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ISEE Upper Level Math : Geometry

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6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Triangles

Triangle 2

Refer to the above figure. Express in terms of .

Possible Answers:

y = x - 82

y = x - 98

$y =\frac{1}{3} x - 8$

$y =\frac{1}{3} x + 8$

Correct answer:

y = x - 98

Explanation:

The measure of an interior angle of a triangle is equal to 180 degrees minus that of its adjacent exterior angle, so

$m \angle ABC = (180 - 82)^{\circ } = 98^{\circ }$

and

$m \angle ACB = (180 - x)^{\circ }$ .

The sum of the degree measures of the three interior angles is 180, so

$m \angle BAC + m \angle ABC +m \angle ACB = 180 ^{\circ }$

y + 98 +(180 -x) = 180

y -x + 278 = 180

y -x + 278 +x - 278 = 180 +x - 278

y = 180 +x - 278

y = x - 98

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Example Question #91 : Plane Geometry

Triangle 3

In the above figure, $\overline{AC} \cong \overline{AD}$ .

Give the measure of $\angle CAD$ .

Possible Answers:

$36 ^{\circ }$

$32 ^{\circ }$

$42^{\circ }$

$40 ^{\circ }$

Correct answer:

$36 ^{\circ }$

Explanation:

$\angle BCA$ and $\angle ACD$ form a linear pair, so their degree measures total $180 ^{\circ }$ ; consequently,

$m \angle ACD = 180 ^{\circ } - m \angle CAD$

$= 180 ^{\circ } - 108 ^{\circ }$

$= 72 ^{\circ }$

$\overline{AC} \cong \overline{AD}$ , so by the Isosceles Triangle Theorem,

$m \angle ADC = m \angle ACD = 72 ^{\circ }$

The sum of the degree measures of a triangle is $180^{\circ }$ , so

$\angle CAD + m \angle ADC + m \angle ACD = 180 ^{\circ }$

$\angle CAD + 72^{\circ } +72^{\circ } = 180 ^{\circ }$

$\angle CAD +144^{\circ } = 180 ^{\circ }$

$\angle CAD = 36 ^{\circ }$

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Example Question #10 : How To Find An Angle In An Acute / Obtuse Triangle

Triangle

Figure NOT drawn to scale.

Refer to the above figure. Evaluate $m \angle ACB$ .

Possible Answers:

$112^{\circ }$

$102 ^{\circ }$

$107^{\circ }$

$97^{\circ }$

Correct answer:

$102 ^{\circ }$

Explanation:

The measure of an exterior angle of a triangle, which here is $\angle ACD$ , is equal to the sum of the measures of its remote interior angles, which here are $\angle A$ and $\angle B$ . Consequently,

$m \angle A + m \angle B = m \angle ACD$

(4x-2)+(2x+8) = 6.5x

4x+2x-2+8 = 6.5x

6x+6 = 6.5x

6x+6 - 6x = 6.5x - 6x

6 = 0.5x

$6 \cdot 2 = 0.5x \cdot 2$

12 = x

$m \angle ACD =( 6.5 x) ^{\circ }= (6.5 \cdot 12 )^{\circ }= 78^{\circ }$

$\angle ACB$ and $\angle ACD$ form a linear pair and, therefore,

$m \angle ACB=180 ^{\circ } - m \angle ACD = 180 ^{\circ } -78^{\circ } = 102 ^{\circ }$ .

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Example Question #91 : Plane Geometry

Exterior_angle

Note: Figure NOT drawn to scale.

What is the measure of angle

Possible Answers:

y = 124

y = 134

y = 152

y = 129

y = 142

Correct answer:

y = 124

Explanation:

The two angles at bottom are marked as congruent. One forms a linear pair with a $152 ^{\circ }$ angle, so it is supplementary to that angle, making its measure $(180-152)^{\circ } = 28^{\circ }$ . Therefore, each marked angle measures $28^{\circ }$ .

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so:

y + 28 + 28 = 180

y +56= 180

y +56-56 = 180-56

y = 124

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Example Question #91 : Geometry

Which of the following is true about a triangle with two angles that measure $120^{\circ }$ and $90^{\circ }$ ?

Possible Answers:

This triangle is scalene and right.

This triangle is scalene and obtuse.

This triangle cannot exist.

This triangle is isosceles and obtuse.

This triangle is isosceles and right.

Correct answer:

This triangle cannot exist.

Explanation:

A triangle must have at least two acute angles; however, a triangle with angles that measure $120^{\circ }$ and $90^{\circ }$ could have at most one acute angle, an impossible situation. Therefore, this triangle is nonexistent.

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Example Question #92 : Geometry

Two sides of a scalene triangle measure 4 centimeters and 7 centimeters, and their corresponding angle measures 30 degrees. Find the area of the triangle.

Possible Answers:

$5\ cm^2$

$7\ cm^2$

$8\ cm^2$

$10\ cm^2$

Correct answer:

$7\ cm^2$

Explanation:

$Area=\frac{1}{2}absinC$ ,

where and are the lengths of two sides and is the angle measure.

Plug in our given values:

$Area=\frac{1}{2}absinC=\frac{1}{2}\times 4\times 7\times sin30^{\circ}$

$sin30^{\circ}=\frac{1}{2}\Rightarrow Area=\frac{1}{2}\times 4\times 7\times \frac{1}{2}=7\ cm^2$

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Example Question #93 : Isee Upper Level (Grades 9 12) Mathematics Achievement

A scalene triangle has a base length and a corresponding altitude of . Give the area of the triangle in terms of .

Possible Answers:

t^2

2t^2

4t^2

Correct answer:

2t^2

Explanation:

$Area =\frac{bh}2{}$ ,

where is the base and is the altitude.

$Area =\frac{bh}2{}=\frac{t\times 4t}{2}=2t^2$

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Example Question #92 : Isee Upper Level (Grades 9 12) Mathematics Achievement

What is the area of a triangle on the coordinate plane with its vertices on the points (0,-4) ,(0,9), (6,7) ?

Possible Answers:

$40 \frac{1}{2}$

Correct answer:

Explanation:

The base can be seen as the (vertical) line segment connecting (0,-4) and (0,9) , which has length 9 - (-4) = 13 . The height is the pependicular distance from (6,7) to the segment; since the segment is part of the -axis, this altitude is horizontal and has length equal to -coordinate .

The area of this triangle is therefore

$A = \frac{1}{2} bh = \frac{1}{2} \cdot13 \cdot 6 = 39$ .

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Example Question #94 : Isee Upper Level (Grades 9 12) Mathematics Achievement

What is the area of a triangle on the coordinate plane with its vertices on the points (-5,0), (10,0), (6,8) ?

Possible Answers:

120

Correct answer:

Explanation:

The base can be seen as the (horizontal) line segment connecting $\left (-5,0 \right )$ and $\left (10,0 \right )$ , the length of which is 10 - (-5) = 15 . The height is the pependicular distance from (6,8) to the segment; since the segment is part of the -axis, this altitude is vertical and has a length equal to -coordinate .

The area of this triangle is therefore

$A = \frac{1}{2} bh = \frac{1}{2} \cdot15 \cdot 8 = 60$ .

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Example Question #53 : Triangles

Right triangle

Figure NOT drawn to scale.

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . What percent of $\bigtriangleup ABC$ has been shaded gray?

Choose the closest answer.

Possible Answers:

$30 \%$

$35 \%$

$20 \%$

$25 \%$

Correct answer:

$25 \%$

Explanation:

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other as well as the large triangle.

The similarity ratio of $\bigtriangleup AXB$ to $\bigtriangleup ABC$ is the ratio of the lengths of their hypotenuses. The hypotenuse of the latter is 18; that of the former, from the Pythagorean Theorem, is

$\sqrt{18 ^{2}+ 30 ^{2}} = \sqrt{324+ 900} = \sqrt{1,224}$

The similarity ratio is therefore $\frac{18}{ \sqrt{1,224}}$ . The ratio of their areas is the square of this, or

$\left (\frac{18}{ \sqrt{1,224}} \right )^{2} = \frac{324}{1,224} = \frac{324 \div 36 }{1,224\div 36} = \frac{9}{34}$

The area of $\bigtriangleup AXB$ is

$\frac{9}{34} \times 100 \% \approx 26.5 \%$

of that of $\bigtriangleup ABC$ , so the choice closest to the correct percent is 25%.

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Geometry

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #41 : Triangles

Example Question #91 : Plane Geometry

Example Question #10 : How To Find An Angle In An Acute / Obtuse Triangle

Example Question #91 : Plane Geometry

Example Question #91 : Geometry

Example Question #92 : Geometry

Example Question #93 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #92 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #94 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #53 : Triangles

All ISEE Upper Level Math Resources

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