Geometry and Trigonometry

184 soru

Soru 41Soru →

Which of the following sets contains all the values of xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ that satisfy the trigonometric equation 2cos⁡2x+3sin⁡x−3=02\cos^2 x + 3\sin x - 3 = 0?

Cevabı ve açıklamayı göster

Cevap: 30∘,90∘,150∘30^\circ, 90^\circ, 150^\circ

Cevap

The values of xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ satisfying the equation are 30∘,90∘,30^\circ, 90^\circ, and 150∘150^\circ.
By substituting cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x, the equation reduces to 2sin⁡2x−3sin⁡x+1=02\sin^2 x - 3\sin x + 1 = 0, which factors into (2sin⁡x−1)(sin⁡x−1)=0(2\sin x - 1)(\sin x - 1) = 0. Solving sin⁡x=1/2\sin x = 1/2 gives x=30∘x = 30^\circ and x=150∘x = 150^\circ within the specified domain. Solving sin⁡x=1\sin x = 1 gives x=90∘x = 90^\circ. Combining these yields the set of solutions 30∘,90∘,150∘30^\circ, 90^\circ, 150^\circ.

Adım Adım Çözüm

1
Use the Pythagorean trigonometric identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x to rewrite the equation in terms of sin⁡x\sin x.
2(1−sin⁡2x)+3sin⁡x−3=0  ⟹  2−2sin⁡2x+3sin⁡x−3=02(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.
−2sin⁡2x+3sin⁡x−1=0  ⟹  2sin⁡2x−3sin⁡x+1=0-2\sin^2 x + 3\sin x - 1 = 0 \implies 2\sin^2 x - 3\sin x + 1 = 0
Multiplying by −1-1 simplifies factoring.
3
Factor the quadratic equation (2sin⁡x−1)(sin⁡x−1)=0(2\sin x - 1)(\sin x - 1) = 0 to solve for sin⁡x\sin x.
sin⁡x=12\sin x = \frac{1}{2} or sin⁡x=1\sin x = 1
Setting each linear factor to zero yields the possible values for sin⁡x\sin x.
4
Determine all values of xx in the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ for each case.
For sin⁡x=12\sin x = \frac{1}{2}, x=30∘x = 30^\circ and x=180∘−30∘=150∘x = 180^\circ - 30^\circ = 150^\circ. For sin⁡x=1\sin x = 1, x=90∘x = 90^\circ.
Sine is positive in the first and second quadrants, and equals 1 at 90∘90^\circ.

Anahtar Kavram

Solving quadratic trigonometric equations by using fundamental identities to express the equation in terms of a single trigonometric function.
Tahmini Süre:2m 0s
Soru 42Soru →

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

Cevabı ve açıklamayı göster

Cevap: 135

Cevap

Each interior angle of a regular octagon measures 135 degrees.
The sum of the interior angles of a polygon with nn sides is (n−2)×180∘(n - 2) \times 180^\circ. For a regular octagon (n=8n = 8), the total interior angle sum is (8−2)×180∘=1080∘(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∘135^\circ for each interior angle.

Adım Adım Çözüm

1
Find the sum of all interior angles of the regular octagon.
Sum of interior angles = (8−2)×180∘=6×180∘=1080∘(8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ.
The sum of interior angles for any nn-sided polygon is (n−2)×180∘(n - 2) \times 180^\circ.
2
Calculate the measure of a single interior angle.
Interior angle = 1080∘8=135∘\frac{1080^\circ}{8} = 135^\circ.
In a regular polygon, all interior angles are equal in measure.

Anahtar Kavram

Interior Angle of a Regular Polygon
Soru 43Soru →

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

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The polygon has 12 sides.
The sum of the interior angles of an nn-sided polygon is (n−2)×180∘(n - 2) \times 180^\circ. From the problem, the sum of the angles is 3×120∘+(n−3)×160∘=160∘n−120∘3 \times 120^\circ + (n - 3) \times 160^\circ = 160^\circ n - 120^\circ. Equating 180∘(n−2)180^\circ(n - 2) to 160∘n−120∘160^\circ n - 120^\circ gives 180∘n−360∘=160∘n−120∘180^\circ n - 360^\circ = 160^\circ n - 120^\circ, which simplifies to 20∘n=240∘20^\circ n = 240^\circ, yielding n=12n = 12.

Adım Adım Çözüm

1
Write down the standard formula for the sum of interior angles of an nn-sided convex polygon.
Sum of interior angles = (n−2)×180∘(n - 2) \times 180^\circ.
Any nn-sided convex polygon can be split into (n−2)(n-2) triangles, each having an interior angle sum of 180∘180^\circ.
2
Express the total sum of interior angles using the given values.
Sum = 3(120∘)+(n−3)(160∘)=360∘+160∘n−480∘=160∘n−120∘3(120^\circ) + (n - 3)(160^\circ) = 360^\circ + 160^\circ n - 480^\circ = 160^\circ n - 120^\circ.
Three angles are 120∘120^\circ, so the remaining (n−3)(n - 3) angles must each equal 160∘160^\circ.
3
Equate the theoretical sum to the calculated sum.
180∘(n−2)=160∘n−120∘  ⟹  180∘n−360∘=160∘n−120∘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 nn.
20∘n=240∘  ⟹  n=1220^\circ n = 240^\circ \implies n = 12.
Subtracting 160∘n160^\circ n from both sides and adding 360∘360^\circ yields 20∘n=240∘20^\circ n = 240^\circ.

Anahtar Kavram

Sum of Interior Angles of a Polygon
Soru 44Soru →

A ship departs from a port PP and sails 10 km10\text{ km} on a bearing of 060∘060^\circ to reach a position QQ. From QQ, the ship changes course and sails 10 km10\text{ km} on a bearing of 150∘150^\circ to arrive at point RR. What is the bearing of point RR from point PP?

Cevabı ve açıklamayı göster

Cevap: 105∘105^\circ

Cevap

The bearing of point RR from point PP is 105∘105^\circ.
The back bearing from position QQ to port PP is 240∘240^\circ. The difference between 240∘240^\circ and the new bearing of 150∘150^\circ yields an interior angle of 90∘90^\circ at vertex QQ. Since both distances PQPQ and QRQR are equal to 10 km10\text{ km}, triangle PQRPQR is a 45∘−45∘−90∘45^\circ-45^\circ-90^\circ isosceles right triangle. Adding ∠QPR=45∘\angle QPR = 45^\circ to the initial bearing of 060∘060^\circ yields a bearing of 105∘105^\circ for point RR from point PP.

Adım Adım Çözüm

1
Determine the back bearing of PP from QQ.
The back bearing of PP from QQ is 060∘+180∘=240∘060^\circ + 180^\circ = 240^\circ.
To find the internal angle at QQ, we need the direction of PP relative to QQ.
2
Calculate the interior angle ∠PQR\angle PQR.
∠PQR=240∘−150∘=90∘\angle PQR = 240^\circ - 150^\circ = 90^\circ.
The difference between the line back to PP (240∘240^\circ) and the line to RR (150∘150^\circ) forms the interior angle at QQ.
3
Determine the properties of triangle PQRPQR and angle ∠QPR\angle QPR.
Since PQ=QR=10 kmPQ = QR = 10\text{ km} and ∠PQR=90∘\angle PQR = 90^\circ, triangle PQRPQR is an isosceles right triangle, so ∠QPR=45∘\angle QPR = 45^\circ.
The two equal sides subtend equal acute angles in a right-angled triangle: (180∘−90∘)/2=45∘(180^\circ - 90^\circ) / 2 = 45^\circ.
4
Calculate the total bearing of RR from PP.
Bearing of RR from P=060∘+45∘=105∘P = 060^\circ + 45^\circ = 105^\circ.
Point RR lies clockwise relative to the segment PQPQ, so we add ∠QPR\angle QPR to the initial bearing of PQPQ.

Anahtar Kavram

Three-point bearing calculations using geometry of parallel lines and right-angled triangles.
Soru 45Soru →

A trapezium has parallel sides of lengths 8 cm8\text{ cm} and 12 cm12\text{ cm}. If the perpendicular distance between these parallel sides is 6 cm6\text{ cm}, what is the area of the trapezium?

Cevabı ve açıklamayı göster

Cevap: 60 cm260\text{ cm}^2

Cevap

60 cm260\text{ cm}^2
The area of a trapezium is given by A=12(a+b)hA = \frac{1}{2}(a + b)h, where aa and bb are the parallel sides and hh is the perpendicular distance between them. Substituting a=8a = 8, b=12b = 12, and h=6h = 6 gives A=12(8+12)(6)=12(20)(6)=60 cm2A = \frac{1}{2}(8 + 12)(6) = \frac{1}{2}(20)(6) = 60\text{ cm}^2.

Adım Adım Çözüm

1
Identify the given dimensions and the formula for the area of a trapezium.
Parallel sides a=8 cma = 8\text{ cm}, b=12 cmb = 12\text{ cm}, height h=6 cmh = 6\text{ cm}. Formula: A=12(a+b)hA = \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 cma + 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=12×20×6=60 cm2A = \frac{1}{2} \times 20 \times 6 = 60\text{ cm}^2.
Completing the formula gives the exact plane surface measure in square centimeters.

Anahtar Kavram

Area of a Trapezium
Soru 46Soru →

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

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of cc is 5.
The midpoint of PQPQ is (4,3)(4, 3) and the gradient of PQPQ is 22. The perpendicular bisector has a gradient of −12-\frac{1}{2} and passes through (4,3)(4, 3). Substituting these into the line equation gives y−3=−12(x−4)y - 3 = -\frac{1}{2}(x - 4), which simplifies to y=−12x+5y = -\frac{1}{2}x + 5. The line intersects the yy-axis at (0,5)(0, 5), so c=5c = 5.

Adım Adım Çözüm

1
Find the midpoint of the line segment PQPQ
Midpoint M=(4,3)M = (4, 3)
The perpendicular bisector must pass through the midpoint of the segment.
2
Calculate the gradient of PQPQ
Gradient mPQ=2m_{PQ} = 2
Using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} for points (2,−1)(2, -1) and (6,7)(6, 7).
3
Determine the gradient of the perpendicular line
Perpendicular gradient m⊥=−12m_{\perp} = -\frac{1}{2}
Perpendicular lines have gradients that are negative reciprocals (m1m2=−1m_1 m_2 = -1).
4
Formulate the equation of the perpendicular bisector and solve for the yy-intercept
y=−12x+5y = -\frac{1}{2}x + 5, hence c=5c = 5
Using point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (4,3)(4, 3) and m=−12m = -\frac{1}{2}, setting x=0x = 0 gives the yy-intercept.

Anahtar Kavram

Perpendicular Bisector and Line Equations
Soru 47Soru →

A sector of a circle of radius 14 cm14\text{ cm} subtends an angle of 90∘90^\circ at the centre of the circle. What is the area of the sector in cm2\text{cm}^2? (Take π=227\pi = \frac{22}{7})

Cevabı ve açıklamayı göster

Cevap: 154

Cevap

The area of the sector is 154 cm2154\text{ cm}^2.
The area of a circular sector is given by Area=θ360∘×πr2\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2. Substituting θ=90∘\theta = 90^\circ, r=14 cmr = 14\text{ cm}, and π=227\pi = \frac{22}{7} yields 90360×227×142=14×616=154 cm2\frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2.

Adım Adım Çözüm

1
Identify the formula for the area of a circular sector.
Area=θ360∘×πr2\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∘360^\circ).
2
Substitute the given values into the formula.
Area=90∘360∘×227×(14)2\text{Area} = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (14)^2
Given radius r=14 cmr = 14\text{ cm}, angle θ=90∘\theta = 90^\circ, and π=227\pi = \frac{22}{7}.
3
Simplify the fraction and calculate the numerical value.
Area=14×227×196=14×22×28=22×7=154 cm2\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 90360\frac{90}{360} yields 14\frac{1}{4} and dividing 196196 by 77 gives 2828.

Anahtar Kavram

Area of a sector of a circle
Tahmini Süre:45s
Soru 48Soru →

A solid right circular cone has a base radius of 7 cm7\text{ cm} and a height of 12 cm12\text{ cm}. Taking π=227\pi = \frac{22}{7}, what is the volume of the cone?

Cevabı ve açıklamayı göster

Cevap: 616 cm3616\text{ cm}^3

Cevap

616 cm3616\text{ cm}^3
The correct answer is obtained by applying the standard cone volume formula V=13πr2hV = \frac{1}{3}\pi r^2 h. Substituting r=7r = 7, h=12h = 12, and π=227\pi = \frac{22}{7} yields V=616 cm3V = 616\text{ cm}^3.

Adım Adım Çözüm

1
Identify the formula for the volume of a solid cone.
V=13πr2hV = \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=13×227×(7)2×12V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 12
Given r=7 cmr = 7\text{ cm}, h=12 cmh = 12\text{ cm}, and π=227\pi = \frac{22}{7}.
3
Simplify the expression to find the volume.
V=13×227×49×12=22×7×4=616 cm3V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 12 = 22 \times 7 \times 4 = 616\text{ cm}^3
Simplifying 497=7\frac{49}{7} = 7 and 123=4\frac{12}{3} = 4 leaves 22×7×422 \times 7 \times 4.

Anahtar Kavram

Volume of a Right Circular Cone
Tahmini Süre:45s
Soru 49Soru →

A straight line LL passes through the point (4,−2)(4, -2) and is perpendicular to the line defined by the equation 3x−2y+8=03x - 2y + 8 = 0. What is the yy-intercept of line LL?

Cevabı ve açıklamayı göster

Cevap: 23\frac{2}{3}

Cevap

The yy-intercept of line LL is 23\frac{2}{3}.
The given line 3x−2y+8=03x - 2y + 8 = 0 has a gradient of 32\frac{3}{2}. A perpendicular line must have a gradient equal to the negative reciprocal, which is −23-\frac{2}{3}. Substituting the point (4,−2)(4, -2) into y−y1=m(x−x1)y - y_1 = m(x - x_1) gives y+2=−23(x−4)y + 2 = -\frac{2}{3}(x - 4), which simplifies to y=−23x+23y = -\frac{2}{3}x + \frac{2}{3}. Setting x=0x = 0 yields the yy-intercept of 23\frac{2}{3}.

Adım Adım Çözüm

1
Find the gradient of the given line.
Expressing 3x−2y+8=03x - 2y + 8 = 0 in slope-intercept form y=mx+cy = mx + c gives 2y=3x+8  ⟹  y=32x+42y = 3x + 8 \implies y = \frac{3}{2}x + 4. The gradient m1=32m_1 = \frac{3}{2}.
The gradient of the given line is required to determine the slope of line LL.
2
Determine the gradient of line LL.
Since line LL is perpendicular to the given line, mL=−1m1=−13/2=−23m_L = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3}.
Perpendicular lines have gradients whose product is −1-1 (m1⋅mL=−1m_1 \cdot m_L = -1).
3
Find the equation of line LL using point-slope form.
Using (x1,y1)=(4,−2)(x_1, y_1) = (4, -2) and mL=−23m_L = -\frac{2}{3}:
y−y1=mL(x−x1)y - y_1 = m_L(x - x_1)
y−(−2)=−23(x−4)y - (-2) = -\frac{2}{3}(x - 4)
y+2=−23x+83y + 2 = -\frac{2}{3}x + \frac{8}{3}
y=−23x+83−2y = -\frac{2}{3}x + \frac{8}{3} - 2
y=−23x+23y = -\frac{2}{3}x + \frac{2}{3}
To find the yy-intercept, we need the complete equation of line LL.
4
Identify the yy-intercept.
Comparing y=−23x+23y = -\frac{2}{3}x + \frac{2}{3} to y=mx+cy = mx + c, the yy-intercept c=23c = \frac{2}{3}.
The constant term cc in y=mx+cy = mx + c represents the yy-intercept.

Anahtar Kavram

Perpendicular lines in coordinate geometry have gradients that are negative reciprocals of each other (m1⋅m2=−1m_1 \cdot m_2 = -1).
Soru 50Soru →

A point PP lies inside a circle of radius 13 cm13\text{ cm} at a distance of 5 cm5\text{ cm} from the center OO. A chord ABAB passes through point PP such that the ratio of segment APAP to segment PBPB is 1:41:4. Calculate the total length of the chord ABAB in centimeters.

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The total length of chord ABAB is 30 cm30\text{ cm}.
Using the Power of a Point property for an interior point PP, the product of the chord segments is AP⋅PB=R2−OP2=132−52=144AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 144. Given AP:PB=1:4AP : PB = 1 : 4, we write AP=xAP = x and PB=4xPB = 4x, leading to 4x2=144  ⟹  x=6 cm4x^2 = 144 \implies x = 6\text{ cm}. Summing the two segments gives AB=6+24=30 cmAB = 6 + 24 = 30\text{ cm}.

Adım Adım Çözüm

1
Calculate the constant product of chord segments passing through interior point PP.
AP⋅PB=R2−OP2=132−52=169−25=144AP \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 PP on any chord equals (R−d)(R+d)=R2−d2(R - d)(R + d) = R^2 - d^2.
2
Set up an algebraic equation using the segment ratio AP:PB=1:4AP : PB = 1 : 4.
Let AP=xAP = x and PB=4xPB = 4x, giving (x)(4x)=144  ⟹  4x2=144(x)(4x) = 144 \implies 4x^2 = 144.
Expressing both chord segments in terms of a single variable xx allows direct calculation of the segment lengths.
3
Solve for xx to find the individual segment lengths.
x2=36  ⟹  x=6 cmx^2 = 36 \implies x = 6\text{ cm}. Therefore, AP=6 cmAP = 6\text{ cm} and PB=24 cmPB = 24\text{ cm}.
Taking the positive square root gives the scale factor xx since physical distances must be positive.
4
Sum the segment lengths to find the total chord length.
AB=AP+PB=6+24=30 cmAB = AP + PB = 6 + 24 = 30\text{ cm}.
The entire chord length is the sum of its two divided parts.

Anahtar Kavram

Intersecting Chords Theorem and Power of an Interior Point
Soru 51Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

x2+y2=25x^2 + y^2 = 25
By definition, the locus of a point moving at a fixed distance of 5 units from the origin (0,0)(0,0) is a circle centered at (0,0)(0,0) with radius 5. Substituting into the standard circle equation x2+y2=r2x^2 + y^2 = r^2 yields x2+y2=52=25x^2 + y^2 = 5^2 = 25.

Adım Adım Çözüm

1
Apply the distance formula between a general point P(x,y)P(x, y) and the origin (0,0)(0,0).
d=(x−0)2+(y−0)2=x2+y2d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}
The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is (x2−x1)2+(y2−y1)2\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.
x2+y2=5  ⟹  x2+y2=25\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.

Anahtar Kavram

The locus of points at a constant distance rr from a fixed point (h,k)(h, k) forms a circle with equation (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.
Tahmini Süre:45s
Soru 52Soru →

Calculate the area, in square units, of the triangle formed by the straight line 3x+4y−24=03x + 4y - 24 = 0 and the coordinate axes.

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

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 xx-intercept and yy-intercept. Setting y=0y = 0 in 3x+4y−24=03x + 4y - 24 = 0 gives x=8x = 8. Setting x=0x = 0 gives y=6y = 6. The vertices of the right triangle are at (0,0)(0,0), (8,0)(8,0), and (0,6)(0,6). The area is calculated as 12×8×6=24\frac{1}{2} \times 8 \times 6 = 24 square units.

Adım Adım Çözüm

1
Find the xx-intercept of the straight line
x=8x = 8, corresponding to the point (8,0)(8, 0)
Setting y=0y = 0 determines where the line crosses the horizontal axis
2
Find the yy-intercept of the straight line
y=6y = 6, corresponding to the point (0,6)(0, 6)
Setting x=0x = 0 determines where the line crosses the vertical axis
3
Compute the area of the right-angled triangle formed with the origin (0,0)(0,0)
Area=12×8×6=24\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

Anahtar Kavram

Area of a triangle bounded by a straight line and the coordinate axes
Tahmini Süre:1m 0s
Soru 53Soru →

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

Cevabı ve açıklamayı göster

Cevap: (156−36π) cm2(156 - 36\pi)\text{ cm}^2

Cevap

(156−36π) cm2(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 cmAB + CD = AD + BC = 26\text{ cm}), giving slant side length 13 cm13\text{ cm}. Using Pythagoras, the perpendicular distance (height) is 132−52=12 cm\sqrt{13^2 - 5^2} = 12\text{ cm}. The area of the trapezium is 12(18+8)(12)=156 cm2\frac{1}{2}(18 + 8)(12) = 156\text{ cm}^2. The inscribed circle has radius equal to half the height (6 cm6\text{ cm}), so its area is π×62=36π cm2\pi \times 6^2 = 36\pi\text{ cm}^2. Subtracting the circle area from the trapezium area yields (156−36π) cm2(156 - 36\pi)\text{ cm}^2.

Adım Adım Çözüm

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 cmAB + CD = AD + BC = 18 + 8 = 26\text{ cm}. Since trapezium ABCDABCD is isosceles, AD=BC=13 cmAD = BC = 13\text{ cm}.
Tangential quadrilaterals have equal sums of opposite side lengths.
2
Calculate the height hh of the trapezium using the Pythagorean theorem
Dropping vertical altitudes from top vertices CC and DD creates right triangles at the base with horizontal leg 18−82=5 cm\frac{18 - 8}{2} = 5\text{ cm}. Thus, h=132−52=169−25=12 cmh = \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 =12(AB+CD)×h=12(18+8)×12=156 cm2= \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 cmh = 12\text{ cm}, so its radius is r=6 cmr = 6\text{ cm}. Area of circle =πr2=36π cm2= \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 =156−36π cm2= 156 - 36\pi\text{ cm}^2.
The region inside the trapezium but outside the circle is the difference between their areas.

Anahtar Kavram

Perimeter and Area of Composite Figures and Inscribed Shapes
Soru 54Soru →

If the line 3x+py−7=03x + py - 7 = 0 is perpendicular to the line passing through the points (1,−2)(1, -2) and (4,7)(4, 7), what is the value of pp?

Cevabı ve açıklamayı göster

Cevap: 99

Cevap

The value of pp is 99.
The line passing through (1,−2)(1, -2) and (4,7)(4, 7) has a gradient of m1=7−(−2)4−1=3m_1 = \frac{7 - (-2)}{4 - 1} = 3. The equation 3x+py−7=03x + py - 7 = 0 can be rewritten as y=−3px+7py = -\frac{3}{p}x + \frac{7}{p}, giving a gradient of m2=−3pm_2 = -\frac{3}{p}. For perpendicular lines, the product of their gradients must equal −1-1, so 3×(−3p)=−13 \times \left(-\frac{3}{p}\right) = -1, which simplifies to p=9p = 9.

Adım Adım Çözüm

1
Calculate the gradient (m1m_1) of the line passing through (1,−2)(1, -2) and (4,7)(4, 7)
m1=7−(−2)4−1=93=3m_1 = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3
The gradient between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Express the line 3x+py−7=03x + py - 7 = 0 in slope-intercept form (y=mx+cy = mx + c) to find its gradient (m2m_2)
py=−3x+7  ⟹  y=−3px+7ppy = -3x + 7 \implies y = -\frac{3}{p}x + \frac{7}{p}, so m2=−3pm_2 = -\frac{3}{p}
The coefficient of xx when solved for yy represents the gradient of the straight line.
3
Apply the perpendicularity condition m1⋅m2=−1m_1 \cdot m_2 = -1 and solve for pp
3⋅(−3p)=−1  ⟹  −9p=−1  ⟹  p=93 \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 −1-1.

Anahtar Kavram

Perpendicular Lines and Gradients
Tahmini Süre:1m 30s
Soru 55Soru →

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

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of kk is 4.
Using the distance formula, the squared distance PA2=(k−1)2+(3−5)2=(k−1)2+4PA^2 = (k-1)^2 + (3-5)^2 = (k-1)^2 + 4, and PB2=(k−7)2+(3−1)2=(k−7)2+4PB^2 = (k-7)^2 + (3-1)^2 = (k-7)^2 + 4. Equating PA2=PB2PA^2 = PB^2 gives (k−1)2=(k−7)2(k-1)^2 = (k-7)^2. Expanding both sides yields k2−2k+1=k2−14k+49k^2 - 2k + 1 = k^2 - 14k + 49. Subtracting k2k^2 from both sides gives 12k=4812k = 48, which leads to k=4k = 4.

Adım Adım Çözüm

1
Write the expressions for the squared distances PA2PA^2 and PB2PB^2 using the distance formula.
PA2=(k−1)2+4PA^2 = (k - 1)^2 + 4 and PB2=(k−7)2+4PB^2 = (k - 7)^2 + 4
The distance formula between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is d2=(x2−x1)2+(y2−y1)2d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.
2
Equate PA2PA^2 and PB2PB^2 since point PP is equidistant from points AA and BB.
(k−1)2+4=(k−7)2+4  ⟹  (k−1)2=(k−7)2(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 kk to find its numerical value.
k2−2k+1=k2−14k+49  ⟹  12k=48  ⟹  k=4k^2 - 2k + 1 = k^2 - 14k + 49 \implies 12k = 48 \implies k = 4
Canceling k2k^2 terms yields a simple linear equation.

Anahtar Kavram

Equidistant points and the distance formula in coordinate geometry
Tahmini Süre:1m 30s
Soru 56Soru →

A solid trophy consists of a right circular cone mounted on top of a right circular cylinder with the same base radius of 6 cm6\text{ cm}. The height of the cylinder is 10 cm10\text{ cm} and the slant height of the cone is 10 cm10\text{ cm}. What is the total volume of the trophy?

Cevabı ve açıklamayı göster

Cevap: 456π cm3456\pi\text{ cm}^3

Cevap

The total volume of the trophy is 456π cm3456\pi\text{ cm}^3.
First, use the Pythagorean theorem on the cone to find its vertical height hcone=102−62=8 cmh_{\text{cone}} = \sqrt{10^2 - 6^2} = 8\text{ cm}. The volume of the cone is 13π(6)2(8)=96π cm3\frac{1}{3}\pi (6)^2 (8) = 96\pi\text{ cm}^3. The volume of the cylinder is π(6)2(10)=360π cm3\pi (6)^2 (10) = 360\pi\text{ cm}^3. Adding both gives a total volume of 456π cm3456\pi\text{ cm}^3.

Adım Adım Çözüm

1
Calculate the vertical height of the cone
hcone=102−62=100−36=64=8 cmh_{\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 (hh), which forms a right-angled triangle with the radius (r=6 cmr=6\text{ cm}) and slant height (l=10 cml=10\text{ cm}).
2
Calculate the volume of the conical part
Vcone=13πr2hcone=13×π×62×8=96π cm3V_{\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=13πr2hV = \frac{1}{3}\pi r^2 h.
3
Calculate the volume of the cylindrical part
Vcylinder=πr2hcylinder=π×62×10=360π cm3V_{\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=πr2hV = \pi r^2 h.
4
Sum the volumes of both 3D shapes to get the total volume
Vtotal=96π+360π=456π cm3V_{\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.

Anahtar Kavram

Volume of Composite 3D Solids
Tahmini Süre:2m 0s
Soru 57Soru →

For the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ, solve the trigonometric equation 2sin⁡x+3=02\sin x + \sqrt{3} = 0. Which of the following gives the complete set of values for xx?

Cevabı ve açıklamayı göster

Cevap: 240∘240^\circ and 300∘300^\circ

Cevap

x=240∘x = 240^\circ and x=300∘x = 300^\circ
Rearranging 2sin⁡x+3=02\sin x + \sqrt{3} = 0 gives sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}. The reference angle for which sine equals 32\frac{\sqrt{3}}{2} is 60∘60^\circ. Since sine is negative in the third and fourth quadrants, the solutions are 180∘+60∘=240∘180^\circ + 60^\circ = 240^\circ and 360∘−60∘=300∘360^\circ - 60^\circ = 300^\circ.

Adım Adım Çözüm

1
Isolate the trigonometric function in the equation
sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}
Subtract 3\sqrt{3} from both sides and divide by 22.
2
Find the basic reference angle
Reference angle α=60∘\alpha = 60^\circ
sin⁡(60∘)=32\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∘≤x≤360∘0^\circ \le x \le 360^\circ
Quadrant III: x=180∘+60∘=240∘x = 180^\circ + 60^\circ = 240^\circ; Quadrant IV: x=360∘−60∘=300∘x = 360^\circ - 60^\circ = 300^\circ
Apply quadrant reduction formulas for Quadrant III (180∘+α180^\circ + \alpha) and Quadrant IV (360∘−α360^\circ - \alpha).

Anahtar Kavram

Solving simple trigonometric equations using reference angles and quadrant rules
Soru 58Soru →

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

Cevabı ve açıklamayı göster

Cevap: 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

Cevap

The equation of the locus is 3x−2y−4=03x - 2y - 4 = 0 (or 3x−2y=43x - 2y = 4).
The locus of a point equidistant from two fixed points A(−1,3)A(-1, 3) and B(5,−1)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(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2 and simplifying yields the linear equation 3x−2y−4=03x - 2y - 4 = 0.

Adım Adım Çözüm

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\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}
Since point P(x,y)P(x, y) is equidistant from AA and BB, PA=PBPA = PB.
2
Square both sides to remove the radical signs.
(x+1)2+(y−3)2=(x−5)2+(y+1)2(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 x2x^2 and y2y^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 ABAB.

Anahtar Kavram

Locus equidistant from two fixed points (Perpendicular Bisector)
Tahmini Süre:2m 0s
Soru 59Soru →

A sine function is given by the equation y=5sin⁡(23x)−2y = 5\sin\left(\frac{2}{3}x\right) - 2. What is the period of this trigonometric function in degrees?

Cevabı ve açıklamayı göster

Cevap: 540

Cevap

The period of the trigonometric function is 540∘540^\circ.
For a general sine curve of the form y=asin⁡(bx+c)+dy = a\sin(bx + c) + d, the period TT in degrees is calculated as T=360∘∣b∣T = \frac{360^\circ}{|b|}. Given y=5sin⁡(23x)−2y = 5\sin\left(\frac{2}{3}x\right) - 2, the value of bb is 23\frac{2}{3}. Dividing 360∘360^\circ by 23\frac{2}{3} gives 540∘540^\circ.

Adım Adım Çözüm

1
Identify the coefficient bb of xx from the standard sine form y=asin⁡(bx+c)+dy = a\sin(bx + c) + d.
b=23b = \frac{2}{3}
The coefficient of xx 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=360∘∣b∣T = \frac{360^\circ}{|b|}
The standard sine function completes one full wavelength over 360∘360^\circ, so scaling the input by bb changes the period to 360∘b\frac{360^\circ}{b}.
3
Substitute b=23b = \frac{2}{3} into the formula and evaluate.
T=360∘23=360∘×32=540∘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.

Anahtar Kavram

Period of Trigonometric Functions
Tahmini Süre:1m 30s
Soru 60Soru →

The ratio of the measure of an interior angle to an exterior angle of a regular polygon is 5:15 : 1. How many sides does the polygon have?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The polygon has 12 sides.
The interior angle and exterior angle of a regular polygon are supplementary, adding up to 180∘180^\circ. For a ratio of 5:15:1, the exterior angle is 15+1×180∘=30∘\frac{1}{5+1} \times 180^\circ = 30^\circ. Using the exterior angle formula n=360∘exterior anglen = \frac{360^\circ}{\text{exterior angle}}, the number of sides is 360∘30∘=12\frac{360^\circ}{30^\circ} = 12.

Adım Adım Çözüm

1
Set up an equation using the linear pair relationship between interior and exterior angles.
5x+1x=180∘  ⟹  6x=180∘  ⟹  x=30∘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∘180^\circ).
2
Identify the measure of the exterior angle.
Exterior angle = 30∘30^\circ
The exterior angle corresponds to 11 part of the 5:15:1 ratio, which equals x=30∘x = 30^\circ.
3
Calculate the number of sides nn of the polygon.
n=360∘30∘=12n = \frac{360^\circ}{30^\circ} = 12
The sum of the exterior angles of any convex polygon is 360∘360^\circ, so n=360∘exterior anglen = \frac{360^\circ}{\text{exterior angle}}.

Anahtar Kavram

Interior and exterior angles of regular polygons
ÖncekiSayfa 3 / 10Sonraki
Geometry and Trigonometry Alıştırma Soruları — JAMB UTME — Sayfa 3 | Examkin