Geometry and Trigonometry Practice Questions — JAMB UTME — Page 3

Question 41Question →

Which of the following sets contains all the values of $x$ in the interval $0^\circ \le x \le 360^\circ$ that satisfy the trigonometric equation $2\cos^2 x + 3\sin x - 3 = 0$ ?

30^\circ, 90^\circ, 150^\circ

60^\circ, 90^\circ, 120^\circ

30^\circ, 150^\circ

60^\circ, 300^\circ

Show answer & explanation

Answer: $30^\circ, 90^\circ, 150^\circ$

Answer

The values of

x

in the interval

0^\circ \le x \le 360^\circ

satisfying the equation are

30^\circ, 90^\circ,

and

150^\circ

.

By substituting

\cos^2 x = 1 - \sin^2 x

, the equation reduces to

2\sin^2 x - 3\sin x + 1 = 0

, which factors into

(2\sin x - 1)(\sin x - 1) = 0

. Solving

\sin x = 1/2

gives

x = 30^\circ

and

x = 150^\circ

within the specified domain. Solving

\sin x = 1

gives

x = 90^\circ

. Combining these yields the set of solutions

30^\circ, 90^\circ, 150^\circ

.

Step-by-Step Solution

1

Use the Pythagorean trigonometric identity

\cos^2 x = 1 - \sin^2 x

to rewrite the equation in terms of

\sin x

.

2(1 - \sin^2 x) + 3\sin x - 3 = 0 \implies 2 - 2\sin^2 x + 3\sin x - 3 = 0

Converting the equation to involve a single trigonometric function allows it to be solved as a quadratic equation.

2

Simplify and rearrange the equation into standard quadratic form.

-2\sin^2 x + 3\sin x - 1 = 0 \implies 2\sin^2 x - 3\sin x + 1 = 0

Multiplying by

-1

simplifies factoring.

3

Factor the quadratic equation

(2\sin x - 1)(\sin x - 1) = 0

to solve for

\sin x

.

\sin x = \frac{1}{2}

or

\sin x = 1

Setting each linear factor to zero yields the possible values for

\sin x

.

4

Determine all values of

x

in the domain

0^\circ \le x \le 360^\circ

for each case.

For

\sin x = \frac{1}{2}

,

x = 30^\circ

and

x = 180^\circ - 30^\circ = 150^\circ

. For

\sin x = 1

,

x = 90^\circ

.

Sine is positive in the first and second quadrants, and equals 1 at

90^\circ

.

Key Concept

Solving quadratic trigonometric equations by using fundamental identities to express the equation in terms of a single trigonometric function.

Estimated Time:2m 0s

Question 42Question →

What is the measure, in degrees, of each interior angle of a regular octagon (an 8-sided regular polygon)?

Show answer & explanation

Answer: 135

Answer

Each interior angle of a regular octagon measures 135 degrees.

The sum of the interior angles of a polygon with

n

sides is

(n - 2) \times 180^\circ

. For a regular octagon (

n = 8

), the total interior angle sum is

(8 - 2) \times 180^\circ = 1080^\circ

. Since all 8 interior angles of a regular octagon are congruent, dividing the total sum by 8 yields

135^\circ

for each interior angle.

Step-by-Step Solution

1

Find the sum of all interior angles of the regular octagon.

Sum of interior angles =

(8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ

.

The sum of interior angles for any

n

-sided polygon is

(n - 2) \times 180^\circ

.

2

Calculate the measure of a single interior angle.

Interior angle =

\frac{1080^\circ}{8} = 135^\circ

.

In a regular polygon, all interior angles are equal in measure.

Key Concept

Interior Angle of a Regular Polygon

Question 43Question →

Three of the interior angles of a convex polygon are each $120^\circ$ , while the remaining interior angles are each $160^\circ$ . How many sides does the polygon have?

Show answer & explanation

Answer: 12

Answer

The polygon has 12 sides.

The sum of the interior angles of an

n

-sided polygon is

(n - 2) \times 180^\circ

. From the problem, the sum of the angles is

3 \times 120^\circ + (n - 3) \times 160^\circ = 160^\circ n - 120^\circ

. Equating

180^\circ(n - 2)

to

160^\circ n - 120^\circ

gives

180^\circ n - 360^\circ = 160^\circ n - 120^\circ

, which simplifies to

20^\circ n = 240^\circ

, yielding

n = 12

.

Step-by-Step Solution

1

Write down the standard formula for the sum of interior angles of an

n

-sided convex polygon.

Sum of interior angles =

(n - 2) \times 180^\circ

.

Any

n

-sided convex polygon can be split into

(n-2)

triangles, each having an interior angle sum of

180^\circ

.

2

Express the total sum of interior angles using the given values.

Sum =

3(120^\circ) + (n - 3)(160^\circ) = 360^\circ + 160^\circ n - 480^\circ = 160^\circ n - 120^\circ

.

Three angles are

120^\circ

, so the remaining

(n - 3)

angles must each equal

160^\circ

.

3

Equate the theoretical sum to the calculated sum.

180^\circ(n - 2) = 160^\circ n - 120^\circ \implies 180^\circ n - 360^\circ = 160^\circ n - 120^\circ

.

Both expressions represent the total interior angle sum of the same polygon.

4

Solve the equation for the number of sides

n

.

20^\circ n = 240^\circ \implies n = 12

.

Subtracting

160^\circ n

from both sides and adding

360^\circ

yields

20^\circ n = 240^\circ

.

Key Concept

Sum of Interior Angles of a Polygon

Question 44Question →

A ship departs from a port $P$ and sails $10\text{ km}$ on a bearing of $060^\circ$ to reach a position $Q$ . From $Q$ , the ship changes course and sails $10\text{ km}$ on a bearing of $150^\circ$ to arrive at point $R$ . What is the bearing of point $R$ from point $P$ ?

015^\circ

105^\circ

135^\circ

150^\circ

Show answer & explanation

Answer: $105^\circ$

Answer

The bearing of point

R

from point

P

is

105^\circ

.

The back bearing from position

Q

to port

P

is

240^\circ

. The difference between

240^\circ

and the new bearing of

150^\circ

yields an interior angle of

90^\circ

at vertex

Q

. Since both distances

PQ

and

QR

are equal to

10\text{ km}

, triangle

PQR

is a

45^\circ-45^\circ-90^\circ

isosceles right triangle. Adding

\angle QPR = 45^\circ

to the initial bearing of

060^\circ

yields a bearing of

105^\circ

for point

R

from point

P

.

Step-by-Step Solution

1

Determine the back bearing of

P

from

Q

.

The back bearing of

P

from

Q

is

060^\circ + 180^\circ = 240^\circ

.

To find the internal angle at

Q

, we need the direction of

P

relative to

Q

.

2

Calculate the interior angle

\angle PQR

.

\angle PQR = 240^\circ - 150^\circ = 90^\circ

.

The difference between the line back to

P

(

240^\circ

) and the line to

R

(

150^\circ

) forms the interior angle at

Q

.

3

Determine the properties of triangle

PQR

and angle

\angle QPR

.

Since

PQ = QR = 10\text{ km}

and

\angle PQR = 90^\circ

, triangle

PQR

is an isosceles right triangle, so

\angle QPR = 45^\circ

.

The two equal sides subtend equal acute angles in a right-angled triangle:

(180^\circ - 90^\circ) / 2 = 45^\circ

.

4

Calculate the total bearing of

R

from

P

.

Bearing of

R

from

P = 060^\circ + 45^\circ = 105^\circ

.

Point

R

lies clockwise relative to the segment

PQ

, so we add

\angle QPR

to the initial bearing of

PQ

.

Key Concept

Three-point bearing calculations using geometry of parallel lines and right-angled triangles.

Question 45Question →

A trapezium has parallel sides of lengths $8\text{ cm}$ and $12\text{ cm}$ . If the perpendicular distance between these parallel sides is $6\text{ cm}$ , what is the area of the trapezium?

60\text{ cm}^2

120\text{ cm}^2

10\text{ cm}^2

288\text{ cm}^2

Show answer & explanation

Answer: $60\text{ cm}^2$

Answer

60\text{ cm}^2

The area of a trapezium is given by

A = \frac{1}{2}(a + b)h

, where

a

and

b

are the parallel sides and

h

is the perpendicular distance between them. Substituting

a = 8

,

b = 12

, and

h = 6

gives

A = \frac{1}{2}(8 + 12)(6) = \frac{1}{2}(20)(6) = 60\text{ cm}^2

.

Step-by-Step Solution

1

Identify the given dimensions and the formula for the area of a trapezium.

Parallel sides

a = 8\text{ cm}

,

b = 12\text{ cm}

, height

h = 6\text{ cm}

. Formula:

A = \frac{1}{2}(a + b)h

.

The area of a trapezium depends on the sum of its parallel sides and its perpendicular height.

2

Sum the parallel sides.

a + b = 8 + 12 = 20\text{ cm}

.

The average of the parallel sides forms the effective width of an equivalent rectangle.

3

Multiply by half of the height to find the total area.

A = \frac{1}{2} \times 20 \times 6 = 60\text{ cm}^2

.

Completing the formula gives the exact plane surface measure in square centimeters.

Key Concept

Area of a Trapezium

Question 46Question →

The perpendicular bisector of the line segment joining the points $P(2, -1)$ and $Q(6, 7)$ intersects the $y$ -axis at $(0, c)$ . Find the value of $c$ .

Show answer & explanation

Answer: 5

Answer

The value of

c

is 5.

The midpoint of

PQ

is

(4, 3)

and the gradient of

PQ

is

2

. The perpendicular bisector has a gradient of

-\frac{1}{2}

and passes through

(4, 3)

. Substituting these into the line equation gives

y - 3 = -\frac{1}{2}(x - 4)

, which simplifies to

y = -\frac{1}{2}x + 5

. The line intersects the

y

-axis at

(0, 5)

, so

c = 5

.

Step-by-Step Solution

1

Find the midpoint of the line segment

PQ

Midpoint

M = (4, 3)

The perpendicular bisector must pass through the midpoint of the segment.

2

Calculate the gradient of

PQ

Gradient

m_{PQ} = 2

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

for points

(2, -1)

and

(6, 7)

.

3

Determine the gradient of the perpendicular line

Perpendicular gradient

m_{\perp} = -\frac{1}{2}

Perpendicular lines have gradients that are negative reciprocals (

m_1 m_2 = -1

).

4

Formulate the equation of the perpendicular bisector and solve for the

y

-intercept

y = -\frac{1}{2}x + 5

, hence

c = 5

Using point-slope form

y - y_1 = m(x - x_1)

with point

(4, 3)

and

m = -\frac{1}{2}

, setting

x = 0

gives the

y

-intercept.

Key Concept

Perpendicular Bisector and Line Equations

Question 47Question →

A sector of a circle of radius $14\text{ cm}$ subtends an angle of $90^\circ$ at the centre of the circle. What is the area of the sector in $\text{cm}^2$ ? (Take $\pi = \frac{22}{7}$ )

Show answer & explanation

Answer: 154

Answer

The area of the sector is

154\text{ cm}^2

.

The area of a circular sector is given by

\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2

. Substituting

\theta = 90^\circ

,

r = 14\text{ cm}

, and

\pi = \frac{22}{7}

yields

\frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2

.

Step-by-Step Solution

1

Identify the formula for the area of a circular sector.

\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2

The area of a sector is proportional to the central angle it subtends relative to a full circle (

360^\circ

).

2

Substitute the given values into the formula.

\text{Area} = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (14)^2

Given radius

r = 14\text{ cm}

, angle

\theta = 90^\circ

, and

\pi = \frac{22}{7}

.

3

Simplify the fraction and calculate the numerical value.

\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 196 = \frac{1}{4} \times 22 \times 28 = 22 \times 7 = 154\text{ cm}^2

Simplifying

\frac{90}{360}

yields

\frac{1}{4}

and dividing

196

by

7

gives

28

.

Key Concept

Area of a sector of a circle

Estimated Time:45s

Question 48Question →

A solid right circular cone has a base radius of $7\text{ cm}$ and a height of $12\text{ cm}$ . Taking $\pi = \frac{22}{7}$ , what is the volume of the cone?

616\text{ cm}^3

1848\text{ cm}^3

924\text{ cm}^3

308\text{ cm}^3

Show answer & explanation

Answer: $616\text{ cm}^3$

Answer

616\text{ cm}^3

The correct answer is obtained by applying the standard cone volume formula

V = \frac{1}{3}\pi r^2 h

. Substituting

r = 7

,

h = 12

, and

\pi = \frac{22}{7}

yields

V = 616\text{ cm}^3

.

Step-by-Step Solution

1

Identify the formula for the volume of a solid cone.

V = \frac{1}{3}\pi r^2 h

The volume of a cone is one-third the volume of a cylinder with the same base radius and height.

2

Substitute the given values into the formula.

V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 12

Given

r = 7\text{ cm}

,

h = 12\text{ cm}

, and

\pi = \frac{22}{7}

.

3

Simplify the expression to find the volume.

V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 12 = 22 \times 7 \times 4 = 616\text{ cm}^3

Simplifying

\frac{49}{7} = 7

and

\frac{12}{3} = 4

leaves

22 \times 7 \times 4

.

Key Concept

Volume of a Right Circular Cone

Estimated Time:45s

Question 49Question →

A straight line $L$ passes through the point $(4, -2)$ and is perpendicular to the line defined by the equation $3x - 2y + 8 = 0$ . What is the $y$ -intercept of line $L$ ?

\frac{2}{3}

-8

4

-\frac{14}{3}

Show answer & explanation

Answer: $\frac{2}{3}$

Answer

The

y

-intercept of line

L

is

\frac{2}{3}

.

The given line

3x - 2y + 8 = 0

has a gradient of

\frac{3}{2}

. A perpendicular line must have a gradient equal to the negative reciprocal, which is

-\frac{2}{3}

. Substituting the point

(4, -2)

into

y - y_1 = m(x - x_1)

gives

y + 2 = -\frac{2}{3}(x - 4)

, which simplifies to

y = -\frac{2}{3}x + \frac{2}{3}

. Setting

x = 0

yields the

y

-intercept of

\frac{2}{3}

.

Step-by-Step Solution

1

Find the gradient of the given line.

Expressing

3x - 2y + 8 = 0

in slope-intercept form

y = mx + c

gives

2y = 3x + 8 \implies y = \frac{3}{2}x + 4

. The gradient

m_1 = \frac{3}{2}

.

The gradient of the given line is required to determine the slope of line

L

.

2

Determine the gradient of line

L

.

Since line

L

is perpendicular to the given line,

m_L = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3}

.

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_L = -1

).

3

Find the equation of line

L

using point-slope form.

Using

(x_1, y_1) = (4, -2)

and

m_L = -\frac{2}{3}

:

y - y_1 = m_L(x - x_1)

y - (-2) = -\frac{2}{3}(x - 4)

y + 2 = -\frac{2}{3}x + \frac{8}{3}

y = -\frac{2}{3}x + \frac{8}{3} - 2

y = -\frac{2}{3}x + \frac{2}{3}

To find the

y

-intercept, we need the complete equation of line

L

.

4

Identify the

y

-intercept.

Comparing

y = -\frac{2}{3}x + \frac{2}{3}

to

y = mx + c

, the

y

-intercept

c = \frac{2}{3}

.

The constant term

c

in

y = mx + c

represents the

y

-intercept.

Key Concept

Perpendicular lines in coordinate geometry have gradients that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

).

Question 50Question →

A point $P$ lies inside a circle of radius $13\text{ cm}$ at a distance of $5\text{ cm}$ from the center $O$ . A chord $AB$ passes through point $P$ such that the ratio of segment $AP$ to segment $PB$ is $1:4$ . Calculate the total length of the chord $AB$ in centimeters.

Show answer & explanation

Answer: 30

Answer

The total length of chord

AB

is

30\text{ cm}

.

Using the Power of a Point property for an interior point

P

, the product of the chord segments is

AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 144

. Given

AP : PB = 1 : 4

, we write

AP = x

and

PB = 4x

, leading to

4x^2 = 144 \implies x = 6\text{ cm}

. Summing the two segments gives

AB = 6 + 24 = 30\text{ cm}

.

Step-by-Step Solution

1

Calculate the constant product of chord segments passing through interior point

P

.

AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144

.

By the intersecting chords theorem, the product of segments created by an interior point

P

on any chord equals

(R - d)(R + d) = R^2 - d^2

.

2

Set up an algebraic equation using the segment ratio

AP : PB = 1 : 4

.

Let

AP = x

and

PB = 4x

, giving

(x)(4x) = 144 \implies 4x^2 = 144

.

Expressing both chord segments in terms of a single variable

x

allows direct calculation of the segment lengths.

3

Solve for

x

to find the individual segment lengths.

x^2 = 36 \implies x = 6\text{ cm}

. Therefore,

AP = 6\text{ cm}

and

PB = 24\text{ cm}

.

Taking the positive square root gives the scale factor

x

since physical distances must be positive.

4

Sum the segment lengths to find the total chord length.

AB = AP + PB = 6 + 24 = 30\text{ cm}

.

The entire chord length is the sum of its two divided parts.

Key Concept

Intersecting Chords Theorem and Power of an Interior Point

Question 51Question →

What is the equation of the locus of a point $P(x, y)$ that moves in a plane such that its distance from the origin $(0,0)$ is always 5 units?

Show answer & explanation

Answer: x^2 + y^2 = 25; x^2+y^2=25; x²+y²=25

Answer

x^2 + y^2 = 25

By definition, the locus of a point moving at a fixed distance of 5 units from the origin

(0,0)

is a circle centered at

(0,0)

with radius 5. Substituting into the standard circle equation

x^2 + y^2 = r^2

yields

x^2 + y^2 = 5^2 = 25

.

Step-by-Step Solution

1

Apply the distance formula between a general point

P(x, y)

and the origin

(0,0)

.

d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}

The distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

.

2

Equate the distance formula to the given fixed distance of 5 units and square both sides.

\sqrt{x^2 + y^2} = 5 \implies x^2 + y^2 = 25

Squaring both sides eliminates the square root to give the algebraic equation of the locus.

Key Concept

The locus of points at a constant distance

r

from a fixed point

(h, k)

forms a circle with equation

(x-h)^2 + (y-k)^2 = r^2

.

Estimated Time:45s

Question 52Question →

Calculate the area, in square units, of the triangle formed by the straight line $3x + 4y - 24 = 0$ and the coordinate axes.

Show answer & explanation

Answer: 24

Answer

The area of the triangle formed by the line and the coordinate axes is 24 square units.

To find the area of the triangle bounded by a straight line and the coordinate axes, determine the magnitude of the

x

-intercept and

y

-intercept. Setting

y = 0

in

3x + 4y - 24 = 0

gives

x = 8

. Setting

x = 0

gives

y = 6

. The vertices of the right triangle are at

(0,0)

,

(8,0)

, and

(0,6)

. The area is calculated as

\frac{1}{2} \times 8 \times 6 = 24

square units.

Step-by-Step Solution

1

Find the

x

-intercept of the straight line

x = 8

, corresponding to the point

(8, 0)

Setting

y = 0

determines where the line crosses the horizontal axis

2

Find the

y

-intercept of the straight line

y = 6

, corresponding to the point

(0, 6)

Setting

x = 0

determines where the line crosses the vertical axis

3

Compute the area of the right-angled triangle formed with the origin

(0,0)

\text{Area} = \frac{1}{2} \times 8 \times 6 = 24

The coordinate axes are perpendicular, making the triangle right-angled with base length 8 and height 6

Key Concept

Area of a triangle bounded by a straight line and the coordinate axes

Estimated Time:1m 0s

Question 53Question →

A circle is inscribed in an isosceles trapezium $ABCD$ with parallel sides $AB$ and $CD$ . If $AB = 18\text{ cm}$ and $CD = 8\text{ cm}$ , what is the area of the region inside the trapezium that lies outside the circle?

(156 - 144\pi)\text{ cm}^2

(312 - 36\pi)\text{ cm}^2

(156 - 36\pi)\text{ cm}^2

(169 - 36\pi)\text{ cm}^2

Show answer & explanation

Answer: $(156 - 36\pi)\text{ cm}^2$

Answer

(156 - 36\pi)\text{ cm}^2

For an isosceles trapezium with an inscribed circle, the sum of opposite sides must be equal (

AB + CD = AD + BC = 26\text{ cm}

), giving slant side length

13\text{ cm}

. Using Pythagoras, the perpendicular distance (height) is

\sqrt{13^2 - 5^2} = 12\text{ cm}

. The area of the trapezium is

\frac{1}{2}(18 + 8)(12) = 156\text{ cm}^2

. The inscribed circle has radius equal to half the height (

6\text{ cm}

), so its area is

\pi \times 6^2 = 36\pi\text{ cm}^2

. Subtracting the circle area from the trapezium area yields

(156 - 36\pi)\text{ cm}^2

.

Step-by-Step Solution

1

Apply the property of a tangential quadrilateral to find the non-parallel sides

For a quadrilateral with an inscribed circle, the sum of opposite sides is equal:

AB + CD = AD + BC = 18 + 8 = 26\text{ cm}

. Since trapezium

ABCD

is isosceles,

AD = BC = 13\text{ cm}

.

Tangential quadrilaterals have equal sums of opposite side lengths.

2

Calculate the height

h

of the trapezium using the Pythagorean theorem

Dropping vertical altitudes from top vertices

C

and

D

creates right triangles at the base with horizontal leg

\frac{18 - 8}{2} = 5\text{ cm}

. Thus,

h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\text{ cm}

.

The height of the trapezium forms the perpendicular leg of the right-angled side triangle.

3

Find the area of the trapezium and the inscribed circle

Area of trapezium

= \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(18 + 8) \times 12 = 156\text{ cm}^2

. The diameter of the inscribed circle equals the height

h = 12\text{ cm}

, so its radius is

r = 6\text{ cm}

. Area of circle

= \pi r^2 = 36\pi\text{ cm}^2

.

The diameter of a circle inscribed between parallel bases equals the vertical height between those bases.

4

Subtract the area of the circle from the area of the trapezium

Remaining Area

= 156 - 36\pi\text{ cm}^2

.

The region inside the trapezium but outside the circle is the difference between their areas.

Key Concept

Perimeter and Area of Composite Figures and Inscribed Shapes

Question 54Question →

If the line $3x + py - 7 = 0$ is perpendicular to the line passing through the points $(1, -2)$ and $(4, 7)$ , what is the value of $p$ ?

9

-1

-9

3

Show answer & explanation

Answer: $9$

Answer

The value of

p

is

9

.

The line passing through

(1, -2)

and

(4, 7)

has a gradient of

m_1 = \frac{7 - (-2)}{4 - 1} = 3

. The equation

3x + py - 7 = 0

can be rewritten as

y = -\frac{3}{p}x + \frac{7}{p}

, giving a gradient of

m_2 = -\frac{3}{p}

. For perpendicular lines, the product of their gradients must equal

-1

, so

3 \times \left(-\frac{3}{p}\right) = -1

, which simplifies to

p = 9

.

Step-by-Step Solution

1

Calculate the gradient (

m_1

) of the line passing through

(1, -2)

and

(4, 7)

m_1 = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3

The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Express the line

3x + py - 7 = 0

in slope-intercept form (

y = mx + c

) to find its gradient (

m_2

)

py = -3x + 7 \implies y = -\frac{3}{p}x + \frac{7}{p}

, so

m_2 = -\frac{3}{p}

The coefficient of

x

when solved for

y

represents the gradient of the straight line.

3

Apply the perpendicularity condition

m_1 \cdot m_2 = -1

and solve for

p

3 \cdot \left(-\frac{3}{p}\right) = -1 \implies -\frac{9}{p} = -1 \implies p = 9

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

.

Key Concept

Perpendicular Lines and Gradients

Estimated Time:1m 30s

Question 55Question →

Find the value of $k$ if the point $P(k, 3)$ is equidistant from the points $A(1, 5)$ and $B(7, 1)$ .

Show answer & explanation

Answer: 4

Answer

The value of

k

is 4.

Using the distance formula, the squared distance

PA^2 = (k-1)^2 + (3-5)^2 = (k-1)^2 + 4

, and

PB^2 = (k-7)^2 + (3-1)^2 = (k-7)^2 + 4

. Equating

PA^2 = PB^2

gives

(k-1)^2 = (k-7)^2

. Expanding both sides yields

k^2 - 2k + 1 = k^2 - 14k + 49

. Subtracting

k^2

from both sides gives

12k = 48

, which leads to

k = 4

.

Step-by-Step Solution

1

Write the expressions for the squared distances

PA^2

and

PB^2

using the distance formula.

PA^2 = (k - 1)^2 + 4

and

PB^2 = (k - 7)^2 + 4

The distance formula between two points

(x_1, y_1)

and

(x_2, y_2)

is

d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2

.

2

Equate

PA^2

and

PB^2

since point

P

is equidistant from points

A

and

B

.

(k - 1)^2 + 4 = (k - 7)^2 + 4 \implies (k - 1)^2 = (k - 7)^2

Subtracting 4 from both sides simplifies the equality of squared distances.

3

Expand both sides and isolate

k

to find its numerical value.

k^2 - 2k + 1 = k^2 - 14k + 49 \implies 12k = 48 \implies k = 4

Canceling

k^2

terms yields a simple linear equation.

Key Concept

Equidistant points and the distance formula in coordinate geometry

Estimated Time:1m 30s

Question 56Question →

A solid trophy consists of a right circular cone mounted on top of a right circular cylinder with the same base radius of $6\text{ cm}$ . The height of the cylinder is $10\text{ cm}$ and the slant height of the cone is $10\text{ cm}$ . What is the total volume of the trophy?

216\pi\text{ cm}^3

456\pi\text{ cm}^3

480\pi\text{ cm}^3

648\pi\text{ cm}^3

Show answer & explanation

Answer: $456\pi\text{ cm}^3$

Answer

The total volume of the trophy is

456\pi\text{ cm}^3

.

First, use the Pythagorean theorem on the cone to find its vertical height

h_{\text{cone}} = \sqrt{10^2 - 6^2} = 8\text{ cm}

. The volume of the cone is

\frac{1}{3}\pi (6)^2 (8) = 96\pi\text{ cm}^3

. The volume of the cylinder is

\pi (6)^2 (10) = 360\pi\text{ cm}^3

. Adding both gives a total volume of

456\pi\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the vertical height of the cone

h_{\text{cone}} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}

The volume formula for a cone requires the vertical height (

h

), which forms a right-angled triangle with the radius (

r=6\text{ cm}

) and slant height (

l=10\text{ cm}

).

2

Calculate the volume of the conical part

V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \times \pi \times 6^2 \times 8 = 96\pi\text{ cm}^3

The formula for the volume of a right circular cone is

V = \frac{1}{3}\pi r^2 h

.

3

Calculate the volume of the cylindrical part

V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} = \pi \times 6^2 \times 10 = 360\pi\text{ cm}^3

The formula for the volume of a right circular cylinder is

V = \pi r^2 h

.

4

Sum the volumes of both 3D shapes to get the total volume

V_{\text{total}} = 96\pi + 360\pi = 456\pi\text{ cm}^3

The total volume of a composite solid is the sum of the volumes of its constituent parts.

Key Concept

Volume of Composite 3D Solids

Estimated Time:2m 0s

Question 57Question →

For the domain $0^\circ \le x \le 360^\circ$ , solve the trigonometric equation $2\sin x + \sqrt{3} = 0$ . Which of the following gives the complete set of values for $x$ ?

60^\circ

and

120^\circ

120^\circ

and

240^\circ

240^\circ

and

300^\circ

210^\circ

and

330^\circ

Show answer & explanation

Answer: $240^\circ$ and $300^\circ$

Answer

x = 240^\circ

and

x = 300^\circ

Rearranging

2\sin x + \sqrt{3} = 0

gives

\sin x = -\frac{\sqrt{3}}{2}

. The reference angle for which sine equals

\frac{\sqrt{3}}{2}

is

60^\circ

. Since sine is negative in the third and fourth quadrants, the solutions are

180^\circ + 60^\circ = 240^\circ

and

360^\circ - 60^\circ = 300^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric function in the equation

\sin x = -\frac{\sqrt{3}}{2}

Subtract

\sqrt{3}

from both sides and divide by

2

.

2

Find the basic reference angle

Reference angle

\alpha = 60^\circ

\sin(60^\circ) = \frac{\sqrt{3}}{2}

.

3

Determine the relevant quadrants

Quadrants III and IV

The sine function is negative in the third and fourth quadrants.

4

Calculate the solutions within the interval

0^\circ \le x \le 360^\circ

Quadrant III:

x = 180^\circ + 60^\circ = 240^\circ

; Quadrant IV:

x = 360^\circ - 60^\circ = 300^\circ

Apply quadrant reduction formulas for Quadrant III (

180^\circ + \alpha

) and Quadrant IV (

360^\circ - \alpha

).

Key Concept

Solving simple trigonometric equations using reference angles and quadrant rules

Question 58Question →

Find the equation of the locus of a point $P(x, y)$ that moves such that it is equidistant from the fixed points $A(-1, 3)$ and $B(5, -1)$ .

Show answer & explanation

Answer: 3x - 2y - 4 = 0; 3x-2y-4=0; 3x - 2y = 4; 3x-2y=4; 12x - 8y - 16 = 0; y = (3/2)x - 2; y = 1.5x - 2

Answer

The equation of the locus is

3x - 2y - 4 = 0

(or

3x - 2y = 4

).

The locus of a point equidistant from two fixed points

A(-1, 3)

and

B(5, -1)

is the perpendicular bisector of the line segment joining them. Equating the squared distances

(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2

and simplifying yields the linear equation

3x - 2y - 4 = 0

.

Step-by-Step Solution

1

Set up the distance equality condition using the distance formula.

\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}

Since point

P(x, y)

is equidistant from

A

and

B

,

PA = PB

.

2

Square both sides to remove the radical signs.

(x + 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2

Squaring both sides eliminates square roots and simplifies polynomial expansion.

3

Expand all squared terms on both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 + 2y + 1

Expanding allows gathering like terms.

4

Cancel

x^2

and

y^2

from both sides and collect all terms on one side.

(2x + 10x) + (-6y - 2y) + (10 - 26) = 0 \\Rightarrow 12x - 8y - 16 = 0

Combining like terms simplifies the locus equation into standard linear form.

5

Divide the entire equation by the common factor of 4.

3x - 2y - 4 = 0

Expressing the linear equation in its simplest form gives the perpendicular bisector of line segment

AB

.

Key Concept

Locus equidistant from two fixed points (Perpendicular Bisector)

Estimated Time:2m 0s

Question 59Question →

A sine function is given by the equation $y = 5\sin\left(\frac{2}{3}x\right) - 2$ . What is the period of this trigonometric function in degrees?

Show answer & explanation

Answer: 540

Answer

The period of the trigonometric function is

540^\circ

.

For a general sine curve of the form

y = a\sin(bx + c) + d

, the period

T

in degrees is calculated as

T = \frac{360^\circ}{|b|}

. Given

y = 5\sin\left(\frac{2}{3}x\right) - 2

, the value of

b

is

\frac{2}{3}

. Dividing

360^\circ

by

\frac{2}{3}

gives

540^\circ

.

Step-by-Step Solution

1

Identify the coefficient

b

of

x

from the standard sine form

y = a\sin(bx + c) + d

.

b = \frac{2}{3}

The coefficient of

x

determines the angular frequency and affects the horizontal compression or stretch of the graph.

2

State the period formula in degrees for a sine function.

T = \frac{360^\circ}{|b|}

The standard sine function completes one full wavelength over

360^\circ

, so scaling the input by

b

changes the period to

\frac{360^\circ}{b}

.

3

Substitute

b = \frac{2}{3}

into the formula and evaluate.

T = \frac{360^\circ}{\frac{2}{3}} = 360^\circ \times \frac{3}{2} = 540^\circ

Dividing by a fraction is performed by multiplying by its reciprocal.

Key Concept

Period of Trigonometric Functions

Estimated Time:1m 30s

Question 60Question →

The ratio of the measure of an interior angle to an exterior angle of a regular polygon is $5 : 1$ . How many sides does the polygon have?

10

12

6

14

Show answer & explanation

Answer: 12

Answer

The polygon has 12 sides.

The interior angle and exterior angle of a regular polygon are supplementary, adding up to

180^\circ

. For a ratio of

5:1

, the exterior angle is

\frac{1}{5+1} \times 180^\circ = 30^\circ

. Using the exterior angle formula

n = \frac{360^\circ}{\text{exterior angle}}

, the number of sides is

\frac{360^\circ}{30^\circ} = 12

.

Step-by-Step Solution

1

Set up an equation using the linear pair relationship between interior and exterior angles.

5x + 1x = 180^\circ \implies 6x = 180^\circ \implies x = 30^\circ

At any vertex of a regular polygon, the interior angle and exterior angle lie on a straight line and are supplementary (

180^\circ

).

2

Identify the measure of the exterior angle.

Exterior angle =

30^\circ

The exterior angle corresponds to

1

part of the

5:1

ratio, which equals

x = 30^\circ

.

3

Calculate the number of sides

n

of the polygon.

n = \frac{360^\circ}{30^\circ} = 12

The sum of the exterior angles of any convex polygon is

360^\circ

, so

n = \frac{360^\circ}{\text{exterior angle}}

.

Key Concept

Interior and exterior angles of regular polygons

Geometry and Trigonometry

Answer

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