Problem 1
In the 30°-60°-90° right triangle below, the hypotenuse is 10. Find the length \(x\) of the side opposite the 30° angle.
A) \(5\)
B) \(5\sqrt{3}\)
C) \(10\)
D) \(5\sqrt{2}\)
E) \(2.5\)
Solution:
In a 30-60-90 triangle the sides are in ratio \(1 : \sqrt{3} : 2\).
The side opposite 30° is the shortest = half the hypotenuse.
\(x = \dfrac{10}{2} = \boxed{5}\)
Problem 2
In the 45°-45°-90° right triangle below, each leg is 7. Find the hypotenuse \(x\).
A) \(14\)
B) \(7\sqrt{2}\)
C) \(7\sqrt{3}\)
D) \(7\)
E) \(14\sqrt{2}\)
Solution:
In a 45-45-90 triangle the sides are in ratio \(1 : 1 : \sqrt{2}\).
Hypotenuse = leg \(\times \sqrt{2}\).
\(x = \boxed{7\sqrt{2}}\)
Problem 3
A right triangle has legs 9 and 12. Find the length \(h\) of the altitude drawn to the hypotenuse.
A) \(6\)
B) \(7.5\)
C) \(7.2\)
D) \(10.8\)
E) \(5.4\)
Solution:
Hypotenuse \(= \sqrt{9^2 + 12^2} = \sqrt{225} = 15\).
Area two ways: \(\tfrac12(9)(12) = \tfrac12(15)(h)\).
\(h = \dfrac{9 \cdot 12}{15} = \dfrac{108}{15} = \boxed{7.2}\)
Problem 4
In triangle \(ABC\), \(DE \parallel BC\) with \(AD = 4\), \(DB = 6\), and \(AE = 6\). Find \(EC\).
A) \(4\)
B) \(6\)
C) \(10\)
D) \(9\)
E) \(8\)
Solution:
Side-splitter theorem: \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\).
\(\dfrac{4}{6} = \dfrac{6}{EC} \Rightarrow 4 \cdot EC = 36\).
\(EC = \boxed{9}\)
Problem 5
The inscribed angle \(x\) at vertex \(V\) intercepts an arc of 80°. Find \(x\).
V
x
arc = 80°
A) \(80^\circ\)
B) \(160^\circ\)
C) \(20^\circ\)
D) \(100^\circ\)
E) \(40^\circ\)
Solution:
An inscribed angle equals half its intercepted arc.
\(x = \tfrac12 (80^\circ)\).
\(x = \boxed{40^\circ}\)
Problem 6
A circle has radius 6. Find the length \(s\) of the arc cut off by a central angle of 60°.
A) \(2\pi\)
B) \(\pi\)
C) \(4\pi\)
D) \(6\pi\)
E) \(12\pi\)
Solution:
Arc length \(= \dfrac{\theta}{360^\circ}\cdot 2\pi r\).
\(= \dfrac{60}{360}\cdot 2\pi(6) = \dfrac16 \cdot 12\pi\).
\(s = \boxed{2\pi}\)
Problem 7
A sector of a circle has radius 8 and central angle 90°. Find its area.
A) \(8\pi\)
B) \(16\pi\)
C) \(64\pi\)
D) \(4\pi\)
E) \(32\pi\)
Solution:
Sector area \(= \dfrac{\theta}{360^\circ}\cdot \pi r^2\).
\(= \dfrac{90}{360}\cdot \pi(8)^2 = \dfrac14 \cdot 64\pi\).
\(A = \boxed{16\pi}\)
Problem 8
A circle has circumference \(10\pi\). Find its area.
C = 10π
A) \(100\pi\)
B) \(10\pi\)
C) \(25\pi\)
D) \(50\pi\)
E) \(5\pi\)
Solution:
\(C = 2\pi r = 10\pi \Rightarrow r = 5\).
Area \(= \pi r^2 = \pi(5)^2\).
\(= \boxed{25\pi}\)
Problem 9
Find the measure of one interior angle of a regular hexagon.
A) \(108^\circ\)
B) \(135^\circ\)
C) \(144^\circ\)
D) \(120^\circ\)
E) \(90^\circ\)
Solution:
Interior angle \(= \dfrac{(n-2)\cdot 180^\circ}{n}\) with \(n = 6\).
\(= \dfrac{4 \cdot 180^\circ}{6} = \dfrac{720^\circ}{6}\).
\(= \boxed{120^\circ}\)
Problem 10
Find the sum of the interior angles of a regular octagon.
A) \(1440^\circ\)
B) \(900^\circ\)
C) \(720^\circ\)
D) \(1260^\circ\)
E) \(1080^\circ\)
Solution:
Sum of interior angles \(= (n-2)\cdot 180^\circ\) with \(n = 8\).
\(= 6 \cdot 180^\circ\).
\(= \boxed{1080^\circ}\)
Problem 11
Each exterior angle of a regular polygon measures 30°. How many sides does it have?
A) \(12\)
B) \(10\)
C) \(6\)
D) \(30\)
E) \(8\)
Solution:
Exterior angles of any polygon sum to \(360^\circ\).
Number of sides \(= \dfrac{360^\circ}{30^\circ}\).
\(= \boxed{12}\)
Problem 12
Find the distance between the points \((1, 2)\) and \((4, 6)\).
(1, 2)
(4, 6)
A) \(7\)
B) \(5\)
C) \(\sqrt{7}\)
D) \(25\)
E) \(\sqrt{13}\)
Solution:
\(d = \sqrt{(4-1)^2 + (6-2)^2}\).
\(= \sqrt{3^2 + 4^2} = \sqrt{25}\).
\(d = \boxed{5}\)
Problem 13
Find the midpoint \(M\) of the segment joining \((-2, 3)\) and \((4, 7)\).
(-2, 3)
(4, 7)
M
A) \((2, 10)\)
B) \((3, 2)\)
C) \((1, 5)\)
D) \((1, 2)\)
E) \((6, 4)\)
Solution:
\(M = \left(\dfrac{-2+4}{2}, \dfrac{3+7}{2}\right)\).
\(= \left(\dfrac{2}{2}, \dfrac{10}{2}\right)\).
\(M = \boxed{(1, 5)}\)
Problem 14
Write the equation of the circle with center \((2, -3)\) and radius 4.
(2, -3)
r = 4
A) \((x+2)^2 + (y-3)^2 = 16\)
B) \((x-2)^2 + (y+3)^2 = 4\)
C) \((x-2)^2 + (y-3)^2 = 16\)
D) \((x-2)^2 + (y+3)^2 = 16\)
E) \((x+2)^2 + (y+3)^2 = 4\)
Solution:
Circle: \((x-h)^2 + (y-k)^2 = r^2\) with center \((h,k)\).
\((h,k) = (2,-3)\), \(r^2 = 16\).
\(\boxed{(x-2)^2 + (y+3)^2 = 16}\)
Problem 15
What is the slope of any line perpendicular to the line shown, \(y = 2x + 1\)?
y = 2x + 1
A) \(2\)
B) \(\tfrac{1}{2}\)
C) \(-2\)
D) \(-1\)
E) \(-\tfrac{1}{2}\)
Solution:
Perpendicular slopes are negative reciprocals.
The given slope is \(2\), so we need \(-\dfrac{1}{2}\).
Slope \(= \boxed{-\tfrac{1}{2}}\)
Problem 16
Find the area of the triangle with vertices \((0,0)\), \((6,0)\), and \((0,8)\).
A) \(24\)
B) \(48\)
C) \(12\)
D) \(14\)
E) \(28\)
Solution:
It is a right triangle with legs along the axes: base 6, height 8.
Area \(= \tfrac12 (6)(8)\).
\(= \boxed{24}\)
Problem 17
Find the area of a triangle with sides 13, 14, and 15.
A) \(90\)
B) \(84\)
C) \(42\)
D) \(91\)
E) \(168\)
Solution:
Heron's formula: \(s = \tfrac{13+14+15}{2} = 21\).
Area \(= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21\cdot 8\cdot 7\cdot 6}\).
\(= \sqrt{7056} = \boxed{84}\)
Problem 18
Find the area of an equilateral triangle with side length 6.
A) \(18\sqrt{3}\)
B) \(36\sqrt{3}\)
C) \(9\sqrt{3}\)
D) \(9\)
E) \(6\sqrt{3}\)
Solution:
Equilateral area \(= \dfrac{\sqrt{3}}{4}s^2\).
\(= \dfrac{\sqrt{3}}{4}(6)^2 = \dfrac{\sqrt3}{4}\cdot 36\).
\(= \boxed{9\sqrt{3}}\)
Problem 19
Find the volume of a cylinder with radius 3 and height 10.
A) \(30\pi\)
B) \(900\pi\)
C) \(60\pi\)
D) \(90\pi\)
E) \(180\pi\)
Solution:
Cylinder volume \(= \pi r^2 h\).
\(= \pi (3)^2 (10) = \pi \cdot 9 \cdot 10\).
\(= \boxed{90\pi}\)
Problem 20
Find the volume of a cone with radius 6 and height 8.
r = 6
h = 8
A) \(288\pi\)
B) \(48\pi\)
C) \(96\)
D) \(32\pi\)
E) \(96\pi\)
Solution:
Cone volume \(= \tfrac13 \pi r^2 h\).
\(= \tfrac13 \pi (6)^2 (8) = \tfrac13 \pi \cdot 36 \cdot 8\).
\(= \boxed{96\pi}\)
Problem 21
Find the volume of a sphere with radius 3.
r = 3
A) \(36\pi\)
B) \(27\pi\)
C) \(108\pi\)
D) \(12\pi\)
E) \(9\pi\)
Solution:
Sphere volume \(= \tfrac43 \pi r^3\).
\(= \tfrac43 \pi (3)^3 = \tfrac43 \pi \cdot 27\).
\(= \boxed{36\pi}\)
Problem 22
Find the surface area of a sphere with radius 5.
r = 5
A) \(50\pi\)
B) \(100\pi\)
C) \(500\pi\)
D) \(25\pi\)
E) \(20\pi\)
Solution:
Sphere surface area \(= 4\pi r^2\).
\(= 4\pi (5)^2 = 4\pi \cdot 25\).
\(= \boxed{100\pi}\)
Problem 23
A rectangular box has dimensions 3 by 4 by 12. Find the length of its space (interior) diagonal \(d\).
A) \(12\)
B) \(\sqrt{160}\)
C) \(13\)
D) \(19\)
E) \(14\)
Solution:
Space diagonal \(d = \sqrt{\ell^2 + w^2 + h^2}\).
\(= \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169}\).
\(d = \boxed{13}\)
Problem 24
Lines \(\ell_1 \parallel \ell_2\) are cut by a transversal. The two marked alternate interior angles are \((2x+10)^\circ\) and \((3x-20)^\circ\). Find \(x\).
(2x+10)°
(3x−20)°
▸
▸
A) \(50\)
B) \(40\)
C) \(35\)
D) \(30\)
E) \(25\)
Solution:
Alternate interior angles are equal: \(2x + 10 = 3x - 20\).
\(10 + 20 = 3x - 2x\).
\(x = \boxed{30}\)
Problem 25
From external point \(P\), a tangent touches a circle of radius 5 at \(T\). If \(OP = 13\), find the tangent length \(PT\).
O
P
T
5
?
13
A) \(8\)
B) \(\sqrt{194}\)
C) \(18\)
D) \(6\)
E) \(12\)
Solution:
A tangent is perpendicular to the radius at \(T\), so \(\triangle OTP\) is right-angled at \(T\).
\(PT = \sqrt{OP^2 - OT^2} = \sqrt{13^2 - 5^2} = \sqrt{144}\).
\(PT = \boxed{12}\)
Problem 26
Two chords intersect inside a circle. One is divided into segments of length 3 and 8; the other into 4 and \(x\). Find \(x\).
3
8
4
x
A) \(6\)
B) \(24\)
C) \(12\)
D) \(5\)
E) \(2\)
Solution:
Intersecting chords: products of the segments are equal.
\(3 \cdot 8 = 4 \cdot x \Rightarrow 24 = 4x\).
\(x = \boxed{6}\)
Problem 27
Triangle \(ABC\) is inscribed in a circle with \(AB\) a diameter. If \(\angle A = 35^\circ\), find \(\angle B\).
A) \(90^\circ\)
B) \(55^\circ\)
C) \(35^\circ\)
D) \(45^\circ\)
E) \(65^\circ\)
Solution:
An angle inscribed in a semicircle is \(90^\circ\), so \(\angle C = 90^\circ\) (Thales).
Angles of a triangle sum to \(180^\circ\): \(\angle B = 180 - 90 - 35\).
\(\angle B = \boxed{55^\circ}\)
Problem 28
In the right triangle below the sides are 3, 4, 5. Find \(\cos\theta\).
A) \(\tfrac{3}{5}\)
B) \(\tfrac{3}{4}\)
C) \(\tfrac{4}{5}\)
D) \(\tfrac{5}{4}\)
E) \(\tfrac{4}{3}\)
Solution:
\(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\).
The side adjacent to \(\theta\) is 4; the hypotenuse is 5.
\(\cos\theta = \boxed{\tfrac{4}{5}}\)
Problem 29
Find the area of a regular hexagon with side length 4.
A) \(48\sqrt{3}\)
B) \(96\sqrt{3}\)
C) \(16\sqrt{3}\)
D) \(24\sqrt{3}\)
E) \(12\sqrt{3}\)
Solution:
Regular hexagon area \(= \dfrac{3\sqrt{3}}{2}s^2\).
\(= \dfrac{3\sqrt{3}}{2}(4)^2 = \dfrac{3\sqrt3}{2}\cdot 16\).
\(= \boxed{24\sqrt{3}}\)
Problem 30
Which graph shows the line \(y = -x + 2\)?
Solution:
\(y = -x + 2\) has negative slope (\(-1\)) and \(y\)-intercept \(+2\).
It passes through \((0, 2)\) and \((2, 0)\), falling from upper-left to lower-right.
The correct graph crosses the \(y\)-axis above the origin and slopes downward.