Geometry and Trigonometry

184 soru

Soru 81Soru

In a convex polygon of nn sides, three of its interior angles are right angles, and each of the remaining interior angles is equal to 150150^\circ. What is the total number of diagonals in this polygon?

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Cevap: 9

Cevap

9 diagonals
The total sum of interior angles of an nn-sided polygon is (n2)×180(n - 2) \times 180^\circ. Setting this equal to 3(90)+(n3)(150)3(90^\circ) + (n - 3)(150^\circ) simplifies to 180n360=150n180180n - 360 = 150n - 180, which solves to n=6n = 6. Substituting n=6n = 6 into the diagonal formula D=n(n3)2D = \frac{n(n - 3)}{2} yields 6×32=9\frac{6 \times 3}{2} = 9 diagonals.

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1
Formulate the interior angle sum equation for the nn-sided convex polygon.
The total interior angle sum is (n2)×180(n - 2) \times 180^\circ. Summing the individual angles gives 3×90+(n3)×1503 \times 90^\circ + (n - 3) \times 150^\circ. Thus: (n2)×180=270+(n3)×150(n - 2) \times 180^\circ = 270^\circ + (n - 3) \times 150^\circ.
The sum of all interior angles of any nn-sided polygon is given by (n2)×180(n - 2) \times 180^\circ.
2
Solve the equation for the number of sides nn.
180n360=270+150n450    180n360=150n180    30n=180    n=6180n - 360 = 270 + 150n - 450 \implies 180n - 360 = 150n - 180 \implies 30n = 180 \implies n = 6.
Expanding terms and isolating nn reveals that the polygon has 6 sides (a hexagon).
3
Calculate the total number of diagonals using the formula D=n(n3)2D = \frac{n(n - 3)}{2}.
D=6(63)2=6×32=9D = \frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9.
Each of the nn vertices can connect to n3n - 3 non-adjacent vertices, and dividing by 2 avoids double-counting each diagonal line segment.

Anahtar Kavram

Polygon interior angle sum theorem and diagonal calculation
Soru 82Soru

If θ\theta is an acute angle such that tanθ=2\tan \theta = 2, calculate the exact numerical value of 3sinθ+cosθsinθ+2cosθ\frac{3\sin \theta + \cos \theta}{\sin \theta + 2\cos \theta}.

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Cevap: 1.75

Cevap

The exact numerical value of the trigonometric expression is 1.75.
Dividing every term in the expression by cosθ\cos \theta converts sinθ\sin \theta terms into tanθ\tan \theta. The quotient becomes 3tanθ+1tanθ+2\frac{3\tan \theta + 1}{\tan \theta + 2}. Substituting tanθ=2\tan \theta = 2 gives 3(2)+12+2=74=1.75\frac{3(2) + 1}{2 + 2} = \frac{7}{4} = 1.75.

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1
Divide numerator and denominator by cosθ\cos \theta
The expression becomes 3(sinθcosθ)+1sinθcosθ+2=3tanθ+1tanθ+2\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 1}{\frac{\sin \theta}{\cos \theta} + 2} = \frac{3\tan \theta + 1}{\tan \theta + 2}.
Using the trigonometric identity tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta} simplifies the calculation without needing to evaluate the hypotenuse.
2
Substitute tanθ=2\tan \theta = 2
3(2)+12+2=74\frac{3(2) + 1}{2 + 2} = \frac{7}{4}
Replace tanθ\tan \theta with its given numerical value of 2.
3
Convert fraction to decimal
1.75
Decimal representation of the exact fraction 74\frac{7}{4}.

Anahtar Kavram

Basic Trigonometric Ratios and Quotient Identity
Soru 83Soru

A point P(x,y)P(x, y) moves in a Cartesian plane such that the sum of the squares of its distances from two fixed points A(0,0)A(0, 0) and B(8,0)B(8, 0) is equal to 8282, defining a locus L1L_1. A second locus L2L_2 is the set of all points equidistant from the parallel lines y=1y = -1 and y=7y = 7. Given that L1L_1 and L2L_2 intersect at two distinct points MM and NN, what is the length of the line segment MNMN?

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Cevap: 8

Cevap

The length of the line segment MNMN is 8 units.
The locus L1L_1 simplifies to the circle (x4)2+y2=25(x - 4)^2 + y^2 = 25 with center (4,0)(4,0) and radius 55. The locus L2L_2 is the horizontal line y=3y = 3. Substituting y=3y = 3 into the circle equation yields (x4)2=16(x - 4)^2 = 16, giving intersection points at (0,3)(0,3) and (8,3)(8,3). The distance between these points is 88 units.

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1
Formulate the algebraic expression for locus L1L_1
The distance squared PA2=x2+y2PA^2 = x^2 + y^2 and PB2=(x8)2+y2PB^2 = (x - 8)^2 + y^2. Setting PA2+PB2=82PA^2 + PB^2 = 82 gives x2+y2+(x8)2+y2=82x^2 + y^2 + (x - 8)^2 + y^2 = 82.
By the distance formula, the sum of squared distances from P(x,y)P(x, y) to A(0,0)A(0,0) and B(8,0)B(8,0) equals 82.
2
Simplify and complete the square for locus L1L_1
Expanding yields 2x216x+64+2y2=82    2x216x+2y2=18    x28x+y2=92x^2 - 16x + 64 + 2y^2 = 82 \implies 2x^2 - 16x + 2y^2 = 18 \implies x^2 - 8x + y^2 = 9. Completing the square gives (x4)2+y2=25(x - 4)^2 + y^2 = 25.
This shows L1L_1 is a circle with center (4,0)(4, 0) and radius R=5R = 5.
3
Find the equation for locus L2L_2
The locus of points equidistant from two parallel lines y=1y = -1 and y=7y = 7 is the line y=1+72=3y = \frac{-1 + 7}{2} = 3.
The locus equidistant from two parallel lines is a parallel line running midway between them.
4
Find the intersection points MM and NN
Substitute y=3y = 3 into (x4)2+y2=25(x - 4)^2 + y^2 = 25: (x4)2+32=25    (x4)2=16    x4=±4(x - 4)^2 + 3^2 = 25 \implies (x - 4)^2 = 16 \implies x - 4 = \pm 4, so x=8x = 8 or x=0x = 0. Thus, M=(0,3)M = (0, 3) and N=(8,3)N = (8, 3).
Solving the circle and line equations simultaneously gives the xx-coordinates of the intersection points.
5
Compute the distance between MM and NN
Since both points share the ordinate y=3y = 3, distance MN=80=8MN = 8 - 0 = 8.
The length of a horizontal segment is the difference between its endpoints' xx-coordinates.

Anahtar Kavram

Intersection of loci involving circles and parallel lines
Soru 84Soru

A point P(x,y)P(x, y) moves such that it is equidistant from two parallel lines 3x4y+11=03x - 4y + 11 = 0 and 3x4y1=03x - 4y - 1 = 0, defining locus L1L_1. A second locus L2L_2 consists of all points that are at a constant distance of 55 units from the fixed point (1,7)(1, 7). Calculate the distance between the two points of intersection of locus L1L_1 and locus L2L_2.

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Cevap: 6

Cevap

The distance between the two points of intersection of locus L1 and locus L2 is 6 units.
Locus L1 is the line midway between the two given parallel lines, having the equation 3x - 4y + 5 = 0. Locus L2 is a circle centered at (1, 7) with radius 5. The perpendicular distance from the center (1, 7) to line L1 is 4 units. Applying Pythagoras' theorem to the right triangle formed by the radius, perpendicular distance, and half-chord gives a half-chord length of sqrt(5^2 - 4^2) = 3. Therefore, the distance between the two intersection points (the full chord length) is 2 * 3 = 6 units.

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1
Determine the equation of locus L1
3x - 4y + 5 = 0
The locus of points equidistant from two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is a line parallel to both and midway between them, given by ax + by + (c1 + c2)/2 = 0.
2
Identify the geometric representation of locus L2
A circle centered at (1, 7) with radius r = 5
The locus of points at a constant distance r from a fixed point (x0, y0) is a circle.
3
Calculate the perpendicular distance d from the circle's center (1, 7) to line L1
d = 4 units
Using the perpendicular distance formula d = |ax0 + by0 + c| / sqrt(a^2 + b^2), we get d = |3(1) - 4(7) + 5| / sqrt(3^2 + (-4)^2) = |-20| / 5 = 4.
4
Compute the total distance between the two intersection points
6 units
The line intersects the circle along a chord. Using Pythagoras' theorem, half the chord length is sqrt(r^2 - d^2) = sqrt(5^2 - 4^2) = 3. Multiplying by 2 gives the full distance of 6 units.

Anahtar Kavram

Intersection of Loci (Parallel Line Bisector and Circle)
Tahmini Süre:2m 30s
Soru 85Soru

The measures of the exterior angles of an convex hexagon are given as xx^\circ, (x+10)(x + 10)^\circ, (2x5)(2x - 5)^\circ, (x+25)(x + 25)^\circ, (2x+15)(2x + 15)^\circ, and (x5)(x - 5)^\circ. What is the measure of the largest interior angle of the hexagon?

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Cevap: 145145^\circ

Cevap

145145^\circ
The sum of all exterior angles of a convex polygon is 360360^\circ. Summing the given expressions yields 8x+40=3608x + 40 = 360^\circ, which gives x=40x = 40^\circ. Evaluating each exterior angle shows that the smallest exterior angle is (405)=35(40 - 5)^\circ = 35^\circ. Since interior and exterior angles on a straight line are supplementary, the largest interior angle is 18035=145180^\circ - 35^\circ = 145^\circ.

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1
Set up equation using the sum of exterior angles
x+(x+10)+(2x5)+(x+25)+(2x+15)+(x5)=360x + (x + 10) + (2x - 5) + (x + 25) + (2x + 15) + (x - 5) = 360^\circ
The sum of the exterior angles of any convex polygon is always 360360^\circ.
2
Simplify and solve for xx
8x+40=360    8x=320    x=408x + 40 = 360 \implies 8x = 320 \implies x = 40^\circ
Combining like terms gives 8x+40=3608x + 40 = 360.
3
Find the smallest exterior angle
x5=405=35x - 5 = 40 - 5 = 35^\circ
The largest interior angle corresponds to the smallest exterior angle because an interior angle and its adjacent exterior angle add up to 180180^\circ.
4
Calculate the largest interior angle
18035=145180^\circ - 35^\circ = 145^\circ
Subtracting the smallest exterior angle from 180180^\circ yields the largest interior angle.

Anahtar Kavram

Exterior and Interior Angle Relationship in Polygons
Tahmini Süre:1m 30s
Soru 86Soru

Given that sinθ+cosθ=62\sin \theta + \cos \theta = \frac{\sqrt{6}}{2} where θ\theta is an acute angle, what is the exact value of tanθ+cotθ\tan \theta + \cot \theta?

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Cevap: 4

Cevap

The exact value of tanθ+cotθ\tan \theta + \cot \theta is 4.
By squaring both sides of sinθ+cosθ=62\sin \theta + \cos \theta = \frac{\sqrt{6}}{2}, we obtain sin2θ+cos2θ+2sinθcosθ=64=32\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{6}{4} = \frac{3}{2}. Applying the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 simplifies this to 1+2sinθcosθ=321 + 2\sin \theta \cos \theta = \frac{3}{2}, which yields sinθcosθ=14\sin \theta \cos \theta = \frac{1}{4}. Since tanθ+cotθ=sinθcosθ+cosθsinθ=sin2θ+cos2θsinθcosθ=1sinθcosθ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}, the exact value is 11/4=4\frac{1}{1/4} = 4.

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1
Square both sides of the given equation sinθ+cosθ=62\sin \theta + \cos \theta = \frac{\sqrt{6}}{2}.
(sinθ+cosθ)2=(62)2    sin2θ+2sinθcosθ+cos2θ=64=32(\sin \theta + \cos \theta)^2 = \left(\frac{\sqrt{6}}{2}\right)^2 \implies \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = \frac{6}{4} = \frac{3}{2}.
Squaring allows the application of the fundamental Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 to isolate the product sinθcosθ\sin \theta \cos \theta.
2
Substitute sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 into the equation and solve for sinθcosθ\sin \theta \cos \theta.
1+2sinθcosθ=32    2sinθcosθ=12    sinθcosθ=141 + 2\sin \theta \cos \theta = \frac{3}{2} \implies 2\sin \theta \cos \theta = \frac{1}{2} \implies \sin \theta \cos \theta = \frac{1}{4}.
Simplifying the algebraic equation isolates the product term.
3
Express tanθ+cotθ\tan \theta + \cot \theta in terms of sinθ\sin \theta and cosθ\cos \theta.
tanθ+cotθ=sinθcosθ+cosθsinθ=sin2θ+cos2θsinθcosθ=1sinθcosθ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}.
Using basic trigonometric identities simplifies the target sum into a reciprocal product.
4
Substitute sinθcosθ=14\sin \theta \cos \theta = \frac{1}{4} into the simplified expression.
tanθ+cotθ=11/4=4\tan \theta + \cot \theta = \frac{1}{1/4} = 4.
Evaluating the reciprocal fraction gives the final numerical answer.

Anahtar Kavram

Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 and reciprocal relation tanθ+cotθ=1sinθcosθ\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}.
Soru 87Soru

A solid hemisphere of radius 6 cm6\text{ cm} has the same total surface area as a solid right circular cone with a base radius of 6 cm6\text{ cm}. What is the slant height of the cone?

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Cevap: 12 cm12\text{ cm}

Cevap

The slant height of the cone is 12 cm12\text{ cm}.
The correct answer is derived by setting the total surface area of the solid hemisphere (3πR2=108π cm23\pi R^2 = 108\pi\text{ cm}^2) equal to the total surface area of the solid cone (πr2+πrl=36π+6πl\pi r^2 + \pi r l = 36\pi + 6\pi l). Solving 36π+6πl=108π36\pi + 6\pi l = 108\pi gives l=12 cml = 12\text{ cm}.

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1
Calculate the total surface area of the solid hemisphere.
TSAhemisphere=3πR2=3π(6)2=108π cm2\text{TSA}_{\text{hemisphere}} = 3\pi R^2 = 3\pi(6)^2 = 108\pi\text{ cm}^2
A solid hemisphere consists of a curved surface (2πR22\pi R^2) plus its flat circular base (πR2\pi R^2).
2
Express the total surface area of the solid cone in terms of slant height ll.
TSAcone=πr2+πrl=π(6)2+π(6)l=36π+6πl\text{TSA}_{\text{cone}} = \pi r^2 + \pi r l = \pi(6)^2 + \pi(6)l = 36\pi + 6\pi l
A solid cone has a base area of πr2\pi r^2 and a curved surface area of πrl\pi r l.
3
Equate the two surface area expressions and solve for ll.
36π+6πl=108π    6πl=72π    l=12 cm36\pi + 6\pi l = 108\pi \implies 6\pi l = 72\pi \implies l = 12\text{ cm}
The question states that the total surface area of both solids is equal.

Anahtar Kavram

Total Surface Area of Composite 3D Solids (Hemisphere and Cone)
Soru 88Soru

A solid right circular cylinder has a height of 10 cm10\text{ cm} and a total surface area of 192π cm2192\pi\text{ cm}^2. What is the volume of the cylinder in cm3\text{cm}^3 expressed in terms of π\pi?

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Cevap: 360

Cevap

360
The total surface area of a closed cylinder is A=2πr2+2πrhA = 2\pi r^2 + 2\pi r h. Given A=192πA = 192\pi and h=10 cmh = 10\text{ cm}, we have 2πr2+20πr=192π2\pi r^2 + 20\pi r = 192\pi. Dividing by 2π2\pi yields r2+10r96=0r^2 + 10r - 96 = 0. Factoring the quadratic gives (r+16)(r6)=0(r + 16)(r - 6) = 0, so r=6 cmr = 6\text{ cm} (since radius must be positive). Substituting r=6 cmr = 6\text{ cm} and h=10 cmh = 10\text{ cm} into the volume formula V=πr2hV = \pi r^2 h yields V=π(62)(10)=360π cm3V = \pi (6^2)(10) = 360\pi\text{ cm}^3. The numerical coefficient of π\pi is 360360.

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1
Set up the equation for the total surface area of the cylinder
2πr2+2πr(10)=192π2\pi r^2 + 2\pi r(10) = 192\pi
The total surface area of a solid cylinder comprises the areas of the top and bottom circular bases (2πr22\pi r^2) plus the curved surface area (2πrh2\pi r h).
2
Simplify and solve the resulting quadratic equation for radius rr
r2+10r96=0    (r+16)(r6)=0    r=6 cmr^2 + 10r - 96 = 0 \implies (r + 16)(r - 6) = 0 \implies r = 6\text{ cm}
Dividing the equation by 2π2\pi yields r2+10r96=0r^2 + 10r - 96 = 0. Since radius must be positive, r=6 cmr = 6\text{ cm}.
3
Calculate the volume of the cylinder using the radius and height
V=πr2h=π(6)2(10)=360π cm3V = \pi r^2 h = \pi (6)^2 (10) = 360\pi\text{ cm}^3
The volume of a cylinder is calculated using the formula V=πr2hV = \pi r^2 h.

Anahtar Kavram

Surface Area and Volume of 3D Solids
Soru 89Soru

A piece of wire of length 88 cm88\text{ cm} is bent to form the perimeter of a sector of a circle of radius 21 cm21\text{ cm}. What is the area of the sector in cm2\text{cm}^2?

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Cevap: 483

Cevap

The area of the sector is 483 cm2483\text{ cm}^2.
The total length of the wire forms the perimeter of the sector, which consists of two radii and the arc length (P=2r+lP = 2r + l). With P=88 cmP = 88\text{ cm} and r=21 cmr = 21\text{ cm}, the arc length is l=882(21)=46 cml = 88 - 2(21) = 46\text{ cm}. Using the sector area formula A=12rlA = \frac{1}{2} r l, the area is 12×21×46=483 cm2\frac{1}{2} \times 21 \times 46 = 483\text{ cm}^2.

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1
Determine the arc length of the sector from the given perimeter
l=46 cml = 46\text{ cm}
The total perimeter of a sector includes two straight radii and the curved arc length: P=2r+lP = 2r + l. Subtracting twice the radius (2×21=42 cm2 \times 21 = 42\text{ cm}) from the total wire length of 88 cm88\text{ cm} gives an arc length of 46 cm46\text{ cm}.
2
Calculate the area of the sector
A=483 cm2A = 483\text{ cm}^2
The area of a sector can be computed using the formula A=12rlA = \frac{1}{2} r l. Substituting r=21 cmr = 21\text{ cm} and l=46 cml = 46\text{ cm} yields A=12×21×46=483 cm2A = \frac{1}{2} \times 21 \times 46 = 483\text{ cm}^2.

Anahtar Kavram

Perimeter and Area of a Sector of a Circle
Soru 90Soru

A solid wooden block is shaped as a frustum of a right circular cone with a top base radius of 4 cm4\text{ cm}, a bottom base radius of 12 cm12\text{ cm}, and a vertical height of 15 cm15\text{ cm}. A cylindrical hole of radius 4 cm4\text{ cm} is drilled vertically through the center of the frustum from the top base straight down to the bottom base. What is the volume of the remaining wooden solid in cm3\text{cm}^3?

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Cevap: 800π cm3800\pi\text{ cm}^3

Cevap

The volume of the remaining wooden solid is 800π cm3800\pi\text{ cm}^3.
The volume of a frustum of a cone with radii R=12 cmR=12\text{ cm}, r=4 cmr=4\text{ cm}, and height h=15 cmh=15\text{ cm} is 13π(15)(122+42+12×4)=1040π cm3\frac{1}{3}\pi(15)(12^2 + 4^2 + 12 \times 4) = 1040\pi\text{ cm}^3. Subtracting the volume of the central cylindrical hole of radius 4 cm4\text{ cm} and height 15 cm15\text{ cm}, which is π(42)(15)=240π cm3\pi(4^2)(15) = 240\pi\text{ cm}^3, yields 1040π240π=800π cm31040\pi - 240\pi = 800\pi\text{ cm}^3.

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1
Calculate the total volume of the conical frustum before drilling.
Vfrustum=13πh(R2+r2+Rr)=13π(15)(122+42+12×4)=5π(144+16+48)=1040π cm3V_{\text{frustum}} = \frac{1}{3}\pi h (R^2 + r^2 + R r) = \frac{1}{3}\pi (15)(12^2 + 4^2 + 12 \times 4) = 5\pi(144 + 16 + 48) = 1040\pi\text{ cm}^3.
The total volume of a frustum of a cone is determined by its vertical height and the radii of its top and bottom circular bases.
2
Calculate the volume of the cylindrical hole drilled through the solid.
Vcylinder=πrhole2h=π(42)(15)=240π cm3V_{\text{cylinder}} = \pi r_{\text{hole}}^2 h = \pi (4^2)(15) = 240\pi\text{ cm}^3.
The drilled hole forms a right circular cylinder of radius 4 cm4\text{ cm} and height equal to the full height of the frustum (15 cm15\text{ cm}).
3
Subtract the cylinder's volume from the frustum's volume to find the remaining volume.
Vremaining=VfrustumVcylinder=1040π240π=800π cm3V_{\text{remaining}} = V_{\text{frustum}} - V_{\text{cylinder}} = 1040\pi - 240\pi = 800\pi\text{ cm}^3.
Removing material by drilling decreases the overall volume of the original solid by the exact volume of the cylindrical bore.

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Volume of Composite Solids and Conical Frustums
Tahmini Süre:2m 30s
Soru 91Soru

Each exterior angle of a regular polygon measures 2424^\circ. How many sides does this polygon have?

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Cevap: 15

Cevap

The polygon has 15 sides.
The sum of all exterior angles of any convex polygon is always 360360^\circ. Because the polygon is regular, all its exterior angles are equal. Dividing 360360^\circ by the measure of one exterior angle (2424^\circ) yields exactly 15 sides.

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1
Identify the formula relating exterior angle and number of sides
The sum of exterior angles of any convex polygon is 360360^\circ. For a regular polygon with nn sides, each exterior angle equals 360n\frac{360^\circ}{n}.
All exterior angles in a regular polygon are equal.
2
Rearrange the formula to solve for the number of sides nn
n=360Exterior Angle=36024=15n = \frac{360^\circ}{\text{Exterior Angle}} = \frac{360^\circ}{24^\circ} = 15.
Dividing the total sum of exterior angles by the measure of one exterior angle yields the total number of sides.

Anahtar Kavram

Exterior angle property of regular polygons
Soru 92Soru

The ratio of the measure of each interior angle to each exterior angle of a regular convex polygon is 7:27 : 2. What is the total number of diagonals of this polygon?

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Cevap: 27

Cevap

The total number of diagonals of the polygon is 27.
Since interior and exterior angles are supplementary, their ratio 7:27 : 2 divides 180180^\circ into 9 equal parts of 2020^\circ. The exterior angle is 2×20=402 \times 20^\circ = 40^\circ. The number of sides nn is 360/40=9360^\circ / 40^\circ = 9. Using the formula for diagonals D=n(n3)2D = \frac{n(n-3)}{2}, we obtain D=9×62=27D = \frac{9 \times 6}{2} = 27.

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1
Calculate the measure of the exterior angle
Exterior angle =27+2×180=40= \frac{2}{7 + 2} \times 180^\circ = 40^\circ
Interior and exterior angles at a vertex are supplementary, summing to 180180^\circ.
2
Determine the number of sides nn
n=36040=9n = \frac{360^\circ}{40^\circ} = 9
The sum of exterior angles of any convex polygon is 360360^\circ, so n=360exterior anglen = \frac{360^\circ}{\text{exterior angle}}.
3
Calculate the number of diagonals using D=n(n3)2D = \frac{n(n - 3)}{2}
D=9(93)2=9×62=27D = \frac{9(9 - 3)}{2} = \frac{9 \times 6}{2} = 27
Each vertex connects to n3n - 3 non-adjacent vertices, and dividing by 2 avoids double-counting.

Anahtar Kavram

Interior and exterior angles of regular polygons and the polygon diagonal count formula.
Tahmini Süre:1m 30s
Soru 93Soru

A composite plane figure is formed from a rectangle ABCDABCD measuring 20 cm20\text{ cm} by 14 cm14\text{ cm}. A semicircle with diameter BC=14 cmBC = 14\text{ cm} is attached externally along side BCBC. At the opposite end, an isosceles triangle AEDAED with base AD=14 cmAD = 14\text{ cm} and perpendicular height 6 cm6\text{ cm} is cut out from the interior of the rectangle. Taking π=227\pi = \frac{22}{7}, what is the total area of the figure in cm2\text{cm}^2?

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Cevap: 315

Cevap

The total area of the composite figure is 315 cm².
The area of the complete shape is found by adding the area of the external semicircle (77 cm277\text{ cm}^2) to the area of the rectangle (280 cm2280\text{ cm}^2) and subtracting the area of the removed triangle (42 cm242\text{ cm}^2), yielding a net area of 315 cm2315\text{ cm}^2.

Adım Adım Çözüm

1
Calculate the area of the rectangular portion ABCDABCD
Area of rectangle =20 cm×14 cm=280 cm2= 20\text{ cm} \times 14\text{ cm} = 280\text{ cm}^2
The main section of the figure before modifications is a rectangle with length 20 cm20\text{ cm} and width 14 cm14\text{ cm}.
2
Calculate the area of the externally attached semicircle
Area of semicircle =12×227×72=77 cm2= \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2
The diameter of the semicircle is BC=14 cmBC = 14\text{ cm}, giving a radius r=7 cmr = 7\text{ cm}. Area is given by 12πr2\frac{1}{2}\pi r^2.
3
Calculate the area of the removed internal triangle AEDAED
Area of triangle =12×14 cm×6 cm=42 cm2= \frac{1}{2} \times 14\text{ cm} \times 6\text{ cm} = 42\text{ cm}^2
The triangle has base AD=14 cmAD = 14\text{ cm} and height 6 cm6\text{ cm}. Area is calculated as 12×base×height\frac{1}{2} \times \text{base} \times \text{height}.
4
Combine the areas to find the total remaining area of the figure
Total Area =280 cm2+77 cm242 cm2=315 cm2= 280\text{ cm}^2 + 77\text{ cm}^2 - 42\text{ cm}^2 = 315\text{ cm}^2
The external semicircle increases the total area while the internal cut-out triangle reduces it.

Anahtar Kavram

Area of Composite Plane Figures
Tahmini Süre:2m 0s
Soru 94Soru

Two parallel lines, L1L_1 and L2L_2, are situated in a plane with L1L_1 above L2L_2. A regular polygon of nn sides has one of its sides, ABAB, lying entirely on line L1L_1. An adjacent side, BCBC, extends downwards into the region between L1L_1 and L2L_2. A line segment BDBD is drawn from vertex BB perpendicular to line L2L_2, meeting L2L_2 at point DD. The segment BDBD lies inside the interior angle ABC\angle ABC of the polygon and divides ABC\angle ABC into two angles, ABD\angle ABD and DBC\angle DBC, such that ABD:DBC=3:2\angle ABD : \angle DBC = 3 : 2. What is the total number of diagonals of this regular polygon?

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Cevap: 54

Cevap

54
Because line L_1 is parallel to line L_2 and BD is perpendicular to L_2, BD is also perpendicular to L_1, making angle ABD equal to 90°. Based on the 3:2 ratio, angle DBC equals 60°, giving a total interior angle of 150°. Solving (n - 2) × 180° / n = 150° yields n = 12 sides. Substituting n = 12 into the diagonal formula n(n - 3) / 2 gives 54 diagonals.

Adım Adım Çözüm

1
Determine the measure of angle ABD using parallel line properties
ABD=90\angle ABD = 90^\circ
Because line L_1 is parallel to line L_2 and line segment BD is perpendicular to L_2, BD is also perpendicular to L_1.
2
Calculate the measure of angle DBC from the given angle ratio
DBC=60\angle DBC = 60^\circ
With \angle ABD : \angle DBC = 3 : 2 and \angle ABD = 90^\circ, setting 3 ratio units equal to 90^\circ gives 1 unit = 30^\circ, so 2 units = 60^\circ.
3
Find the total interior angle measure of the regular polygon
ABC=150\angle ABC = 150^\circ
The interior angle \angle ABC is the sum of \angle ABD and \angle DBC, which gives 90^\circ + 60^\circ = 150^\circ.
4
Determine the number of sides n of the regular polygon
n = 12
Using the regular polygon interior angle formula \frac{(n - 2) \times 180^\circ}{n} = 150^\circ, solving for n yields 180n - 360 = 150n \implies 30n = 360 \implies n = 12.
5
Calculate the total number of diagonals
54
Applying the polygon diagonal formula D = \frac{n(n - 3)}{2} with n = 12 gives D = \frac{12 \times (12 - 3)}{2} = \frac{12 \times 9}{2} = 54.

Anahtar Kavram

Combining parallel line perpendicularity properties, angle partitioning, regular polygon interior angle measure, and diagonal count calculation
Soru 95Soru

A point P(x,y)P(x, y) moves in the Cartesian plane such that its distance from the origin O(0,0)O(0, 0) is always half of its distance from the fixed point Q(6,0)Q(6, 0). Which of the following equations represents the locus of PP?

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Cevap: x2+y2+4x12=0x^2 + y^2 + 4x - 12 = 0

Cevap

The equation of the locus of PP is x2+y2+4x12=0x^2 + y^2 + 4x - 12 = 0.
The correct equation x2+y2+4x12=0x^2 + y^2 + 4x - 12 = 0 is derived by expressing the condition PO=12PQPO = \frac{1}{2} PQ as 2PO=PQ2 \cdot PO = PQ, squaring both sides to get 4(x2+y2)=(x6)2+y24(x^2 + y^2) = (x - 6)^2 + y^2, and simplifying to standard circle form.

Adım Adım Çözüm

1
Express the distance condition algebraically.
The distance from P(x,y)P(x,y) to O(0,0)O(0,0) is PO=x2+y2PO = \sqrt{x^2 + y^2}, and the distance from P(x,y)P(x,y) to Q(6,0)Q(6,0) is PQ=(x6)2+y2PQ = \sqrt{(x-6)^2 + y^2}. Given PO=12PQPO = \frac{1}{2} PQ, we have 2PO=PQ2 \cdot PO = PQ.
Translate the geometric distance description into algebraic expressions.
2
Square both sides of the equation to eliminate square roots.
4(x2+y2)=(x6)2+y24(x^2 + y^2) = (x - 6)^2 + y^2.
Squaring removes the radicals; note that (2PO)2=4PO2(2 \cdot PO)^2 = 4 \cdot PO^2.
3
Expand and group like terms.
4x2+4y2=x212x+36+y2    3x2+12x+3y236=04x^2 + 4y^2 = x^2 - 12x + 36 + y^2 \implies 3x^2 + 12x + 3y^2 - 36 = 0.
Expand (x6)2=x212x+36(x - 6)^2 = x^2 - 12x + 36 and collect terms on one side.
4
Divide the entire equation by 3 to standard form.
x2+y2+4x12=0x^2 + y^2 + 4x - 12 = 0.
Simplifying by the common factor of 3 gives the equation of a circle representing the Circle of Apollonius.

Anahtar Kavram

Locus of a point with a constant ratio of distances from two fixed points (Circle of Apollonius)
Tahmini Süre:2m 0s
Soru 96Soru

Given that tanθ+cotθ=4\tan \theta + \cot \theta = 4 for an acute angle θ\theta, what is the exact decimal value of sin4θ+cos4θ\sin^4 \theta + \cos^4 \theta?

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Cevap: 0.875

Cevap

The exact decimal value of sin4θ+cos4θ\sin^4 \theta + \cos^4 \theta is 0.875.
By writing tanθ+cotθ\tan \theta + \cot \theta as sinθcosθ+cosθsinθ=sin2θ+cos2θsinθcosθ=4\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = 4, we find that sinθcosθ=14\sin \theta \cos \theta = \frac{1}{4}. Squaring the fundamental identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 gives sin4θ+2sin2θcos2θ+cos4θ=1\sin^4 \theta + 2\sin^2 \theta \cos^2 \theta + \cos^4 \theta = 1. Isolating sin4θ+cos4θ\sin^4 \theta + \cos^4 \theta yields 12(sinθcosθ)2=12(116)=118=78=0.8751 - 2(\sin \theta \cos \theta)^2 = 1 - 2\left(\frac{1}{16}\right) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875.

Adım Adım Çözüm

1
Rewrite the expression tanθ+cotθ=4\tan \theta + \cot \theta = 4 using sine and cosine ratios
\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = 4
Applies the fundamental quotient identities for tangent and cotangent.
2
Combine fractions over a common denominator and apply the Pythagorean identity
\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = 4 \implies \sin \theta \cos \theta = \frac{1}{4}
Utilizes the identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1.
3
Relate sin4θ+cos4θ\sin^4 \theta + \cos^4 \theta to (sin2θ+cos2θ)2(\sin^2 \theta + \cos^2 \theta)^2
\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2(\sin \theta \cos \theta)^2
Uses the algebraic identity a2+b2=(a+b)22aba^2 + b^2 = (a+b)^2 - 2ab where a=sin2θa = \sin^2 \theta and b=cos2θb = \cos^2 \theta.
4
Substitute sinθcosθ=14\sin \theta \cos \theta = \frac{1}{4} into the algebraic relation
1 - 2\left(\frac{1}{4}\right)^2 = 1 - 2\left(\frac{1}{16}\right) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875
Evaluates the expression to obtain the final decimal result.

Anahtar Kavram

Trigonometric Identities and Algebraic Polynomial Expansion
Tahmini Süre:2m 0s
Soru 97Soru

In ΔABC\Delta ABC, the length of side aa is 10 cm10\text{ cm}, angle A=30\angle A = 30^\circ, and angle B=45\angle B = 45^\circ. What is the length of side bb?

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Cevap: 102 cm10\sqrt{2}\text{ cm}

Cevap

The length of side bb is 102 cm10\sqrt{2}\text{ cm}.
The option specifying 102 cm10\sqrt{2}\text{ cm} is correct because applying the Sine Rule gives 10sin30=bsin45\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}, which simplifies directly to b=102 cmb = 10\sqrt{2}\text{ cm}.

Adım Adım Çözüm

1
State the Sine Rule formula relating sides aa, bb and their opposite angles AA, BB.
asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B}
The Sine Rule connects the ratio of side lengths to the sines of their opposite angles.
2
Substitute the known values (a=10a = 10, A=30A = 30^\circ, B=45B = 45^\circ) into the formula.
10sin30=bsin45\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}
Isolating the variable bb requires substituting given numeric angle and side measures.
3
Evaluate exact values of trigonometric functions and solve for bb.
b=10×2212=102 cmb = \frac{10 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 10\sqrt{2}\text{ cm}
Simplifying the algebraic fraction yields the exact length of side bb.

Anahtar Kavram

Application of the Sine Rule to find an unknown side length in a non-right-angled triangle.
Tahmini Süre:45s
Soru 98Soru

A solid right circular cylinder of radius 5 cm5\text{ cm} and height 12 cm12\text{ cm} has a conical cavity of the same radius and height hollowed out from its top face. What is the total surface area of the remaining solid in cm2\text{cm}^2, expressed as a multiple of π\pi (that is, find the value of KK where the total surface area is Kπ cm2K\pi\text{ cm}^2)?

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Cevap: 210

Cevap

The total surface area of the remaining solid is 210π cm2210\pi\text{ cm}^2, so the required numerical coefficient KK is 210210.
The total surface area consists of three parts: the flat circular base at the bottom (25π cm225\pi\text{ cm}^2), the outer curved surface of the cylinder (120π cm2120\pi\text{ cm}^2), and the newly created inner curved surface of the conical cavity (65π cm265\pi\text{ cm}^2). Adding these together yields 25π+120π+65π=210π cm225\pi + 120\pi + 65\pi = 210\pi\text{ cm}^2, giving the coefficient K=210K = 210.

Adım Adım Çözüm

1
Calculate the slant height (ll) of the conical cavity.
l=52+122=169=13 cml = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ cm}.
The slant height forms the hypotenuse of the right triangle whose leg lengths are the base radius and the vertical height of the cone.
2
Determine the area of all exposed boundary surfaces of the remaining solid.
Flat bottom base area =π(5)2=25π cm2= \pi(5)^2 = 25\pi\text{ cm}^2; Outer curved cylindrical surface area =2π(5)(12)=120π cm2= 2\pi(5)(12) = 120\pi\text{ cm}^2; Inner curved conical surface area =π(5)(13)=65π cm2= \pi(5)(13) = 65\pi\text{ cm}^2.
Hollowing out the cone creates an internal curved boundary while leaving the outer cylindrical boundary and the flat bottom base exposed.
3
Sum the areas of all exposed surfaces to calculate the total surface area.
Total Surface Area =25π+120π+65π=210π cm2= 25\pi + 120\pi + 65\pi = 210\pi\text{ cm}^2.
The total surface area is the sum of all external and internal exposed surface areas.

Anahtar Kavram

Surface Area of Hollowed and Composite 3D Solids
Soru 99Soru

The sum of the interior angles of a convex polygon is three times the sum of its exterior angles. Calculate the number of sides of the polygon.

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Cevap: 8

Cevap

The polygon has 8 sides.
The sum of the exterior angles of any convex polygon is always 360360^\circ. Since the sum of the interior angles is three times this value, the interior angle sum is 3×360=10803 \times 360^\circ = 1080^\circ. Setting (n2)×180=1080(n - 2) \times 180^\circ = 1080^\circ gives n2=6n - 2 = 6, which yields n=8n = 8 sides.

Adım Adım Çözüm

1
Determine the exterior angle sum
The sum of exterior angles for any convex polygon is 360360^\circ.
The exterior angles of any convex polygon sum to a full turn (360360^\circ).
2
Compute the sum of the interior angles
Sum of interior angles = 3×360=10803 \times 360^\circ = 1080^\circ.
The question states that the interior angle sum is three times the exterior angle sum.
3
Apply the interior angle sum formula
(n2)×180=1080(n - 2) \times 180^\circ = 1080^\circ.
The interior angle sum of an nn-sided convex polygon is given by (n2)×180(n - 2) \times 180^\circ.
4
Solve for nn
n2=1080180=6    n=8n - 2 = \frac{1080^\circ}{180^\circ} = 6 \implies n = 8.
Dividing 10801080^\circ by 180180^\circ gives 66, and adding 22 gives n=8n = 8.

Anahtar Kavram

Relationship between the sum of interior and exterior angles of a convex polygon
Tahmini Süre:1m 0s
Soru 100Soru

If θ\theta is an acute angle such that 1tanθ1+tanθ=23\frac{1 - \tan \theta}{1 + \tan \theta} = 2 - \sqrt{3}, what is the exact value of 2sinθ+3cosθ2\sin \theta + \sqrt{3}\cos \theta?

Cevabı ve açıklamayı göster

Cevap: 52\frac{5}{2}

Cevap

The exact value of 2sinθ+3cosθ2\sin \theta + \sqrt{3}\cos \theta is 52\frac{5}{2}.
Cross-multiplying the equation 1tanθ1+tanθ=23\frac{1 - \tan \theta}{1 + \tan \theta} = 2 - \sqrt{3} yields tanθ=13\tan \theta = \frac{1}{\sqrt{3}}, which identifies θ\theta as 3030^\circ. Evaluating 2sin30+3cos30=2(12)+3(32)=1+32=522\sin 30^\circ + \sqrt{3}\cos 30^\circ = 2\left(\frac{1}{2}\right) + \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = 1 + \frac{3}{2} = \frac{5}{2}.

Adım Adım Çözüm

1
Solve the given equation 1tanθ1+tanθ=23\frac{1 - \tan \theta}{1 + \tan \theta} = 2 - \sqrt{3} for tanθ\tan \theta.
1tanθ=(23)(1+tanθ)    1tanθ=23+(23)tanθ1 - \tan \theta = (2 - \sqrt{3})(1 + \tan \theta) \implies 1 - \tan \theta = 2 - \sqrt{3} + (2 - \sqrt{3})\tan \theta. Rearranging terms gives (31)=(33)tanθ    tanθ=3133=313(31)=13(\sqrt{3} - 1) = (3 - \sqrt{3})\tan \theta \implies \tan \theta = \frac{\sqrt{3} - 1}{3 - \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} - 1)} = \frac{1}{\sqrt{3}}.
Isolating tanθ\tan \theta allows determination of the specific angle θ\theta.
2
Determine the acute angle θ\theta corresponding to tanθ=13\tan \theta = \frac{1}{\sqrt{3}}.
Since θ\theta is acute and tan30=13\tan 30^\circ = \frac{1}{\sqrt{3}}, θ=30\theta = 30^\circ.
Special angle identities state that tan30=13\tan 30^\circ = \frac{1}{\sqrt{3}}.
3
Evaluate the target trigonometric expression 2sinθ+3cosθ2\sin \theta + \sqrt{3}\cos \theta at θ=30\theta = 30^\circ.
2sin30+3cos30=2(12)+3(32)=1+32=522\sin 30^\circ + \sqrt{3}\cos 30^\circ = 2\left(\frac{1}{2}\right) + \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = 1 + \frac{3}{2} = \frac{5}{2}.
Substituting the exact surd values sin30=12\sin 30^\circ = \frac{1}{2} and cos30=32\cos 30^\circ = \frac{\sqrt{3}}{2} simplifies directly to the final fraction.

Anahtar Kavram

Trigonometric ratio deduction using special angles and surd simplification.
Tahmini Süre:2m 0s
ÖncekiSayfa 5 / 10Sonraki
Geometry and Trigonometry Alıştırma Soruları — JAMB UTME — Sayfa 5 | Examkin