Geometry and Trigonometry

184 soru

Soru 61Soru

The sum of the interior angles of a regular polygon is 14401440^\circ. What is the measure, in degrees, of one exterior angle of this polygon?

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Cevap: 36

Cevap

The measure of one exterior angle of the polygon is 3636^\circ.
The sum of the interior angles of an nn-sided polygon is given by (n2)×180(n-2) \times 180^\circ. Setting (n2)×180=1440(n-2) \times 180^\circ = 1440^\circ gives n2=8n-2 = 8, so the polygon has n=10n = 10 sides (a decagon). The measure of each exterior angle of a regular polygon is 360n=36010=36\frac{360^\circ}{n} = \frac{360^\circ}{10} = 36^\circ.

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1
Set up the equation for the sum of interior angles of an nn-sided polygon.
(n2)×180=1440(n - 2) \times 180^\circ = 1440^\circ
The sum of interior angles of any convex nn-sided polygon is (n2)×180(n - 2) \times 180^\circ.
2
Solve for nn, the number of sides.
n2=1440180=8    n=10n - 2 = \frac{1440}{180} = 8 \implies n = 10
Dividing the interior angle sum by 180180^\circ gives n2n - 2.
3
Calculate the measure of one exterior angle.
Exterior angle =36010=36= \frac{360^\circ}{10} = 36^\circ
The sum of exterior angles of any convex polygon is 360360^\circ, so each exterior angle of a regular polygon with nn sides is 360n\frac{360^\circ}{n}.

Anahtar Kavram

Relationship between interior angle sum, number of sides, and exterior angles of a regular polygon.
Soru 62Soru

The interior angles of a convex pentagon are given as (2x+10)(2x + 10)^\circ, (x+25)(x + 25)^\circ, (3x15)(3x - 15)^\circ, (2x+30)(2x + 30)^\circ, and (2x+10)(2x + 10)^\circ. What is the value of xx?

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Cevap: 48

Cevap

The value of xx is 48.
For a pentagon (n=5n = 5), the sum of interior angles is (52)×180=540(5 - 2) \times 180^\circ = 540^\circ. Summing the five given interior angle expressions yields 10x+6010x + 60. Equating 10x+60=54010x + 60 = 540 gives 10x=48010x = 480, which solves to x=48x = 48.

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1
Calculate the sum of interior angles of a 5-sided polygon (pentagon)
Sum =(52)×180=3×180=540= (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ
The formula for the sum of interior angles of a polygon with nn sides is (n2)×180(n - 2) \times 180^\circ.
2
Sum the given algebraic expressions for the interior angles
(2x+10)+(x+25)+(3x15)+(2x+30)+(2x+10)=10x+60(2x + 10) + (x + 25) + (3x - 15) + (2x + 30) + (2x + 10) = 10x + 60
Combine like terms for the xx terms and constant terms.
3
Equate the sum of expressions to 540540^\circ and solve for xx
10x+60=540    10x=480    x=4810x + 60 = 540 \implies 10x = 480 \implies x = 48
Subtract 60 from both sides and divide by 10 to isolate xx.

Anahtar Kavram

Interior Angle Sum of Polygons
Tahmini Süre:1m 15s
Soru 63Soru

In a regular polygon, the measure of each interior angle is 132132^\circ greater than the measure of each exterior angle. How many sides does this polygon have?

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Cevap: 15

Cevap

The polygon has 15 sides.
Since the interior angle II and exterior angle EE of a regular polygon sum to 180180^\circ (I+E=180I + E = 180^\circ) and their given difference is IE=132I - E = 132^\circ, subtracting the difference equation from the sum equation gives 2E=482E = 48^\circ, which simplifies to E=24E = 24^\circ. The number of sides is n=360E=36024=15n = \frac{360^\circ}{E} = \frac{360^\circ}{24^\circ} = 15.

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1
Set up the linear pair equation for interior and exterior angles
I+E=180I + E = 180^\circ
An interior angle and its adjacent exterior angle at any vertex of a polygon lie on a straight line and sum to 180180^\circ.
2
Set up the given condition equation
IE=132I - E = 132^\circ
The question states that each interior angle is 132132^\circ greater than each exterior angle.
3
Solve for the exterior angle EE
E=24E = 24^\circ
Subtracting IE=132I - E = 132^\circ from I+E=180I + E = 180^\circ yields 2E=482E = 48^\circ, giving E=24E = 24^\circ.
4
Calculate the number of sides nn
n=15n = 15
The sum of all exterior angles of any convex polygon is 360360^\circ, so n=360E=36024=15n = \frac{360^\circ}{E} = \frac{360^\circ}{24^\circ} = 15.

Anahtar Kavram

Interior and Exterior Angle Properties of Regular Polygons
Soru 64Soru

Determine the equation of the locus of a point P(x,y)P(x, y) that is equidistant from the two parallel lines 2x3y+6=02x - 3y + 6 = 0 and 2x3y4=02x - 3y - 4 = 0.

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Cevap: 2x - 3y + 1 = 0; 2x-3y+1=0; 2x - 3y = -1; 2x-3y=-1

Cevap

2x3y+1=02x - 3y + 1 = 0
The locus of points equidistant from two parallel lines ax+by+c1=0ax + by + c_1 = 0 and ax+by+c2=0ax + by + c_2 = 0 is a parallel line midway between them, defined by ax+by+c1+c22=0ax + by + \frac{c_1 + c_2}{2} = 0. Substituting c1=6c_1 = 6 and c2=4c_2 = -4 yields 6+(4)2=1\frac{6 + (-4)}{2} = 1, giving the equation 2x3y+1=02x - 3y + 1 = 0.

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1
Identify the geometric principle for the locus between two parallel lines.
The locus of points equidistant from two parallel lines ax+by+c1=0ax + by + c_1 = 0 and ax+by+c2=0ax + by + c_2 = 0 is a third parallel line given by ax+by+c1+c22=0ax + by + \frac{c_1 + c_2}{2} = 0.
Points equidistant from two parallel lines lie on a parallel line midway between them.
2
Calculate the average of the constant terms c1=6c_1 = 6 and c2=4c_2 = -4.
cmid=6+(4)2=22=1c_{mid} = \frac{6 + (-4)}{2} = \frac{2}{2} = 1.
The midpoint constant term is the arithmetic mean of the two original constants.
3
Construct the equation of the locus line.
2x3y+1=02x - 3y + 1 = 0.
Combining the common linear coefficients 2x3y2x - 3y with the calculated midpoint constant 11 gives the required equation.

Anahtar Kavram

Locus equidistant from two parallel lines
Tahmini Süre:1m 30s
Soru 65Soru

An open storage container is designed in the shape of a frustum of a right circular cone. The top radius of the container is 4 m4\text{ m}, the bottom radius is 1 m1\text{ m}, and its vertical height is 7 m7\text{ m}. Taking π=227\pi = \frac{22}{7}, what is the volume of the container in cubic metres (m3\text{m}^3)?

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Cevap: 154

Cevap

The volume of the container is 154 m3154\text{ m}^3.
Using the volume formula for a frustum of a cone V=13πh(R2+Rr+r2)V = \frac{1}{3}\pi h (R^2 + Rr + r^2) with R=4 mR = 4\text{ m}, r=1 mr = 1\text{ m}, h=7 mh = 7\text{ m}, and π=227\pi = \frac{22}{7} yields V=13×227×7×(16+4+1)=22×7=154 m3V = \frac{1}{3} \times \frac{22}{7} \times 7 \times (16 + 4 + 1) = 22 \times 7 = 154\text{ m}^3.

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1
Identify the given dimensions of the frustum of the cone.
Top radius R=4 mR = 4\text{ m}, bottom radius r=1 mr = 1\text{ m}, height h=7 mh = 7\text{ m}, and π=227\pi = \frac{22}{7}.
Establishing known parameter values for the frustum volume formula.
2
Apply the standard formula for the volume of a frustum of a circular cone.
V=13πh(R2+Rr+r2)V = \frac{1}{3}\pi h (R^2 + Rr + r^2)
This formula accounts for the non-uniform cross-sectional area of a truncated cone.
3
Calculate the sum of the squares and the product of the radii.
R2+Rr+r2=42+(4)(1)+12=16+4+1=21R^2 + Rr + r^2 = 4^2 + (4)(1) + 1^2 = 16 + 4 + 1 = 21
Evaluating the quadratic radius factor in the frustum equation.
4
Substitute all numeric values and simplify.
V=13×227×7×21=22×7=154 m3V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 21 = 22 \times 7 = 154\text{ m}^3
Simplifying by cancelling 77 in the numerator and denominator, then dividing 2121 by 33 yields 22×7=15422 \times 7 = 154.

Anahtar Kavram

Volume of a Frustum of a Cone
Soru 66Soru

A point P(x,y)P(x, y) moves in a plane such that the line segment joining the fixed points A(1,2)A(1, 2) and B(5,6)B(5, 6) subtends a right angle at PP. Which of the following equations represents the locus of PP?

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Cevap: x2+y26x8y+17=0x^2 + y^2 - 6x - 8y + 17 = 0

Cevap

The equation of the locus of PP is x2+y26x8y+17=0x^2 + y^2 - 6x - 8y + 17 = 0.
The locus of a point PP that subtends a 9090^\circ angle at two fixed points A(1,2)A(1, 2) and B(5,6)B(5, 6) forms a circle with ABAB as diameter. Using the gradient condition for perpendicular lines, y2x1×y6x5=1\frac{y-2}{x-1} \times \frac{y-6}{x-5} = -1, which simplifies to x2+y26x8y+17=0x^2 + y^2 - 6x - 8y + 17 = 0.

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1
Apply the perpendicularity condition for the line segments APAP and BPBP.
Gradient of AP=y2x1AP = \frac{y - 2}{x - 1} and gradient of BP=y6x5BP = \frac{y - 6}{x - 5}. Since APB=90\angle APB = 90^\circ, their product must be 1-1: (y2x1)(y6x5)=1\left(\frac{y - 2}{x - 1}\right) \cdot \left(\frac{y - 6}{x - 5}\right) = -1.
Two perpendicular line segments have gradients whose product is 1-1.
2
Multiply out the denominators and numerators.
(y - 2)(y - 6) = -(x - 1)(x - 5) \implies y^2 - 8y + 12 = -(x^2 - 6x + 5).
Algebraic expansion of the equation obtained from the gradient product.
3
Rearrange all terms to one side to express in standard second-degree form.
x^2 + y^2 - 6x - 8y + 17 = 0.
Rearranging yields the Cartesian equation of the locus.

Anahtar Kavram

Locus of a point subtending a right angle at two fixed points
Tahmini Süre:1m 30s
Soru 67Soru

An isosceles trapezium has parallel sides of lengths 14 cm14\text{ cm} and 8 cm8\text{ cm}. If each of the non-parallel sides has a length of 5 cm5\text{ cm}, calculate the area of the trapezium in cm2\text{cm}^2.

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Cevap: 44

Cevap

The area of the trapezium is 44 cm244\text{ cm}^2.
Projecting the top base of length 8 cm8\text{ cm} onto the bottom base of length 14 cm14\text{ cm} leaves a difference of 6 cm6\text{ cm}, which is divided equally into two 3 cm3\text{ cm} segments on either side. Using the Pythagorean theorem with the non-parallel side (5 cm5\text{ cm}) and the base segment (3 cm3\text{ cm}) gives a height of 4 cm4\text{ cm}. The area is then calculated as 12×(14+8)×4=44 cm2\frac{1}{2} \times (14 + 8) \times 4 = 44\text{ cm}^2.

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1
Determine the projection segment length on the longer base
x=1482=3 cmx = \frac{14 - 8}{2} = 3\text{ cm}
Since the trapezium is isosceles, dropping perpendiculars from both ends of the top base creates two identical right-angled triangles at the sides.
2
Calculate the perpendicular height using the Pythagorean theorem
h=5232=16=4 cmh = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{ cm}
The slant side (5 cm5\text{ cm}), the height (hh), and the projection segment (3 cm3\text{ cm}) form a right-angled triangle.
3
Calculate the area of the trapezium
\text{Area} = \frac{1}{2}(14 + 8) \times 4 = 44\text{ cm}^2
The area of a trapezium is given by half the sum of its parallel sides multiplied by its perpendicular height.

Anahtar Kavram

Perimeter and Area of Plane Shapes - Area of Isosceles Trapezium
Soru 68Soru

Given that θ\theta is an acute angle satisfying secθ+tanθ=3\sec \theta + \tan \theta = 3, what is the exact numerical value of 10sinθ10 \sin \theta?

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Cevap: 8

Cevap

The numerical value of 10sinθ10 \sin \theta is 8.
Using the standard fundamental identity sec2θtan2θ=1\sec^2 \theta - \tan^2 \theta = 1, we factor it into (secθ+tanθ)(secθtanθ)=1(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1. Substituting the given value secθ+tanθ=3\sec \theta + \tan \theta = 3 gives secθtanθ=13\sec \theta - \tan \theta = \frac{1}{3}. Solving these linear equations yields secθ=53\sec \theta = \frac{5}{3} (so cosθ=35\cos \theta = \frac{3}{5}) and tanθ=43\tan \theta = \frac{4}{3}. Using sinθ=tanθcosθ\sin \theta = \tan \theta \cdot \cos \theta, we find sinθ=45=0.8\sin \theta = \frac{4}{5} = 0.8. Multiplying by 10 gives the exact result 8.

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1
Apply the trigonometric identity sec2θtan2θ=1\sec^2 \theta - \tan^2 \theta = 1
(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1
Difference of squares factorization links secθ+tanθ\sec \theta + \tan \theta and secθtanθ\sec \theta - \tan \theta as reciprocals.
2
Substitute secθ+tanθ=3\sec \theta + \tan \theta = 3 to find secθtanθ\sec \theta - \tan \theta
\sec \theta - \tan \theta = \frac{1}{3}
Dividing both sides of 3(secθtanθ)=13(\sec \theta - \tan \theta) = 1 by 3.
3
Solve the system of equations for secθ\sec \theta and tanθ\tan \theta
\sec \theta = \frac{5}{3} \text{ and } \tan \theta = \frac{4}{3}
Adding equations gives 2secθ=1032\sec \theta = \frac{10}{3}; subtracting gives 2tanθ=832\tan \theta = \frac{8}{3}.
4
Calculate sinθ\sin \theta and evaluate 10sinθ10 \sin \theta
\sin \theta = \frac{4}{5} \implies 10 \sin \theta = 8
Since cosθ=35\cos \theta = \frac{3}{5} and tanθ=43\tan \theta = \frac{4}{3}, sinθ=tanθcosθ=45\sin \theta = \tan \theta \cdot \cos \theta = \frac{4}{5}.

Anahtar Kavram

Trigonometric identities relating secant and tangent ratios
Soru 69Soru

Which set contains all the solutions to the trigonometric equation sin2x=cosx\sin 2x = \cos x for 0x1800^\circ \le x \le 180^\circ?

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Cevap: {30,90,150}\{30^\circ, 90^\circ, 150^\circ\}

Cevap

The correct set of solutions is \{30^\circ, 90^\circ, 150^\circ\}.
Using the identity sin2x=2sinxcosx\sin 2x = 2\sin x \cos x, the equation becomes 2sinxcosxcosx=02\sin x \cos x - \cos x = 0. Factoring out cosx\cos x gives cosx(2sinx1)=0\cos x(2\sin x - 1) = 0. Setting each factor to zero yields cosx=0\cos x = 0 (giving x=90x = 90^\circ) and sinx=12\sin x = \frac{1}{2} (giving x=30x = 30^\circ and x=150x = 150^\circ). Thus, the complete set of solutions in the given interval is \{30^\circ, 90^\circ, 150^\circ\}.

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1
Apply the double-angle identity for sine.
Substitute sin2x=2sinxcosx\sin 2x = 2\sin x \cos x into the equation to get 2sinxcosx=cosx2\sin x \cos x = \cos x.
This expresses the equation in terms of single angle xx.
2
Rearrange and factor the equation.
2sinxcosxcosx=0    cosx(2sinx1)=02\sin x \cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0.
Factoring prevents losing valid roots that occur when a variable factor equals zero.
3
Set each factor to zero and solve for xx in the interval 0x1800^\circ \le x \le 180^\circ.
First factor: cosx=0    x=90\cos x = 0 \implies x = 90^\circ.
Second factor: 2sinx1=0    sinx=12    x=302\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = 30^\circ or x=18030=150x = 180^\circ - 30^\circ = 150^\circ.
Finding all principal and secondary angles within the specified domain.
4
Combine all valid solutions into a set.
x{30,90,150}x \in \{30^\circ, 90^\circ, 150^\circ\}.
All three values satisfy the original equation and lie within 0x1800^\circ \le x \le 180^\circ.

Anahtar Kavram

Solving trigonometric equations using identities and factoring
Tahmini Süre:1m 30s
Soru 70Soru

Which of the following sets contains all values of xx in the interval 0x3600^\circ \le x \le 360^\circ that satisfy the trigonometric equation 3sinx+cosx=0\sqrt{3}\sin x + \cos x = 0?

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Cevap: 150 and 330150^\circ \text{ and } 330^\circ

Cevap

150 and 330150^\circ \text{ and } 330^\circ
The given equation 3sinx+cosx=0\sqrt{3}\sin x + \cos x = 0 simplifies to tanx=13\tan x = -\frac{1}{\sqrt{3}}. Since tangent is negative in the second and fourth quadrants with a reference angle of 3030^\circ, the solutions in the domain 0x3600^\circ \le x \le 360^\circ are 18030=150180^\circ - 30^\circ = 150^\circ and 36030=330360^\circ - 30^\circ = 330^\circ.

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1
Rearrange the trigonometric equation into single ratio form
3sinx=cosx    sinxcosx=13    tanx=13\sqrt{3}\sin x = -\cos x \implies \frac{\sin x}{\cos x} = -\frac{1}{\sqrt{3}} \implies \tan x = -\frac{1}{\sqrt{3}}
Dividing both sides by cosx\cos x converts the sum of sine and cosine terms into a simple tangent equation.
2
Determine the reference angle
Reference angle α=30\text{Reference angle } \alpha = 30^\circ
The acute angle whose tangent is 13\frac{1}{\sqrt{3}} is 3030^\circ.
3
Identify the quadrants and find all solutions in 0x3600^\circ \le x \le 360^\circ
x=18030=150x = 180^\circ - 30^\circ = 150^\circ (Quadrant II) and x=36030=330x = 360^\circ - 30^\circ = 330^\circ (Quadrant IV)
The tangent function is negative in Quadrants II and IV.

Anahtar Kavram

Solving simple trigonometric equations by reducing to a basic ratio and finding all solutions within a given domain.
Tahmini Süre:1m 30s
Soru 71Soru

In triangle XYZXYZ, side length x=3 cmx = 3\text{ cm}, side length y=8 cmy = 8\text{ cm}, and the included angle Z=60\angle Z = 60^\circ. What is the length of side zz in cm?

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Cevap: 7

Cevap

The length of side zz is 7 cm7\text{ cm}.
Using the Cosine Rule z2=x2+y22xycosZz^2 = x^2 + y^2 - 2xy \cos Z, substituting x=3x = 3, y=8y = 8, and Z=60\angle Z = 60^\circ yields z2=32+822(3)(8)(0.5)=9+6424=49z^2 = 3^2 + 8^2 - 2(3)(8)(0.5) = 9 + 64 - 24 = 49. Taking the positive square root gives z=7 cmz = 7\text{ cm}.

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1
Identify known triangle components and select the appropriate rule
Two sides and the included angle (SAS) are given: x=3x = 3, y=8y = 8, Z=60\angle Z = 60^\circ, requiring the Cosine Rule.
When given two sides and the included angle (SAS), the Cosine Rule is used to find the third side.
2
Substitute values into the Cosine Rule formula z2=x2+y22xycosZz^2 = x^2 + y^2 - 2xy \cos Z
z2=32+822(3)(8)cos60z^2 = 3^2 + 8^2 - 2(3)(8)\cos 60^\circ
Direct algebraic substitution of side lengths and angle measure into the Cosine Rule.
3
Evaluate the trigonometric term and simplify the arithmetic expression
z2=9+6448(0.5)=7324=49z^2 = 9 + 64 - 48(0.5) = 73 - 24 = 49
The exact value of cos60\cos 60^\circ is 0.50.5.
4
Solve for side length zz by taking the principal square root
z=49=7 cmz = \sqrt{49} = 7\text{ cm}
Side length must be positive.

Anahtar Kavram

Applying the Cosine Rule to find the third side of a non-right-angled triangle given two sides and an included angle (SAS).
Tahmini Süre:1m 30s
Soru 72Soru

A circle has a circumference of 44 cm44\text{ cm}. What is the area of the circle? (Take π=227\pi = \frac{22}{7})

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Cevap: 154 cm2154\text{ cm}^2

Cevap

The area of the circle is 154 cm2154\text{ cm}^2.
First, find the radius from the circumference: C=2πr    44=2×227×r    r=7 cmC = 2\pi r \implies 44 = 2 \times \frac{22}{7} \times r \implies r = 7\text{ cm}. Then compute the area: A=πr2=227×72=154 cm2A = \pi r^2 = \frac{22}{7} \times 7^2 = 154\text{ cm}^2. Therefore, the value 154 cm2154\text{ cm}^2 is correct.

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1
Calculate the radius of the circle using the circumference formula
r=7 cmr = 7\text{ cm}
The formula for circumference is C=2πrC = 2\pi r. Substituting C=44C = 44 and π=227\pi = \frac{22}{7} yields 44=2×227×r44 = 2 \times \frac{22}{7} \times r, which simplifies to 447r=44\frac{44}{7}r = 44, so r=7 cmr = 7\text{ cm}.
2
Calculate the area using the area formula for a circle
Area = 154 cm2154\text{ cm}^2
The area formula is A=πr2A = \pi r^2. Substituting r=7r = 7 and π=227\pi = \frac{22}{7} gives A=227×72=22×7=154 cm2A = \frac{22}{7} \times 7^2 = 22 \times 7 = 154\text{ cm}^2.

Anahtar Kavram

Relationship between circumference and area of a circle
Soru 73Soru

Two regular polygons, P1P_1 and P2P_2, have nn sides and 2n2n sides respectively. If the sum of one interior angle of P1P_1 and one exterior angle of P2P_2 is 150150^\circ, what is the total number of diagonals of polygon P2P_2?

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Cevap: 54

Cevap

The total number of diagonals of polygon P2P_2 is 54.
The interior angle of an nn-sided polygon is 180360n180^\circ - \frac{360^\circ}{n} and the exterior angle of a 2n2n-sided polygon is 180n\frac{180^\circ}{n}. Adding these gives 180180n=150180^\circ - \frac{180^\circ}{n} = 150^\circ, which yields n=6n = 6. Polygon P2P_2 therefore has 1212 sides, and the number of diagonals is 12×92=54\frac{12 \times 9}{2} = 54.

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1
Express the interior angle of P1P_1 and the exterior angle of P2P_2 in terms of nn.
Interior angle of P1=180360nP_1 = 180^\circ - \frac{360^\circ}{n}; Exterior angle of P2=3602n=180nP_2 = \frac{360^\circ}{2n} = \frac{180^\circ}{n}.
The interior angle of an nn-sided regular polygon is 180360n180^\circ - \frac{360^\circ}{n}, and the exterior angle of a 2n2n-sided regular polygon is 3602n\frac{360^\circ}{2n}.
2
Set up and solve the angle sum equation.
\left(180^\circ - \frac{360^\circ}{n}\right) + \frac{180^\circ}{n} = 150^\circ \implies 180^\circ - \frac{180^\circ}{n} = 150^\circ \implies \frac{180^\circ}{n} = 30^\circ \implies n = 6$.
The sum of the two angles is given as 150150^\circ.
3
Determine the number of sides of polygon P2P_2.
Polygon P2P_2 has 2n=2(6)=122n = 2(6) = 12 sides.
Polygon P2P_2 has 2n2n sides.
4
Calculate the total number of diagonals for a polygon with 12 sides using D=k(k3)2D = \frac{k(k-3)}{2}.
D = \frac{12(12 - 3)}{2} = \frac{12 \times 9}{2} = 54.
The formula for the number of diagonals in a polygon with kk sides is k(k3)2\frac{k(k-3)}{2}.

Anahtar Kavram

Interior and exterior angles of regular polygons and the polygon diagonal count formula
Soru 74Soru

A composite plane figure is formed by joining a major sector of a circle of radius 14 cm14\text{ cm} (having a central angle of 270270^\circ and center OO) to a square OACBOACB. The vertices AA and BB of the square lie on the circle such that OAOA and OBOB serve as the bounding radii of the major sector. What is the total area of the combined figure in cm2\text{cm}^2? (Take π=227\pi = \frac{22}{7})

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Cevap: 658 cm2658\text{ cm}^2

Cevap

658 cm2658\text{ cm}^2
The total area of the composite figure is the sum of the non-overlapping regions: the major sector of central angle 270270^\circ (462 cm2462\text{ cm}^2) and the square of side length 14 cm14\text{ cm} (196 cm2196\text{ cm}^2), giving 462+196=658 cm2462 + 196 = 658\text{ cm}^2.

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1
Calculate the area of the major sector of central angle 270270^\circ and radius 14 cm14\text{ cm}.
\text{Area}_{\text{sector}} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times \frac{22}{7} \times 14^2 = \frac{3}{4} \times 616 = 462\text{ cm}^2
The major sector covers 270360=34\frac{270}{360} = \frac{3}{4} of the full circle.
2
Calculate the area of the square OACBOACB with side length s=OA=14 cms = OA = 14\text{ cm}.
\text{Area}_{\text{square}} = s^2 = 14^2 = 196\text{ cm}^2
The radii OAOA and OBOB form two adjacent sides of the square OACBOACB of length 14 cm14\text{ cm}.
3
Sum the areas of the major sector and the square to find the total composite area.
\text{Total Area} = 462\text{ cm}^2 + 196\text{ cm}^2 = 658\text{ cm}^2
The major sector and square share boundaries OAOA and OBOB without interior overlap.

Anahtar Kavram

Perimeter and Area of Plane Shapes
Tahmini Süre:2m 0s
Soru 75Soru

Find the equation of the locus of a point P(x,y)P(x, y) that moves such that its distance from the fixed point (3,0)(3, 0) is always equal to its perpendicular distance from the vertical line x=3x = -3.

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Cevap: y^2 = 12x; y^2 - 12x = 0; y^2=12x; y^2 - 12x = 0

Cevap

The equation of the locus is y2=12xy^2 = 12x (or y212x=0y^2 - 12x = 0).
Equating the distance from P(x,y)P(x, y) to (3,0)(3, 0), which is (x3)2+y2\sqrt{(x - 3)^2 + y^2}, and the distance from P(x,y)P(x, y) to x=3x = -3, which is x+3|x + 3|, squaring both sides gives x26x+9+y2=x2+6x+9x^2 - 6x + 9 + y^2 = x^2 + 6x + 9. Subtracting x2+9x^2 + 9 from both sides yields y2=12xy^2 = 12x.

Adım Adım Çözüm

1
Express the distance from P(x,y)P(x, y) to the point (3,0)(3, 0) using the distance formula.
d1=(x3)2+(y0)2=(x3)2+y2d_1 = \sqrt{(x - 3)^2 + (y - 0)^2} = \sqrt{(x - 3)^2 + y^2}
The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by (x2x1)2+(y2y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
2
Express the perpendicular distance from P(x,y)P(x, y) to the line x=3x = -3.
d2=x(3)=x+3d_2 = |x - (-3)| = |x + 3|
The perpendicular distance from a point (x,y)(x, y) to a vertical line x=kx = k is given by xk|x - k|.
3
Set the two distance expressions equal according to the locus condition.
(x3)2+y2=x+3\sqrt{(x - 3)^2 + y^2} = |x + 3|
The locus condition states that the distance to (3,0)(3, 0) is equal to the distance to line x=3x = -3.
4
Square both sides and simplify to obtain the Cartesian equation.
(x3)2+y2=(x+3)2    x26x+9+y2=x2+6x+9    y2=12x(x - 3)^2 + y^2 = (x + 3)^2 \implies x^2 - 6x + 9 + y^2 = x^2 + 6x + 9 \implies y^2 = 12x
Squaring eliminates the square root and absolute value signs, leading to the algebraic representation of the locus.

Anahtar Kavram

Definition and equation of a parabola as the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix)

Alternatif Yöntem

Recognize that the definition of a parabola is the locus of points equidistant from a focus (a,0)(a, 0) and a directrix x=ax = -a. Here a=3a = 3, so the standard equation y2=4axy^2 = 4ax directly gives y2=4(3)x=12xy^2 = 4(3)x = 12x.
Tahmini Süre:1m 30s
Soru 76Soru

A sector of a circle of radius 6 cm6\text{ cm} has a total perimeter whose numerical value is equal to the numerical value of its area. What is the area of the sector, in cm2\text{cm}^2?

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Cevap: 18

Cevap

The area of the sector is 18 cm218\text{ cm}^2.
For a sector of radius r=6 cmr = 6\text{ cm}, its total perimeter is P=2(6)+s=12+sP = 2(6) + s = 12 + s, where ss is the arc length. Its area is A=12(6)s=3sA = \frac{1}{2}(6)s = 3s. Setting P=AP = A gives 12+s=3s    2s=12    s=6 cm12 + s = 3s \implies 2s = 12 \implies s = 6\text{ cm}. Substituting s=6s = 6 into the area expression gives A=3(6)=18 cm2A = 3(6) = 18\text{ cm}^2.

Adım Adım Çözüm

1
Express the perimeter and area of the sector in terms of the arc length ss
P=2r+s=12+sP = 2r + s = 12 + s and A=12rs=3sA = \frac{1}{2}rs = 3s
The total perimeter of a sector includes two straight radii plus the arc length, while its area is given by 12rs\frac{1}{2}rs.
2
Equate the numerical values of perimeter and area
12+s=3s12 + s = 3s
The question specifies that the numerical value of the total perimeter equals the numerical value of its area.
3
Solve for the arc length ss
2s=12    s=6 cm2s = 12 \implies s = 6\text{ cm}
Subtracting ss from both sides yields 2s=122s = 12, so s=6 cms = 6\text{ cm}.
4
Calculate the sector area
A=3(6)=18 cm2A = 3(6) = 18\text{ cm}^2
Substituting s=6s = 6 into A=3sA = 3s yields an area of 18 cm218\text{ cm}^2.

Anahtar Kavram

Perimeter and Area of a Circular Sector
Soru 77Soru

A rhombus has diagonals of lengths 12 cm12\text{ cm} and 16 cm16\text{ cm}. What is the perimeter of the rhombus in cm\text{cm}?

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Cevap: 40

Cevap

The perimeter of the rhombus is 40 cm40\text{ cm}.
The diagonals of a rhombus bisect each other at right angles, dividing the rhombus into four congruent right-angled triangles. Each triangle has legs measuring 6 cm6\text{ cm} and 8 cm8\text{ cm}. Applying the Pythagorean theorem, the hypotenuse (which is the side length of the rhombus) is 62+82=100=10 cm\sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}. Since all four sides of a rhombus are equal, the perimeter is 4×10=40 cm4 \times 10 = 40\text{ cm}.

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1
Calculate the lengths of the semi-diagonals
The semi-diagonals are 6 cm6\text{ cm} and 8 cm8\text{ cm}.
The diagonals of a rhombus bisect each other perpendicularly.
2
Determine the side length of the rhombus using the Pythagorean theorem
Side length s=62+82=100=10 cms = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}.
Each side of the rhombus forms the hypotenuse of a right-angled triangle formed by the semi-diagonals.
3
Calculate the total perimeter
Perimeter P=4×10=40 cmP = 4 \times 10 = 40\text{ cm}.
All four sides of a rhombus are equal in length.

Anahtar Kavram

Perimeter of a rhombus derived from diagonal lengths using right-triangle properties
Tahmini Süre:1m 0s
Soru 78Soru

Line ABAB is parallel to line CDCD. A transversal line EFEF intersects line ABAB at point PP and line CDCD at point QQ. If APQ=(4x10)\angle APQ = (4x - 10)^\circ and PQD=(2x+30)\angle PQD = (2x + 30)^\circ are alternate interior angles, what is the measure of BPQ\angle BPQ?

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Cevap: 110110^\circ

Cevap

110110^\circ
Since alternate interior angles are equal, 4x10=2x+304x - 10 = 2x + 30, yielding x=20x = 20. Thus, APQ=70\angle APQ = 70^\circ. Because APQ\angle APQ and BPQ\angle BPQ are adjacent angles on straight line ABAB, their sum is 180180^\circ, giving BPQ=110\angle BPQ = 110^\circ.

Adım Adım Çözüm

1
Set up the equation for alternate interior angles.
4x10=2x+304x - 10 = 2x + 30
Alternate interior angles formed by a transversal cutting parallel lines are equal.
2
Solve the linear equation for xx.
2x=40    x=202x = 40 \implies x = 20
Subtract 2x2x and add 1010 to both sides.
3
Calculate the measure of APQ\angle APQ.
\angle APQ = 4(20) - 10 = 70^\circ$
Substitute x=20x = 20 into the expression (4x10)(4x - 10)^\circ.
4
Determine BPQ\angle BPQ using the straight line angle property.
\angle BPQ = 180^\circ - 70^\circ = 110^\circ$
Angles APQ\angle APQ and BPQ\angle BPQ form a linear pair on straight line ABAB, summing to 180180^\circ.

Anahtar Kavram

Alternate interior angles of parallel lines and angles on a straight line
Soru 79Soru

A rectangular garden measuring 12 m12\text{ m} by 8 m8\text{ m} has a paved path of uniform width x mx\text{ m} constructed inside it along its perimeter. If the area of the remaining inner garden is 60 m260\text{ m}^2, what is the width xx of the path?

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Cevap: 1 m1\text{ m}

Cevap

The width of the path is 1 m1\text{ m}.
The total outer garden area is 12×8=96 m212 \times 8 = 96\text{ m}^2. Subtracting the path width xx from both ends gives inner dimensions (122x)(12 - 2x) and (82x)(8 - 2x). Equating the inner area (122x)(82x)=60(12 - 2x)(8 - 2x) = 60 yields x210x+9=0x^2 - 10x + 9 = 0. The realistic physical solution is x=1 mx = 1\text{ m}.

Adım Adım Çözüm

1
Express the inner dimensions in terms of path width xx
Length = (122x) m(12 - 2x)\text{ m}, Width = (82x) m(8 - 2x)\text{ m}
The path reduces each side dimension by xx at both ends.
2
Set up the area equation for the inner rectangular garden
(122x)(82x)=60(12 - 2x)(8 - 2x) = 60
The area of a rectangle is length multiplied by width.
3
Expand and simplify the quadratic equation
9640x+4x2=60    4x240x+36=0    x210x+9=096 - 40x + 4x^2 = 60 \implies 4x^2 - 40x + 36 = 0 \implies x^2 - 10x + 9 = 0
Divide the whole equation by 4 to simplify quadratic terms.
4
Solve for xx by factoring
(x1)(x9)=0    x=1(x - 1)(x - 9) = 0 \implies x = 1 or x=9x = 9
Since the width xx cannot exceed half of the smaller side (x<4 mx < 4\text{ m}), x=9x = 9 is extraneous.

Anahtar Kavram

Perimeter and Area of Rectangles with Uniform Borders
Tahmini Süre:1m 30s
Soru 80Soru

What is the exact simplified value of sin60+cos45\sin 60^\circ + \cos 45^\circ?

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Cevap: 3+22\frac{\sqrt{3} + \sqrt{2}}{2}

Cevap

3+22\frac{\sqrt{3} + \sqrt{2}}{2}
Substituting the exact values sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2} and cos45=22\cos 45^\circ = \frac{\sqrt{2}}{2} gives 32+22=3+22\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} = \frac{\sqrt{3} + \sqrt{2}}{2}.

Adım Adım Çözüm

1
Identify the exact trigonometric values for the special angles.
sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2} and cos45=22\cos 45^\circ = \frac{\sqrt{2}}{2}
Standard special angle values in trigonometry.
2
Substitute the values into the given expression.
32+22\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}
Direct substitution of known ratio values.
3
Combine the fractions over the common denominator of 22.
3+22\frac{\sqrt{3} + \sqrt{2}}{2}
Adding numerators over a shared common denominator.

Anahtar Kavram

Special Angles and Exact Trigonometric Values
ÖncekiSayfa 4 / 10Sonraki
Geometry and Trigonometry Alıştırma Soruları — JAMB UTME — Sayfa 4 | Examkin