A sphere is a perfectly round solid — every point on its surface is the same distance (the radius) from the center. Cutting a sphere exactly in half gives a hemisphere, with volume $\frac{1}{2}\cdot\frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3$. Composite solids are real-world shapes built from simpler pieces: a grain silo might be a cylinder topped by a hemisphere, or a bullet is a cylinder with a cone tip. The key strategy is always to identify the component shapes, compute each volume separately, then add or subtract. A sphere's surface area is exactly 4 times the area of a great circle — a useful fact to remember.
★★★ Cone Removed from Cylinder (Ice Cream Problem)
An ice cream cup is a cylinder of radius 3 cm and height 8 cm with a conical bottom of the same radius and height 5 cm. Find the volume of ice cream the cup can hold.
The ice cream fills the cylindrical portion only. The conical bottom holds the stick/wafer, not ice cream. Actually — let me restate: find the total internal volume = cylinder + cone.
A solid sphere is inscribed in a cube of side length 10 cm. Find (a) the volume of the sphere and (b) the volume of space between the sphere and the cube walls.
A sphere inscribed in a cube touches all six faces. Its diameter equals the side length, so $r=5$ cm.
The sphere fills only $\approx52.4\%$ of the cube. Final answers: V (sphere) = 523.6 cm³, V (gap) = 476.4 cm³.
★★★ How Many Steel Balls Fit?
A cylindrical container with diameter 20 cm and height 30 cm is to be filled with solid steel balls each of diameter 4 cm. Assuming random packing at 64% efficiency, how many balls fit and what is their total volume?
Volume of one ball: $V_{\text{ball}}=\frac{4}{3}\pi(2)^3=\frac{32\pi}{3}=33.51\text{ cm}^3$.
Domingo decides to ride a Ferris wheel at the local carnival. When he gets into the seat that is at the bottom of the Ferris wheel, he is 4 feet above the ground. If the radius of the Ferris wheel is 30 feet, how far above the ground is Domingo after the Ferris wheel rotates 300° counterclockwise?
Answer:
19ft
22ft
45ft
60ft
The bottom seat is 4 ft above the ground, so the center is $4+30=34$ ft high. After a 300° counterclockwise rotation from the bottom, the vertical displacement from the center is: $30\sin(270^\circ+300^\circ)=30\sin570^\circ=-15$ ft Height above ground: $34-15=19$ ft $\boxed{19\text{ ft}}$
Question Bank: t444
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
How many circles of radius 4 cm can be cut from a rectangular sheet measuring 16 cm $\times$ 24 cm?
4
12
6
8
Each circle needs a square of side equal to its diameter, $2(4)=8$ cm. Along 16 cm: $16/8=2$ circles; along 24 cm: $24/8=3$ circles. Total $=2\times3=6$. $\boxed{6}$
Question Bank: t467
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
Two wheels, one 60 cm in diameter and the other 40 cm in diameter are rolling along a straight line with the same linear speed. How many revolutions has the larger wheel made if the smaller wheel has made 360 revolutions?
200
240
280
300
Both wheels cover the same distance. Distance $=360\times\pi(40)$ cm. Larger wheel revolutions $=\dfrac{360\times\pi(40)}{\pi(60)}=360\times\dfrac{40}{60}=240$. $\boxed{240}$
Question Bank: t468
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
Two speed boats simultaneously sailed out from port A, on a 10-km radius circular lake, towards port B directly opposite of port A. The first boat took the shortest route and reached the destination in 100 minutes. The second boat has to pass by port C, 8 km from A, before proceeding to port B.
What is the speed of the first boat in miles per hour?
10.74
7.46
6.32
9.21
What is the distance CB in kilometers?
15.67
12
12.35
18.33
With what speed (in kph) should the second boat sail in order to arrive port B at the same time with the first boat?
19.8
14.5
15.8
16.3
Part 1.
The shortest route is the diameter $AB=2(10)=20$ km, covered in 100 min $=\tfrac{5}{3}$ h, so speed $=\dfrac{20}{5/3}=12$ kph. Converting: $12\div1.609=7.46$ mph. $\boxed{7.46}$
Part 2.
Port C lies on the circle, and $AB$ is a diameter, so angle $ACB=90^\circ$ (inscribed in a semicircle). $CB=\sqrt{AB^2-AC^2}=\sqrt{20^2-8^2}=\sqrt{336}=18.33$ km. $\boxed{18.33}$
Part 3.
The second boat travels $A\to C\to B=8+18.33=26.33$ km in the same $\tfrac{5}{3}$ h. Speed $=\dfrac{26.33}{5/3}=15.8$ kph. $\boxed{15.8}$
Question Bank: t479
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A circle with radius 10 has half its area removed by cutting off a border of uniform width. Find the width of the border.
2.93
1.65
2.21
3.21
Removing a border of width $w$ leaves an inner circle of radius $10-w$ with half the area. $\pi(10-w)^2=\tfrac12\pi(10)^2\Rightarrow(10-w)^2=50\Rightarrow10-w=7.07$. $w=2.93$. $\boxed{2.93}$
Question Bank: t494
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The area of one face of a cube is 9x square units. What is the volume of the cube?
27√x cubic units
27x cubic units
27x√x cubic units
27x² cubic units
One face area $=s^2=9x\Rightarrow s=3\sqrt{x}$. Volume $=s^3=(3\sqrt{x})^3=27x^{3/2}=27x\sqrt{x}$. $\boxed{27x\sqrt{x}\text{ cubic units}}$
Question Bank: t495
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A cube of side 2 cm is to be made. However, due to erroneous measurement, the actual side of the cube is 2.1 cm. What is the relative error in the volume of the cube?
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The entire surface of a solid cube with edge of length 12 cm, is painted. The cube is then cut into smaller cubes each with edge of length 1 cm. How many of the smaller cubes have paint on exactly one face?
600
620
580
480
Cubes with exactly one painted face lie in the interior of each face: an $(12-2)\times(12-2)=10\times10=100$ block per face. With 6 faces: $6\times100=600$. $\boxed{600}$
Question Bank: t501
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The opposite faces of a cube are ABCD and EFGH. If the edge of the cube is 1 m, what is the distance between the midpoint of edge AB and midpoint of edge DH.
1.225 m
1.325 m
1.425 m
1.125 m
Place the cube with $A=(0,0,0)$, $B=(1,0,0)$, $D=(0,1,0)$, $H=(0,1,1)$. Midpoint of $AB=(0.5,0,0)$; midpoint of $DH=(0,1,0.5)$. Distance $=\sqrt{0.5^2+1^2+0.5^2}=\sqrt{1.5}=1.225$ m. $\boxed{1.225\text{ m}}$
Question Bank: t521
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A solid spherical steel ball of radius 6 cm is placed inside a 20-cm diameter cylinder containing 15 cm deep of water. How deep is the water in the cylinder after immersion?
19.5 cm
17.9 cm
18.3 cm
17.2 cm
Ball volume $=\tfrac43\pi(6)^3=904.78$ cm$^3$ displaces water in the cylinder ($r=10$). Rise $=\dfrac{904.78}{\pi(10)^2}=2.88$ cm. New depth $=15+2.88=17.9$ cm. $\boxed{17.9\text{ cm}}$
Question Bank: t534
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The lateral area of a right circular cone is 634 square meters. Its diameter is two-thirds its altitude.
Determine the base radius of the cone in meters.
7.5
8
8.5
9
Determine the volume of the cone in cubic meters?
1,986
1,747
1,606
1,852
What is the volume of the largest sphere that can be inscribed in the cone in m³?
856.5
965.3
801.7
1021.4
Part 1.
Diameter $=\tfrac23$ altitude means $2r=\tfrac23h$, so $h=3r$ and slant $l=\sqrt{r^2+9r^2}=r\sqrt{10}$. Lateral area $=\pi rl=\pi r^2\sqrt{10}=634\Rightarrow r^2=\dfrac{634}{\pi\sqrt{10}}=63.8$. $r=8$ m. $\boxed{8}$
Part 2.
With $r=8$, $h=3r=24$. $V=\tfrac13\pi r^2 h=\tfrac13\pi(64)(24)=512\pi\approx1{,}606$ m$^3$. $\boxed{1{,}606}$
Part 3.
The inscribed sphere's radius equals the inradius of the axial-section triangle (base $2r=16$, sides $l=\sqrt{8^2+24^2}=25.3$). Area $=\tfrac12(16)(24)=192$, $s=33.3$, so $r_s=\dfrac{192}{33.3}=5.77$ m. $V=\tfrac43\pi(5.77)^3=801.7$ m$^3$. $\boxed{801.7}$
Question Bank: t555
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A pit 6 m deep was dug out from the ground. The pit is 3 m by 4 m at the top and 2 m by 3 m at the bottom. What is the volume of earth removed in cubic meter?
64
63
87
53
The pit is a frustum. $A_1=3\times4=12$, $A_2=2\times3=6$, $\sqrt{A_1A_2}=\sqrt{72}=8.49$. $V=\tfrac{h}{3}(A_1+A_2+\sqrt{A_1A_2})=\tfrac{6}{3}(12+6+8.49)=2(26.49)=53$ m$^3$. $\boxed{53}$
Question Bank: t561
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A right circular cylinder is inscribed in a right circular cone of altitude 12 cm, and radius of base 5 cm. Find the radius of the cylinder if its lateral area is equal to the lateral area of the small cone which surmounts the cylinder.
3.51 cm
2.43 cm
3.24 cm
1.53 cm
Let the cylinder radius be $r$. The small cone above the cylinder is similar to the full cone (base radius 5, altitude 12), so its height is $h_s = \frac{12}{5}r$ and its slant height is: $L_s = \sqrt{r^2 + \left(\tfrac{12}{5}r\right)^2} = \frac{13}{5}r$ Cylinder height: $H = 12 - \frac{12}{5}r$. Equate lateral areas: $2\pi r H = \pi r L_s$ $2\left(12 - \tfrac{12}{5}r\right) = \frac{13}{5}r$ $24 = \frac{24}{5}r + \frac{13}{5}r = \frac{37}{5}r$ $r = \frac{120}{37}$ $\boxed{r = 3.24 \text{ cm}}$
Question Bank: t563
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
If the radius "r" of a sphere is increased by "x", then its surface area is increased by:
$4\pi rx + 8\pi x^2$
$8\pi rx + 4\pi x^2$
$\pi rx + \pi x^2$
$4\pi(r + x)^2$
Surface area of a sphere is $S = 4\pi r^2$. The increase when the radius grows from $r$ to $r+x$ is: $\Delta S = 4\pi(r+x)^2 - 4\pi r^2$ $\Delta S = 4\pi\left(r^2 + 2rx + x^2 - r^2\right)$ $\Delta S = 4\pi\left(2rx + x^2\right)$ $\boxed{\Delta S = 8\pi rx + 4\pi x^2}$
Question Bank: t565
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A soft clay spherical in shape, has a radius of 10 cm. If cubes 2 cm on each side are to be made out of this clay, how many cubes can be made?
522
524
523
525
Volume of the clay sphere ($r = 10$ cm): $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (10)^3 = 4188.79$ cm3 Volume of each cube ($2$ cm side): $V_c = 2^3 = 8$ cm3 Number of whole cubes: $n = \frac{4188.79}{8} = 523.6$ Only complete cubes count, so: $\boxed{n = 523 \text{ cubes}}$
Question Bank: t567
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A metal sphere weighing 24 kg is melted down and recast into a solid cone of base radius 8.0 cm. If the density of the metal is 8000 kg/m³ determine the height of the metal cone, assuming that 15% of the metal is lost in the process.
35.87 cm
41.23 cm
32.89 cm
38.05 cm
Volume of the sphere from mass and density: $V = \frac{m}{\rho} = \frac{24}{8000} = 0.003$ m3 $= 3000$ cm3 With 15% lost, the cone volume is: $V_c = 0.85(3000) = 2550$ cm3 Cone height ($r = 8$ cm): $V_c = \frac{1}{3}\pi r^2 h \Rightarrow h = \frac{3V_c}{\pi r^2} = \frac{3(2550)}{\pi (8)^2}$ $\boxed{h = 38.05 \text{ cm}}$
Question Bank: t570
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The total volume of two spheres is 100$\pi$ cubic cm. The ratio of their areas is 4:9. What is the volume of the larger sphere in cubic cm?
75.85
314.16
71.79
242.22
Surface areas are in ratio $4:9$, so radii are in ratio $2:3$ and volumes in ratio $2^3:3^3 = 8:27$. Total volume $= 100\pi$ split into $8 + 27 = 35$ parts. Volume of the larger sphere: $V_L = \frac{27}{35}(100\pi)$ $\boxed{V_L \approx 242.22 \text{ cm}^3}$
Question Bank: t571
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The volume of a spherical sector with an angle of "x" in a sphere of radius 40 cm is 47,881 cc. Find x in degrees.
80
100
120
110
Volume of a spherical sector: $V = \frac{2}{3}\pi R^2 h$, where $h$ is the zone height. $47{,}881 = \frac{2}{3}\pi (40)^2 h \Rightarrow h = 14.29$ cm The zone height relates to the central angle $x$ by $h = R\left(1 - \cos\frac{x}{2}\right)$: $14.29 = 40\left(1 - \cos\frac{x}{2}\right)$ $\cos\frac{x}{2} = 0.6428 \Rightarrow \frac{x}{2} = 50^\circ$ $\boxed{x = 100^\circ}$
Question Bank: t577
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A frustum of a sphere have base diameters of 18.2 cm and 32.9 cm. The thickness of the frustum is 2.6 cm.
What is the volume of frustum?
1539.3 cm³
1624.5 cm³
1342.7 cm³
1452.6 cm³
What is the radius of the sphere?
32.54 cm
38.51 cm
34.86 cm
40.25 cm
What is the curved surface area of the frustum?
629.02 cm²
732.65 cm²
569.43 cm²
601.47 cm²
Part 1.
Base radii: $a = 18.2/2 = 9.1$ cm and $b = 32.9/2 = 16.45$ cm, thickness $h = 2.6$ cm. Volume of a spherical frustum (segment with two bases): $V = \frac{\pi h}{6}\left(3a^2 + 3b^2 + h^2\right)$ $V = \frac{\pi (2.6)}{6}\left(3(9.1)^2 + 3(16.45)^2 + (2.6)^2\right)$ $V = \frac{\pi (2.6)}{6}(1067.0)$ $\boxed{V = 1452.6 \text{ cm}^3}$
Part 2.
Each base circle lies at a distance from the center given by $\sqrt{R^2 - (\text{radius})^2}$. With the larger base nearer the center, the thickness is: $h = \sqrt{R^2 - a^2} - \sqrt{R^2 - b^2}$ $2.6 = \sqrt{R^2 - 9.1^2} - \sqrt{R^2 - 16.45^2}$ Solving gives $R = 38.51$ cm: $\sqrt{38.51^2 - 82.81} - \sqrt{38.51^2 - 270.6} = 37.42 - 34.82 = 2.6$ ✓ $\boxed{R = 38.51 \text{ cm}}$
Part 3.
The curved surface of a spherical frustum is a zone of the sphere: $A = 2\pi R h$ Using $R = 38.51$ cm and $h = 2.6$ cm: $A = 2\pi (38.51)(2.6)$ $\boxed{A = 629.02 \text{ cm}^2}$
Question Bank: t581
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
The frustum of a sphere is 3.8 cm thick. The diameters of its bases are 13.2 cm and 20.5 cm. Determine the following:
The volume of the frustum.
864.87 cm³
923.87 cm³
1023.78 cm³
756.23 cm³
The radius of the sphere.
11.98 cm
12.89 cm
13.37 cm
10.43 cm
The surface area of the zone.
267.43 cm²
421.85 cm²
321.89 cm²
285.92 cm²
Part 2.
Base radii: $a = 13.2/2 = 6.6$ cm and $b = 20.5/2 = 10.25$ cm, thickness $h = 3.8$ cm. With both bases on the same side of the center (larger base nearer): $h = \sqrt{R^2 - a^2} - \sqrt{R^2 - b^2}$ $3.8 = \sqrt{R^2 - 6.6^2} - \sqrt{R^2 - 10.25^2}$ Solving gives $R = 11.98$ cm: $\sqrt{11.98^2 - 43.56} - \sqrt{11.98^2 - 105.06} = 10.0 - 6.2 = 3.8$ ✓ $\boxed{R = 11.98 \text{ cm}}$
Part 3.
The zone is the curved surface of the spherical frustum: $A = 2\pi R h$ Using $R = 11.98$ cm and $h = 3.8$ cm: $A = 2\pi (11.98)(3.8)$ $\boxed{A = 285.92 \text{ cm}^2}$
Question Bank: t585
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A solid sphere 60 cm in diameter is formed of a certain kind of wood whose specific gravity is 3/4. Find the depth to which the floating sphere will sink in water.
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A frustum of a sphere have base radii of 9.8 cm and 18.4 cm and thickness of 4.2 cm.
What is the volume the sphere?
87,455 cm³
182724 cm³
203,846 cm³
143,593 cm³
What is the radius of the sphere?
27.54 cm
35.20 cm
36.51 cm
32.49 cm
What is the area of the zone?
857.26 cm²
726.66 cm²
963.46 cm²
840.49 cm²
Part 1.
First find the sphere radius (see next item): $R = 32.49$ cm. Volume of the sphere: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (32.49)^3$ $\boxed{V = 143{,}593 \text{ cm}^3}$
Part 2.
Base radii $a = 9.8$ cm, $b = 18.4$ cm, thickness $h = 4.2$ cm. With both bases on the same side of the center: $h = \sqrt{R^2 - a^2} - \sqrt{R^2 - b^2}$ $4.2 = \sqrt{R^2 - 9.8^2} - \sqrt{R^2 - 18.4^2}$ Solving gives $R = 32.49$ cm: $\sqrt{32.49^2 - 96.04} - \sqrt{32.49^2 - 338.56} = 30.98 - 26.78 = 4.2$ ✓ $\boxed{R = 32.49 \text{ cm}}$
Part 3.
The zone area is the curved surface of the frustum: $A = 2\pi R h = 2\pi (32.49)(4.2)$ $\boxed{A = 857.26 \text{ cm}^2}$
Question Bank: t594
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
How far from the center of a solid sphere of radius 20 cm must a cutting plane pass in order to remove 10% of its volume?
9.874 cm
12.168 cm
6.578 cm
7.832 cm
Total volume of the sphere ($R = 20$ cm): $V = \frac{4}{3}\pi (20)^3 = 33{,}510.3$ cm3 The removed cap is 10% of this: $V_{cap} = 3351.0$ cm3 Cap of height $h$: $V_{cap} = \frac{\pi h^2(3R - h)}{3}$ $h^2(60 - h) = 3200 \Rightarrow h = 7.832$ cm Distance of the cutting plane from the center: $d = R - h = 20 - 7.832$ $\boxed{d = 12.168 \text{ cm}}$
Question Bank: t602
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
Find the area of the lune of a spherical wedge of radius 12 cm if the volume of the wedge is 502.655 cm³.
287.2 cm²
187.6 cm²
112.1 cm²
125.7 cm²
Both the wedge volume and the lune area are the same fraction $\frac{\theta}{360}$ of the whole sphere, so their ratio is: $\frac{V_{wedge}}{A_{lune}} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3}$ Therefore: $A_{lune} = \frac{3V}{R} = \frac{3(502.655)}{12}$ $\boxed{A_{lune} = 125.7 \text{ cm}^2}$
Question Bank: t620
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3
A square hole 2" $\times$ 2" is cut through a 4-inch diameter log along its diameter and perpendicular to its axis. Find the volume of wood that was removed in in³.
18.65
15.31
16.58
17.32
Log radius $= 2$ in (axis along $z$). The square hole (cross-section $2\times2$) is drilled along a diameter (axis along $x$), so $-1 \le y \le 1$ and $-1 \le z \le 1$. For each $(y,z)$ the hole spans $|x| \le \sqrt{4 - y^2}$, length $2\sqrt{4 - y^2}$. Integrating ($z$ contributes a factor 2): $V = 2\int_{-1}^{1} 2\sqrt{4 - y^2}\,dy = 4\left[\frac{y}{2}\sqrt{4 - y^2} + 2\sin^{-1}\frac{y}{2}\right]_{-1}^{1}$ $\boxed{V = 15.31 \text{ in}^3}$
Question Bank: t2123
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Besavilla CE Pre-Board Math & Surveying
A spherical sector is cut from a sphere whose radius is 12 cm. Find its volume if its central angle is 20°.
60 cu.cm.
57 cu.cm.
55 cu.cm.
52 cu.cm.
50 cu.cm.
A spherical sector of radius $R$ and cap height $h$ has volume $V=\frac{2}{3}\pi R^2h$ For central angle $20^\circ$, the cap height is $h=R(1-\cos10^\circ)$. $h=12(1-\cos10^\circ)=0.1823$ cm $V=\frac{2}{3}\pi(12)^2(0.1823)$ $\boxed{V\approx55\text{ cu.cm.}}$
Question Bank: t2126
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Besavilla CE Pre-Board Math & Surveying
The volume of a frustum of a sphere with two bases is equal to $159\pi$ cu.m. If the radii of the bases are 4 m. and 5 m. respectively, compute the radius of the sphere.
3.12 m.
7.85 m.
6.85 m.
5.48 m.
4.25 m.
For a spherical zone with two base radii $a=4$ m and $b=5$ m, volume is $V=\frac{\pi h}{6}(3a^2+3b^2+h^2)$ Given $V=159\pi$: $159=\frac{h}{6}[3(4^2)+3(5^2)+h^2]$ $954=h(123+h^2)$, so $h=6$ m. Let the distances of the two cutting planes from the sphere center be $x_1$ and $x_2$. Then $x_1-x_2=6$ and $x_1^2-x_2^2=(R^2-4^2)-(R^2-5^2)=9$ $(x_1-x_2)(x_1+x_2)=9 \Rightarrow x_1+x_2=1.5$ Thus $x_1=3.75$, and $R^2=x_1^2+4^2=30.0625$. $\boxed{R=5.48\text{ m}}$
Question Bank: t2148
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Besavilla CE Pre-Board Math & Surveying
What is the curved surface area of a spherical wedge having a radius of 3.5 m. and a central angle of 0.65 radians.
12.3 m2
15.9 m2
20.5 m2
18.2 m2
10.5 m2
The curved surface area of a spherical wedge is $A=2r^2\theta$ where $\theta$ is in radians. $A=2(3.5)^2(0.65)$ $A=15.925\text{ m}^2$ $\boxed{A\approx 15.9\text{ m}^2}$
Question Bank: w29
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / MSTE May 2019
A metal sphere with a specific gravity of 3 originally weighs 34 kg. It underwent a process of recasting and was formed into a solid cone with a base having a radius of 6 cm. Assuming 12% of the metal was lost in the process, what is the perpendicular height of the new cone?
MSTE - Geometry and Trigonometry / Plane and Solid Geometry / MSTE November 2019
A steel storage tank for methane gas is to be constructed in the shape of a right circular cylinder of altitude 10 feet with a hemisphere attached to each of its ends. Express the volume (in cubic feet) of the tank as a function of the tank radius $r$.
$\frac{4}{3}\pi r^2 (15 + 2r)$
$\frac{1}{3}\pi r^2 (15 + 2r)$
$\frac{2}{3}\pi r^2 (15 + 2r)$
$\frac{3}{4}\pi r^2 (15 + 2r)$
Two hemispheres make one full sphere, so: $V = V_{cyl} + V_{sphere} = \pi r^2 L + \tfrac{4}{3}\pi r^3$ $V = \pi r^2(10) + \tfrac{4}{3}\pi r^3 = \dfrac{30\pi r^2 + 4\pi r^3}{3}$ $V = \tfrac{2}{3}\pi r^2(15 + 2r)$ $\boxed{V = \tfrac{2}{3}\pi r^2 (15 + 2r)}$
