Geometry Placement Test Practice with Solutions

Question 1

Which of the following is not correct?

• I) A line tangent to a circle intersects the circle at one point only. A line through the center and the point of tangency is perpendicular to the tangent.
• II) The longer side in a triangle is greater than the sum of the other two sides.
• III) A triangle with two equal angles is an isosceles triangle.
• IV) The opposite sides of a parallelogram are parallel but not congruent.
• V) Two triangles with congruent angles are similar and their corresponding sides are proportional.
• VI) The size of an angle inscribed in a semicircle is greater than \( 90^{\circ} \).
• VII) The sum of all interior angles of a quadrilateral is \( 360^{\circ} \).

1. (II) only
2. (IV) and (V)
3. (V)
4. (II) and (III)
5. (II), (IV) and (VI)

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The incorrect statements are II, IV, and VI. Corrections: (II) The sum of any two sides of a triangle must be greater than the third side. (IV) Opposite sides of a parallelogram are both parallel and congruent. (VI) The angle inscribed in a semicircle is exactly \(90^{\circ}\). Answer: E

Question 2

Find \( x \) in the figure below.

1. \( 35^{\circ} \)
2. \( 25^{\circ} \)
3. \( 26^{\circ} \)
4. \( 27^{\circ} \)
5. \( 34^{\circ} \)

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Angles \( ACB \) and \( BCD \) make a straight angle (\(180^{\circ}\)). \(\angle ACB + 3x + 10 = 180 \implies \angle ACB = 170 - 3x\). The sum of angles in a triangle is \(180^{\circ}\): \(45 + 2x + (170 - 3x) = 180\). Solving for \(x\): \(x = 35^{\circ}\). Answer: A

Question 3

Calculate the area of the isosceles triangle in the figure below.

1. \( 100 \sin (65.45^{\circ}) \)
2. \( 50 \sin (65.45^{\circ}) \)
3. \( 50 \sin (40^{\circ}) \)
4. \( 50 \)
5. \( 100 \sin (40^{\circ}) \)

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The triangle is isosceles, so base angles are equal. Interior angle sum: \(3.5x + 3.5x + 2x = 180 \implies 9x = 180 \implies x = 20\). Angle BAC = \(2x = 40^{\circ}\). Area = \(0.5 \times side \times side \times \sin(40^{\circ}) = 0.5 \times 10 \times 10 \times \sin(40^{\circ}) = 50 \sin(40^{\circ})\). Answer: C

Question 4

Calculate the area of the quadrilateral in the figure below (approximate to 2 decimal places).

1. \( 44.06 \)
2. \( 22.03 \)
3. \( 88.12 \)
4. \( 105.00 \)
5. \( 52.50 \)

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Divide into two triangles, \(ACD\) and \(ABC\). Area of \(ACD = 0.5 \cdot 9 \cdot 7 \cdot \sin(94^{\circ})\). Use the Law of Cosines to find length of diagonal \(AC\), then solve for the interior angle of triangle \(ABC\) and calculate its area. Total area is approximately 44.06. Answer: A

Question 5

Find the area of the rectangle \( AEFG \) given that the sum of the areas of the triangles \( ABC \) and \( CDE \) is equal to 200 square units.

1. \( 800 \)
2. \( 400 \)
3. \( 200 \)
4. \( 600 \)
5. \( 500 \)

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All triangles inside the rectangle share the same height \(h\) as the rectangle. Area of \(\triangle ABC = 0.5 \cdot AC \cdot h\). Area of \(\triangle CDE = 0.5 \cdot CE \cdot h\). Sum = \(0.5 \cdot h \cdot (AC+CE) = 0.5 \cdot h \cdot AE\). Rectangle Area = \(AE \cdot h = 2 \cdot (Sum) = 2 \cdot 200 = 400\). Answer: B

Question 6

In the figure below \( AB \) is parallel to \( CD \). What is the size of \( \angle CHF \)?

1. \( 112^{\circ} \)
2. \( 56^{\circ} \)
3. \( 22^{\circ} \)
4. \( 68^{\circ} \)
5. \( 90^{\circ} \)

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Corresponding angles \(\angle AGE\) and \(\angle CHG\) are equal, so \(\angle CHG = 112^{\circ}\). \(\angle CHG\) and \(\angle CHF\) are supplementary (on a straight line). \(\angle CHF = 180^{\circ} - 112^{\circ} = 68^{\circ}\). Answer: D

Question 7

What is the size of \( \angle GAF \) in the figure below?

1. \( 47^{\circ} \)
2. \( 31^{\circ} \)
3. \( 55^{\circ} \)
4. \( 35^{\circ} \)
5. \( 45^{\circ} \)

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Vertical angles make \(\angle GFA = 102^{\circ}\) and \(\angle FGA = 47^{\circ}\). The sum of interior angles of triangle \(FAG = 180^{\circ}\). \(\angle GAF + 102^{\circ} + 47^{\circ} = 180^{\circ} \implies \angle GAF = 31^{\circ}\). Answer: B

Question 8

ABCD is a rectangle (width 80, length 240). The circle is tangent to AE, EF, FD, and DA. Find the area of the shaded surface.

1. \( 12800 \)
2. \( 19200 + 1600 \pi \)
3. \( 19200 - 1600 \pi \)
4. \( 12800 + 1600 \pi \)
5. \( 12800 - 1600 \pi \)

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Rectangle Area = \(80 \cdot 240 = 19200\). Circle Area = \(\pi \cdot 40^2 = 1600\pi\). Right Triangle EFC Area = \(0.5 \cdot 80 \cdot (240-80) = 6400\). Shaded Area = \(19200 - 1600\pi - 6400 = 12800 - 1600\pi\). Answer: E

Question 9

Which of the following is a right triangle?

(I): \( a=0.3, b=0.4, c=0.5 \)   (II): \( A=23^{\circ}, B=80^{\circ} \)   (III): \( C=34^{\circ}, B=56^{\circ} \)   (IV): \( a=2, b=4, c=5 \)

1. (I), (II), (III) and (IV)
2. (II) and (III) only
3. (I) and (III) only
4. (I) and (II) only
5. (III) only

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(I) \(0.3^2 + 0.4^2 = 0.09+0.16 = 0.25 = 0.5^2\). Yes. (II) Sum angles = 103, third angle is 77. No. (III) \(180 - (34+56) = 90\). Yes. (IV) \(2^2 + 4^2 = 20 \neq 25\). No. Answer: C

Question 10

Find the area of the trapezoid ABCD.

1. \( \frac{39\sqrt{3}}{4} \)
2. \( \frac{39}{4} \)
3. \( 24 \)
4. \( 15 \)
5. \( \frac{39\sqrt{3}}{2} \)

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Height \( h = 3 \sin(60^{\circ}) \). Area = \(\frac{1}{2}(AB + DC) \cdot h = \frac{1}{2}(5+8) \cdot 3 \cdot \frac{\sqrt{3}}{2} = \frac{39\sqrt{3}}{4}\). Answer: A

Question 11

Calculate the area of \(\triangle ODB\).

1. \( 252.87 \)
2. \( 272.72 \)
3. \( 440.25 \)
4. \( 660.12 \)
5. \( 230.66 \)

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Use Intersecting Chords Theorem: \(OD \cdot OC = OA \cdot OB \implies OD = \frac{22 \cdot 20}{22} \dots\) (using triangle area sine formula) Area \(\approx 252.87\). Answer: A

Question 12

In the diagram below, \( AB \) is parallel to \( DC \). Find the length of segment \( OC \).

1. \( 3 \)
2. \( 12 \)
3. \( 24/7 \)
4. \( 21/4 \)
5. \( 4/21 \)

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By Intercept Theorem (Thales): \(\frac{AB}{DC} = \frac{OA}{OC} \implies \frac{8}{6} = \frac{7}{OC} \implies OC = \frac{42}{8} = 21/4\). Answer: D

Question 13

Sphere of radius \(r\) inscribed in a cylinder of height \(2r\). If sphere volume is 3000, find cylinder volume.

1. \( 2000 \)
2. \( 4500 \)
3. \( 5000 \)
4. \( 3500 \)
5. \( 4000 \)

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Cylinder Volume \(V_c = 2\pi r^3\). Sphere Volume \(V_s = \frac{4}{3}\pi r^3\). Ratio \(V_c/V_s = 3/2\). \(V_c = 1.5 \cdot 3000 = 4500\). Answer: B

Question 14

AD and AC are tangent to the circle. Find the size of angle \( CBD \).

1. \( 124^{\circ} \)
2. \( 56^{\circ} \)
3. \( 62^{\circ} \)
4. \( 34^{\circ} \)
5. \( 68^{\circ} \)

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Let O be the center of the circle. Sum of angles in quadrilateral ACOD = \(360^{\circ}\). \( \quad 56 + 90 + 90 + \angle DOC = 360 \implies \angle DOC = 124^{\circ}\). Inscribed angle \(\angle DBC = 0.5 \cdot 124^{\circ} = 62^{\circ}\). Answer: C

Question 15

Find the size of \( \angle ACB \) in the diagram below.

1. \( 30^{\circ} \)
2. \( 60^{\circ} \)
3. \( 41^{\circ} \)
4. \( 49^{\circ} \)
5. \( 31^{\circ} \)

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Angles \(CAE\) and \(CBE\) intercept the same arc, so \(\angle CBE = 54^{\circ}\). Triangle \(CBO\) sum = \(180^{\circ}\). \(\angle ACB = 180 - 95 - 54 = 31^{\circ}\). Answer: E

Question 16

Which of the following is not correct?

• I) A square is a rhombus.
• II) A rectangle is a parallelogram.
• III) A square is not a parallelogram.
• IV) A trapezoid is not a quadrilateral.
• V) A kite has equal adjacent sides.
• VI) A parallelogram has its opposite sides parallel but not necessarily equal.
• VII) A rhombus is not a parallelogram.

1. (III) only
2. (IV) only
3. (III) and (VII) only
4. (III), (IV), (VI) and (VII)
5. (VI) and (VII) only

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Incorrect statements: III (Squares are parallelograms), IV (Trapezoids have 4 sides), VI (Opposite sides of parallelograms are equal), VII (Rhombuses are parallelograms). Answer: D