ISEE Upper Level Math : Geometry

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

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Example Questions

Example Question #1 : How To Find An Angle In An Acute / Obtuse Triangle

Triangle 2

Refer to the above figure. Express \displaystyle y in terms of \displaystyle x.

Possible Answers:

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

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

\displaystyle y = x - 98

\displaystyle y = x - 82

Correct answer:

\displaystyle 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

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

and

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

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

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

\displaystyle y + 98 +(180 -x) = 180

\displaystyle y -x + 278 = 180

\displaystyle y -x + 278 +x - 278 = 180 +x - 278

\displaystyle y = 180 +x - 278

\displaystyle y = x - 98

Example Question #2 : How To Find An Angle In An Acute / Obtuse Triangle

Triangle 3

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

Give the measure of \displaystyle \angle CAD.

Possible Answers:

\displaystyle 40 ^{\circ }

\displaystyle 36 ^{\circ }

\displaystyle 32 ^{\circ }

\displaystyle 42^{\circ }

Correct answer:

\displaystyle 36 ^{\circ }

Explanation:

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

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

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

\displaystyle = 72 ^{\circ }

 

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

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

 

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

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

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

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

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

Example Question #3 : How To Find An Angle In An Acute / Obtuse Triangle

Triangle

Figure NOT drawn to scale.

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

Possible Answers:

\displaystyle 107^{\circ }

\displaystyle 102 ^{\circ }

\displaystyle 112^{\circ }

\displaystyle 97^{\circ }

Correct answer:

\displaystyle 102 ^{\circ }

Explanation:

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

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

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

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

\displaystyle 6x+6 = 6.5x

\displaystyle 6x+6 - 6x = 6.5x - 6x

\displaystyle 6 = 0.5x

\displaystyle 6 \cdot 2 = 0.5x \cdot 2

\displaystyle 12 = x

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

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

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

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

Exterior_angle

Note: Figure NOT drawn to scale.

What is the measure of angle \displaystyle y?

Possible Answers:

\displaystyle y = 142

\displaystyle y = 152

\displaystyle y = 129

\displaystyle y = 134

\displaystyle y = 124

Correct answer:

\displaystyle y = 124

Explanation:

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

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

\displaystyle y + 28 + 28 = 180

\displaystyle y +56= 180

\displaystyle y +56-56 = 180-56

\displaystyle y = 124

Example Question #91 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Possible Answers:

This triangle is scalene and obtuse.

This triangle is scalene and right.

This triangle cannot exist.

This triangle is isosceles and right.

This triangle is isosceles and obtuse.

Correct answer:

This triangle cannot exist.

Explanation:

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

Example Question #91 : Plane 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:

\displaystyle 10\ cm^2

\displaystyle 5\ cm^2

\displaystyle 8\ cm^2

\displaystyle 7\ cm^2

Correct answer:

\displaystyle 7\ cm^2

Explanation:

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

where \displaystyle a and \displaystyle b  are the lengths of two sides and \displaystyle C is the angle measure.

Plug in our given values:

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

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

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

A scalene triangle has a base length \displaystyle t and a corresponding altitude of \displaystyle 4t. Give the area of the triangle in terms of \displaystyle t.

Possible Answers:

\displaystyle t^2

\displaystyle 4t^2

\displaystyle t

\displaystyle 2t

\displaystyle 2t^2

Correct answer:

\displaystyle 2t^2

Explanation:

 \displaystyle Area =\frac{bh}2{},

where \displaystyle b is the base and \displaystyle h is the altitude.

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

Example Question #14 : Acute / Obtuse Triangles

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

Possible Answers:

\displaystyle 42

\displaystyle 91

\displaystyle 78

\displaystyle 40 \frac{1}{2}

\displaystyle 39

Correct answer:

\displaystyle 39

Explanation:

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

The area of this triangle is therefore

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

Example Question #15 : Acute / Obtuse Triangles

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

Possible Answers:

\displaystyle 90

\displaystyle 60

\displaystyle 45

\displaystyle 30

\displaystyle 120

Correct answer:

\displaystyle 60

Explanation:

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

The area of this triangle is therefore

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

Example Question #16 : Acute / Obtuse Triangles

Right triangle

Figure NOT drawn to scale.

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

Choose the closest answer.

Possible Answers:

\displaystyle 30 \%

\displaystyle 25 \%

\displaystyle 20 \%

\displaystyle 35 \%

Correct answer:

\displaystyle 25 \%

Explanation:

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

The similarity ratio of \displaystyle \bigtriangleup AXB to \displaystyle \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

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

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

\displaystyle \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 \displaystyle \bigtriangleup AXB is

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

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

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