $90^{\circ }$ - m \angle 1 = $90^{\circ }$ - 80 ^{\circ } = $10^{\circ }$$

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Question

Note: Figure NOT drawn to scale.

In the above figure, $m \angle 1=\left ( x+17 \right $)^{\circ}$$ and $m \angle 2= \left (7x+11 \right $)^{\circ}$$ . Which of the following is equal to $m \angle 1$ ?

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Answer

$\angle 1$ and $\angle 2$ form a linear pair, so their angle measures total . Set up and solve the following equation:

$m \angle 1+ m \angle 2 = $180^{\circ}$$

$\left ( x+17 \right ) + \left (7x+11 \right ) = 180$

$m \angle 1=\left ( x+17 \right $)^{\circ}$ = \left ( 19+17 \right $)^{\circ}$ = $36^{\circ}$$

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Question

Two angles which form a linear pair have measures and . Which is the lesser of the measures (or the common measure) of the two angles?

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Answer

Two angles that form a linear pair are supplementary - that is, they have measures that total . Therefore, we set and solve for in this equation:

$x = $\frac{233}{7}$= 33 $\frac{2}{7}$$

The two angles have measure

$2 \cdot 33 $\frac{2}{7}$ +72= 138 $\frac{4}{7}$ ^{\circ}$

and

$5 \cdot 33 $\frac{2}{7}$ -125 = $41$\frac{3}{7}$^{\circ}$$

is the lesser of the two measures and is the correct choice.

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Question

Two vertical angles have measures and $\left ( 5x-17\right $)^{\circ }$$ . Which is the lesser of the measures (or the common measure) of the two angles?

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Answer

Two vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure. Therefore, we set up and solve the equation

$x= 15$\frac{1}{2}$$

$3x+14 = 3 \cdot 15$\frac{1}{2}$ + 14 = 46$\frac{1}{2}$ + 14= 60$\frac{1}{2}$ ^{\circ }$

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Question

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 .

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Answer

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,

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

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

Backsolve:

, which is the correct response.

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Question

Note: Figure NOT drawn to scale.

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

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Answer

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

,

or, simplified,

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

Substitute for :

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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Question

Figure NOT drawn to scale

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

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Answer

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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Question

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

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Answer

To find the missing side, use the Pythagorean Theorem . Plug in (remember c is always the hypotenuse!) so that . Simplify and you get Subtract 36 from both sides so that you get Take the square root of both sides. B is 8.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

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

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Answer

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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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

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

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Answer

First, find .

Since $\overline{DE}$ is an altitude of $\Delta CDB$ from its right angle to its hypotenuse,

$\Delta DEC \sim \Delta BED$

$\frac{BE}{DE}$= $\frac{ED}{EC}$

$\frac{BE}{6}$= $\frac{6}{12}$

$BE= $\frac{6}{12}$ \cdot 6 = 3$

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

$\frac{AB}{DE}$ = $\frac{CB}{CE}$

$\frac{AB}{6}$ = $\frac{15}{12}$

$AB= $\frac{15}{12}$ \cdot 6$

$AB = 7$\frac{1}{2}$$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

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Answer

By the Pythagorean Theorem,

$x= 3$\frac{3}{8}$$

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Question

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?

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Answer

By the Pythagorean Theorem,

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Question

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

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Answer

By the Pythagorean Theorem,

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Question

A right triangle $\Delta ABC$ with hypotenuse $\overline{AC}$ is inscribed in $\odot O$ , a circle with radius 26. If , evaluate the length of $\overline{BC}$ .

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Answer

The arcs intercepted by a right angle are both semicircles, so hypotenuse $\overline{AC}$ shares its endpoints with two semicircles. This makes $\overline{AC}$ a diameter of the circle, and $AC = 2r = 2 \cdot 26 = 52$ .

By the Pythagorean Theorem,

$BC = $$\sqrt{(AC)^{2}$$ - $(AB)^{2}$} = $$\sqrt{52^{2}$$ $-20^{2}$} = $\sqrt{2,704-400}$ = $\sqrt{2,304}$ = 48$

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Question

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

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Answer

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

Use the following formula and solve for x:

Begin by dividing over the 100

Then multiply by 360

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Question

If sector AJL covers 45% of circle J, what is the measure of sector AJL's central angle?

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Answer

If sector AJL covers 45% of circle J, what is the measure of sector AJL's central angle?

To find an angle measure from a percentage, simply convert the percentage to a decimal and then multiply it by 360 degrees.

$45%\rightarrow 0.45$

$0.45 * $360^{\circ}$$=162^{\circ}$$

So, our answer is 162 degrees.

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