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Frustums and Surface Area

A frustum is what remains when you slice the top off a cone or pyramid with a cut parallel to the base — like a truncated cone. Water towers, buckets, and hopper bins are frustum shapes. The formula looks unusual at first but notice that $R^2 + Rr + r^2$ averages the three "sizes" of the cross-section — it reduces to the cone formula ($r=0$) and to a cylinder ($R=r$) as special cases. For the lateral (curved) surface area of a conical frustum, you need the slant height $l=\sqrt{h^2+(R-r)^2}$ — the distance measured along the slanted face. A prismatoid (any solid with parallel top and bottom faces) uses Simpson's Rule: $V=\frac{h}{6}(A_{\text{top}}+4A_{\text{mid}}+A_{\text{bot}})$.

$$V_{\text{cone frustum}}=\frac{\pi h}{3}(R^2+Rr+r^2), \quad l=\sqrt{h^2+(R-r)^2}$$
$$LSA_{\text{cone frustum}}=\pi(R+r)l \qquad V_{\text{prismatoid}}=\frac{h}{6}(A_1+4A_m+A_2)$$

Frustum Volume

Find the volume of a conical frustum with $R=5$ m, $r=2$ m, and $h=6$ m.

$$V=\frac{\pi(6)}{3}(5^2+5(2)+2^2)=78\pi=245.04\text{ m}^3$$

Final answer: $245.04\text{ m}^3$.

★ Lateral Surface Area of a Cone

A traffic cone has a base radius of 15 cm and a slant height of 40 cm. Find the lateral surface area of the cone.

$$LSA=\pi rl=\pi(15)(40)=600\pi=1884.96\text{ cm}^2$$

Final answer: 1884.96 cm².

★ Total Surface Area of a Closed Cylinder

A cylindrical water pipe (both ends closed) has radius 0.5 m and length 3 m. How many square meters of steel are needed to fabricate it?

Total surface area = lateral surface + 2 circular ends.

$$TSA=2\pi rh+2\pi r^2=2\pi(0.5)(3)+2\pi(0.5)^2=3\pi+0.5\pi=3.5\pi=10.996\text{ m}^2$$

Final answer: $10.996\text{ m}^2$ (approximately 11.0 m²).

★★ Frustum of a Square Pyramid

A hopper bin is shaped like a frustum of a square pyramid. The bottom is 1.5 m × 1.5 m, the top is 3 m × 3 m, and the height is 2 m. Find the volume.

For a prismatoid (or using the prismatoid/frustum of pyramid formula): $V=\frac{h}{3}(A_1+\sqrt{A_1 A_2}+A_2)$.

$A_1 = 1.5^2 = 2.25\text{ m}^2$, $A_2 = 3^2 = 9\text{ m}^2$, $h=2$ m.

$$V=\frac{2}{3}\left(2.25+\sqrt{2.25\times9}+9\right)=\frac{2}{3}(2.25+4.5+9)=\frac{2}{3}(15.75)=10.5\text{ m}^3$$

Final answer: $10.5\text{ m}^3$.

★★ Frustum: Finding the Slant Height and LSA

A conical frustum has lower radius $R=6$ m, upper radius $r=3$ m, and vertical height $h=4$ m. Find the slant height and the lateral surface area.

The slant height is the hypotenuse of a right triangle with legs $h$ and $(R-r)$.

$$l=\sqrt{h^2+(R-r)^2}=\sqrt{4^2+(6-3)^2}=\sqrt{16+9}=5\text{ m}$$
$$LSA=\pi(R+r)l=\pi(6+3)(5)=45\pi=141.37\text{ m}^2$$

Final answers: $l=5$ m, LSA = 141.37 m².

★★★ How Much Paint to Coat a Tank?

A closed cylindrical tank with radius 2 m and height 5 m needs two coats of anti-corrosion paint. One liter of paint covers 8 m². How many liters are needed?

Find the total external surface area first.

$$TSA=2\pi rh+2\pi r^2=2\pi(2)(5)+2\pi(2)^2=20\pi+8\pi=28\pi=87.96\text{ m}^2$$

Two coats means double the area equivalent:

$$\text{Liters needed}=\frac{2\times87.96}{8}=\frac{175.93}{8}=22.0\text{ liters}$$

Final answer: approximately 22 liters of paint.

★★★ Water Volume in a Frustum-Shaped Tank

A water reservoir is shaped like a conical frustum with lower radius 4 m, upper radius 7 m, and depth 3 m. The water depth is 2 m from the bottom. Find the volume of water.

At depth $y$ from the bottom, the radius varies linearly: $r(y)=R_{\text{bot}}+\frac{(R_{\text{top}}-R_{\text{bot}})}{H}\cdot y=4+\frac{3}{3}y=4+y$.

At $y=0$: $r=4$ m. At $y=2$ m: $r=6$ m. Use frustum formula with these values and $h=2$.

$$V=\frac{\pi(2)}{3}(4^2+4(6)+6^2)=\frac{2\pi}{3}(16+24+36)=\frac{2\pi}{3}(76)=\frac{152\pi}{3}=159.0\text{ m}^3$$

Final answer: $159.0\text{ m}^3$.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t438

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

Town houses 200 m by 300 m are built in a lot containing 81,600 square meters. A parking strip of uniform width surrounds the houses. How wide is the strip?

  1. 20 m
  2. 21 m
  3. 22 m
  4. 19 m
With a uniform strip of width $x$, the whole lot measures $(200+2x)(300+2x)=81{,}600$.
$4x^2+1000x+60{,}000=81{,}600\Rightarrow x^2+250x-5400=0$.
$x=\dfrac{-250+\sqrt{250^2+4(5400)}}{2}=\dfrac{-250+290}{2}=20$ m.
$\boxed{20\text{ m}}$

Question Bank: t510

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A large 15-m-diameter cylindrical tank that sits on the ground is to be painted. If one liter of paint covers 12 square meters, how many liters are required to paint the outside of the tank if it is 10 m high? (Include the top)

  1. 54
  2. 58
  3. 65
  4. 71
Paint the side and top only ($r=7.5$, $h=10$): area $=2\pi rh+\pi r^2=2\pi(7.5)(10)+\pi(7.5)^2=471.2+176.7=647.9$ m$^2$.
Liters $=\dfrac{647.9}{12}=54$.
$\boxed{54}$

Question Bank: t517

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The volume of a closed cylinder is numerically equal to its lateral area. The height is 10 m. Find the total surface area.

  1. $32\pi$
  2. $48\pi$
  3. $52\pi$
  4. $60\pi$
Volume $=$ lateral area: $\pi r^2h=2\pi rh\Rightarrow r=2$.
Total surface $=2\pi r(r+h)=2\pi(2)(2+10)=48\pi$.
$\boxed{48\pi}$

Question Bank: t518

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A model of a boiler is made having an overall height of 75 mm corresponding to an overall height of the actual boiler of 6 m. If the area of metal required for the model is 12,500 mm² determine, in square metres, the area of metal required for the actual boiler.

  1. 90 square meter
  2. 80 square meter
  3. 120 square meter
  4. 40 square meter
Linear scale $=\dfrac{6000\text{ mm}}{75\text{ mm}}=80$. Areas scale by $80^2=6400$.
Actual area $=12{,}500\text{ mm}^2\times6400=8\times10^7\text{ mm}^2=80$ m$^2$.
$\boxed{80\text{ square meter}}$

Question Bank: t520

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

Two vertical cylindrical tanks, one 8 m in diameter and the other 12 m in diameter, stand on the same horizontal plane, and connected by a 150-mm pipe at their bases. When the valve in the pipe is closed, the depth of water in the smaller tank is 15 m and in the other tank is 4 m. If the pipe valve is opened, what it is at equal depth in each tank?

  1. 5.47
  2. 7.85
  3. 6.25
  4. 7.38 m
Water volume is conserved. Areas: $A_1=\pi(4)^2=50.27$, $A_2=\pi(6)^2=113.10$ m$^2$.
$A_1(15)+A_2(4)=(A_1+A_2)h\Rightarrow753.98+452.39=163.36\,h$.
$h=\dfrac{1206.37}{163.36}=7.38$ m.
$\boxed{7.38\text{ m}}$

Question Bank: t525

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A closed right circular cylinder 1 meter in diameter and 1 meter tall is partially filled with water. When lying in a horizontal position, the depth of water is 2/3 the diameter. When the cylinder is in the vertical position, what is the depth of water?

  1. 0.654 m
  2. 0.854 m
  3. 0.708 m
  4. 0.985 m
Horizontal ($r=0.5$): the empty top segment has depth $1-\tfrac23=0.333$. $\cos(\alpha/2)=\dfrac{0.5-0.333}{0.5}=0.333\Rightarrow\alpha=2.4621$ rad. Empty area $=\tfrac12(0.25)(2.4621-0.6285)=0.2292$.
Water cross-section $=\pi(0.25)-0.2292=0.5562$, so water volume $=0.5562$ m$^3$.
Vertical depth $=\dfrac{0.5562}{\pi(0.5)^2}=0.708$ m.
$\boxed{0.708\text{ m}}$

Question Bank: t537

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A conical vessel with lower base circular have a height of 120 cm and a full capacity of 113.097 liters.

What is the radius of the base in centimeter?

  1. 32
  2. 30
  3. 94.9
  4. 9.49

If the vessel is made of steel that is 8 mm thick, what is the dead weight of the tank in kilograms? Density of steel is 7860 kilograms per cubic meter.

  1. 112.4
  2. 145.8
  3. 91.1
  4. 98.7

If 60 liters of water is in the tank, what is the depth of water in centimeters?

  1. 93.27
  2. 35.77
  3. 26.73
  4. 84.23

Part 1.

Capacity $=113.097$ L $=113{,}097$ cm$^3$. $V=\tfrac13\pi r^2 h$ with $h=120$.
$\tfrac13\pi r^2(120)=113{,}097\Rightarrow r^2=\dfrac{113{,}097}{40\pi}=900$.
$r=30$ cm.
$\boxed{30}$

Part 2.

Slant $l=\sqrt{30^2+120^2}=123.7$. Surface (lateral + base) $=\pi rl+\pi r^2=\pi(30)(123.7)+\pi(30)^2=14{,}485$ cm$^2$.
Steel volume $=14{,}485\times0.8\text{ cm}=11{,}588$ cm$^3=0.011588$ m$^3$.
Weight $=0.011588\times7860=91.1$ kg.
$\boxed{91.1}$

Part 3.

With the circular base at the bottom (apex up), the empty top is a similar cone holding $113.097-60=53.097$ L.
$\tfrac13\pi\left(\tfrac{H}{4}\right)^2 H=53{,}097$ (since $r/h=30/120=1/4$) $\Rightarrow H^3=811{,}268\Rightarrow H=93.27$ cm.
Water depth $=120-93.27=26.73$ cm.
$\boxed{26.73}$

Question Bank: t541

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A closed conical tank has a base diameter of 2 m and is 5 m tall. When standing in upright position, the water in the tank is 3 m deep. How deep will the water stand in the tank when in inverted position?

  1. 3.87 m
  2. 4.61 m
  3. 4.89 m
  4. 4.23 m
Total cone volume $=\tfrac13\pi(1)^2(5)=5.236$ m$^3$. Upright (apex up), the empty top cone of height $5-3=2$ has radius $0.4$: volume $=\tfrac13\pi(0.4)^2(2)=0.335$. Water $=5.236-0.335=4.901$ m$^3$.
Inverted (apex down), water forms a cone of depth $d$ with $r=d/5$: $\tfrac13\pi\left(\tfrac{d}{5}\right)^2 d=4.901\Rightarrow d^3=116.97$.
$d=4.89$ m.
$\boxed{4.89\text{ m}}$

Question Bank: t543

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The total surface area of a regular tetrahedron is 400 square cm. Find the volume of the tetrahedron in cc.

  1. 356.2
  2. 413.6
  3. 487.5
  4. 254.2
Four equilateral faces, so each $=\dfrac{400}{4}=100=\dfrac{\sqrt3}{4}a^2\Rightarrow a=15.20$ cm.
$V=\dfrac{a^3}{6\sqrt2}=\dfrac{15.20^3}{8.485}=413.6$ cc.
$\boxed{413.6}$

Question Bank: t545

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

Determine the lateral area in cm² of the following right pyramid:

Base is regular octagon of side 20 cm and altitude of 30 cm.

  1. 3256.32
  2. 2856.21
  3. 3080.62
  4. 2564.58

Base is regular hexagon of side 20 cm and altitude of 30 cm.

  1. 2078.46
  2. 2236.54
  3. 1863.54
  4. 2174.56

Base is regular hexagon of side 30 cm and slant height of 50 cm.

  1. 4200
  2. 4500
  3. 3300
  4. 4400

Part 1.

Base apothem $=\tfrac{20}{2}\cot22.5^\circ=24.14$. Slant height $l=\sqrt{30^2+24.14^2}=38.51$.
Lateral area $=\tfrac12(\text{perimeter})l=\tfrac12(8\times20)(38.51)=3080.62$ cm$^2$.
$\boxed{3080.62}$

Part 2.

Hexagon apothem $=\tfrac{20}{2}\cot30^\circ=17.32$. Slant height $l=\sqrt{30^2+17.32^2}=34.64$.
Lateral area $=\tfrac12(6\times20)(34.64)=2078.46$ cm$^2$.
$\boxed{2078.46}$

Part 3.

Lateral area $=\tfrac12(\text{perimeter})(\text{slant height})=\tfrac12(6\times30)(50)$.
$=\tfrac12(180)(50)=4500$ cm$^2$.
$\boxed{4500}$

Question Bank: t548

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A square pyramid has base width of 6 cm and altitude of 18 cm. Find the volume.

  1. 216 cc
  2. 261 cc
  3. 248 cc
  4. 284 cc
$V=\tfrac13(\text{base area})(\text{height})=\tfrac13(6^2)(18)=\tfrac13(36)(18)=216$ cc.
$\boxed{216\text{ cc}}$

Question Bank: t550

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A square pyramid has base width of 4 m and height of 10 m. A smaller pyramid is cut from this pyramid by passing a plane 6 m from the base of the given pyramid. What is the volume of the small pyramid?

  1. 3.98 m³
  2. 3.41 m³
  3. 4.36 m³
  4. 2.85 m³
The cutting plane is 6 m above the base, so the small top pyramid has height $10-6=4$ m. Scale $=\tfrac{4}{10}=0.4$, so its base width $=4(0.4)=1.6$ m.
$V=\tfrac13(1.6^2)(4)=3.41$ m$^3$.
$\boxed{3.41\text{ m}^3}$

Question Bank: t551

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The frustum of a regular triangular pyramid has equilateral triangles for its bases. The lower and upper base edges are 9m and 3m, respectively. If the volume is 118.2 cubic meters, how far apart are the bases?

  1. 9m
  2. 8m
  3. 7m
  4. 10m
Base areas: $A_1=\tfrac{\sqrt3}{4}(9)^2=35.07$, $A_2=\tfrac{\sqrt3}{4}(3)^2=3.90$, $\sqrt{A_1A_2}=11.69$.
Frustum volume $=\tfrac{h}{3}(A_1+A_2+\sqrt{A_1A_2})=\tfrac{h}{3}(50.66)=118.2$.
$h=7$ m.
$\boxed{7\text{m}}$

Question Bank: t552

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The upper base of a frustum of a pyramid is 2.5 m by 4 m and the lower base is 5 m by 8 m. Find its volume if the distance between the bases is 6 m.

  1. 160 m³
  2. 140 m³
  3. 130 m³
  4. 120 m³
$A_1=5\times8=40$, $A_2=2.5\times4=10$, $\sqrt{A_1A_2}=\sqrt{400}=20$.
$V=\tfrac{h}{3}(A_1+A_2+\sqrt{A_1A_2})=\tfrac{6}{3}(40+10+20)=2(70)=140$ m$^3$.
$\boxed{140\text{ m}^3}$

Question Bank: t554

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The lower and upper bases of a frustum of a pyramid measure 18 cm × 18 cm and 10 cm × 10 cm, respectively. If its lateral surface area is 448 cm², find the altitude of the pyramid.

  1. 9.36 cm
  2. 8 cm
  3. 6.72 cm
  4. 6.93 cm
Lateral area $=\tfrac12(P_1+P_2)l$: $448=\tfrac12(72+40)l=56l\Rightarrow l=8$ (slant height).
The horizontal offset of the slant is $\tfrac{18-10}{2}=4$, so the altitude $h=\sqrt{l^2-4^2}=\sqrt{64-16}=\sqrt{48}=6.93$ cm.
$\boxed{6.93\text{ cm}}$

Question Bank: t556

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A rectangular prism of metal having dimensions 4.3 cm by 7.2 cm by 12.4 cm is melted down and recast into a frustum of a square pyramid, 10% of the metal being lost in the process. If the ends of the frustum are squares of side 3 cm and 8 cm respectively, find the thickness of the frustum.

  1. 11.78 cm
  2. 10.69 cm
  3. 9.43 cm
  4. 12.87 cm
Volume of the rectangular prism:
$V = 4.3 \times 7.2 \times 12.4 = 383.904$ cm3
With 10% lost, the recast volume is:
$V_f = 0.90(383.904) = 345.514$ cm3
For a frustum of a square pyramid with base sides $a_1=3$ cm and $a_2=8$ cm, the areas are $A_1=9$ cm2 and $A_2=64$ cm2:
$V_f = \frac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 A_2}\right)$
$345.514 = \frac{h}{3}\left(9 + 64 + \sqrt{9 \times 64}\right) = \frac{h}{3}(97)$
$h = \frac{3(345.514)}{97}$
$\boxed{h = 10.69 \text{ cm}}$

Question Bank: t557

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 28.0 cm and 6.00 cm and the vertical distance between the ends is 30.0 cm, find the area of material needed to cover the curved surface of the speaker.

  1. 1,836.32 cm²
  2. 1,589.45 cm²
  3. 1,706.52 cm²
  4. 1,645.32 cm²
End diameters give radii $R = 14$ cm and $r = 3$ cm, with vertical height $h = 30$ cm.
Slant height of the frustum:
$L = \sqrt{h^2 + (R-r)^2} = \sqrt{30^2 + 11^2} = \sqrt{1021} = 31.95$ cm
Curved (lateral) surface area:
$A = \pi(R + r)L = \pi(14 + 3)(31.95)$
$\boxed{A = 1{,}706.52 \text{ cm}^2}$

Question Bank: t558

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The upper base diameter of the frustum of a right circular cone is 2 m and the lower base diameter is 4 m. If its altitude is 5 m, find its lateral area.

  1. 55.21 m²
  2. 43.98 m²
  3. 48.06 m²
  4. 32.87 m²
Radii: $r = 1$ m (top), $R = 2$ m (bottom), altitude $h = 5$ m.
Slant height:
$L = \sqrt{h^2 + (R-r)^2} = \sqrt{5^2 + 1^2} = \sqrt{26} = 5.099$ m
Lateral area:
$A = \pi(R + r)L = \pi(2 + 1)(5.099)$
$\boxed{A = 48.06 \text{ m}^2}$

Question Bank: t559

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The upper base of a frustum of a rectangular pyramid is 2.5 m × 4 m and its altitude is 5 m. If its volume is 116.67 m³ what is the area of the lower base?

  1. 38 m²
  2. 40 m²
  3. 44 m²
  4. 64 m²
Upper base area $A_1 = 2.5 \times 4 = 10$ m2, altitude $h = 5$ m, volume $V = 116.67$ m3.
Frustum volume formula:
$V = \frac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 A_2}\right)$
$116.67 = \frac{5}{3}\left(10 + A_2 + \sqrt{10 A_2}\right)$
$70 = 10 + A_2 + \sqrt{10 A_2}$
Solving the equation gives $A_2 = 40$:
$10 + 40 + \sqrt{400} = 10 + 40 + 20 = 70$ ✓
$\boxed{A_2 = 40 \text{ m}^2}$

Question Bank: t560

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A cylindrical water tank is 2 m in diameter and 4 m high. This tank is to be replaced with another tank of the same capacity in the shape of a frustum of a cone with lower base diameter of 2 m and upper base diameter of 1.4 m. What height of tank is required?

  1. 4.87 m
  2. 5.92 m
  3. 5.48 m
  4. 6.12 m
Cylinder capacity ($r = 1$ m, height 4 m):
$V = \pi r^2 h = \pi (1)^2(4) = 12.566$ m3
Frustum cone with $R = 1$ m (lower), $r = 0.7$ m (upper):
$V = \frac{\pi h}{3}\left(R^2 + r^2 + Rr\right) = \frac{\pi h}{3}\left(1 + 0.49 + 0.7\right)$
Set equal to the cylinder volume:
$12.566 = \frac{\pi h}{3}(2.19)$
$h = \frac{3(12.566)}{\pi(2.19)}$
$\boxed{h = 5.48 \text{ m}}$

Question Bank: t566

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

The surface area of a sphere is 2058.87 square centimeter. What is the volume in cubic inches?

  1. 56674
  2. 143953
  3. 1362
  4. 536
Find the radius from the surface area:
$S = 4\pi r^2 = 2058.87 \Rightarrow r^2 = 163.84$, $r = 12.80$ cm
Volume:
$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (12.80)^3 = 8783$ cm3
Convert to cubic inches ($1$ in $= 2.54$ cm, so $1$ in3 $= 16.387$ cm3):
$V = \frac{8783}{16.387}$
$\boxed{V = 536 \text{ in}^3}$

Question Bank: t572

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

If 28.27 cubic meter of water is inside a spherical tank whose radius is 2 m, what is the wetted area of the tank?

  1. 28.7 m²
  2. 58.6 m²
  3. 42.8 m²
  4. 37.7 m²
The water forms a spherical segment (cap) of height $h$ in a sphere of radius $R = 2$ m.
$V = \pi h^2\left(R - \frac{h}{3}\right)$
$28.27 = \pi h^2\left(2 - \frac{h}{3}\right)$
Solving gives $h = 3$ m: $\pi(9)(2 - 1) = 28.27$ ✓
Wetted area is the spherical zone:
$A = 2\pi R h = 2\pi (2)(3)$
$\boxed{A = 37.7 \text{ m}^2}$

Question Bank: t573

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A spherical tank 4.5 m in diameter contains water to a depth of 3 m. Find the volume.

  1. 32.14 m³
  2. 28.57 m³
  3. 35.34 m³
  4. 38.67 m³
Sphere radius $R = 4.5/2 = 2.25$ m, water depth (cap height) $h = 3$ m.
Volume of a spherical segment:
$V = \pi h^2\left(R - \frac{h}{3}\right) = \pi (3)^2\left(2.25 - \frac{3}{3}\right)$
$V = \pi (9)(1.25)$
$\boxed{V = 35.34 \text{ m}^3}$

Question Bank: t589

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A solid metal ball is immersed in a pail of paint and displaces 288$\pi$ cm³ of paint. What is the total area painted?

  1. 136$\pi$
  2. 72$\pi$
  3. 112$\pi$
  4. 144$\pi$
The displaced paint equals the ball's volume:
$\frac{4}{3}\pi r^3 = 288\pi \Rightarrow r^3 = 216 \Rightarrow r = 6$ cm
Painted area is the ball's surface area:
$A = 4\pi r^2 = 4\pi (6)^2$
$\boxed{A = 144\pi}$

Question Bank: t598

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A frustum of a sphere have a thickness of 10 cm. If the base radii are 19 cm and 22 cm, determine the following:

The volume of the frustum.

  1. 1268.7 cm³
  2. 1563.8 cm³
  3. 13796.8 cm³
  4. 14256.2 cm³

The radius of the sphere.

  1. 23.56 cm
  2. 25.87 cm
  3. 22.03 cm
  4. 26.98 cm

The total surface area of the frustum.

  1. 4038.83 cm²
  2. 1384.19 cm²
  3. 4763.54 cm²
  4. 3856.54 cm²

Part 1.

Base radii $a = 19$ cm, $b = 22$ cm, thickness $h = 10$ cm.
Volume of the spherical frustum (segment with two bases):
$V = \frac{\pi h}{6}\left(3a^2 + 3b^2 + h^2\right)$
$V = \frac{\pi (10)}{6}\left(3(361) + 3(484) + 100\right) = \frac{\pi (10)}{6}(2635)$
$\boxed{V = 13796.8 \text{ cm}^3}$

Part 2.

With both bases on the same side of the center (larger base nearer):
$h = \sqrt{R^2 - a^2} - \sqrt{R^2 - b^2}$
$10 = \sqrt{R^2 - 19^2} - \sqrt{R^2 - 22^2}$
Solving gives $R = 22.03$ cm:
$\sqrt{22.03^2 - 361} - \sqrt{22.03^2 - 484} = 11.15 - 1.14 = 10.0$ ✓
$\boxed{R = 22.03 \text{ cm}}$

Part 3.

Total surface = curved zone + both base circles.
Zone: $A_z = 2\pi R h = 2\pi (22.03)(10) = 1384.2$ cm2
Base circles: $\pi a^2 + \pi b^2 = \pi(19^2) + \pi(22^2) = \pi(845) = 2654.6$ cm2
Total:
$A = 1384.2 + 2654.6$
$\boxed{A = 4038.83 \text{ cm}^2}$

Question Bank: t610

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A triangle whose vertices are (1, 1), (3, 7), and (5, 4) is revolved about the line 2x + y + 2 = 0.

What is the area of the triangle in square units?

  1. 8.5
  2. 9
  3. 9.5
  4. 10

What is the coordinate of the centroid of the triangle?

  1. (2, 4)
  2. (3, 6)
  3. (3, 4)
  4. (2, 5)

What is the volume generated by the triangle in cubic units?

  1. 303.5
  2. 312.8
  3. 324.7
  4. 285.2

Part 1.

Area from the vertices $(1,1),(3,7),(5,4)$:
$A = \frac{1}{2}\left|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\right|$
$A = \frac{1}{2}\left|1(7-4) + 3(4-1) + 5(1-7)\right| = \frac{1}{2}|3 + 9 - 30|$
$\boxed{A = 9}$

Part 2.

Centroid is the average of the vertices:
$\bar{x} = \frac{1+3+5}{3} = 3, \quad \bar{y} = \frac{1+7+4}{3} = 4$
$\boxed{(3,\ 4)}$

Part 3.

Distance from the centroid $(3,4)$ to the line $2x + y + 2 = 0$:
$d = \frac{|2(3) + 4 + 2|}{\sqrt{2^2 + 1^2}} = \frac{12}{\sqrt{5}} = 5.367$
By Pappus' theorem:
$V = 2\pi d A = 2\pi (5.367)(9)$
$\boxed{V = 303.5}$

Question Bank: t613

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Gemini mapped Chapter 1 to 3

A square ABCD has the following corners: A(0, 3), B(3, 0), C(0, -3), D(-3, 0). Find the volume if the square is revolved about the y-axis.

  1. 9$\pi$
  2. 15$\pi$
  3. 18$\pi$
  4. 6$\pi$
The square is symmetric about the y-axis, so revolve its right half — triangle $(0,3),(3,0),(0,-3)$ — about the y-axis.
Half-area: $A = \frac{1}{2}(6)(3) = 9$, with centroid at $x = \frac{0+3+0}{3} = 1$.
By Pappus' theorem:
$V = 2\pi \bar{x} A = 2\pi (1)(9)$
$\boxed{V = 18\pi}$

Question Bank: t2094

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Besavilla CE Pre-Board Math & Surveying

Find the volume of the frustum of a regular pyramid if its upper base is 1.5 m. x 1.5 m. square and its lower base is 3.2 m. x 3.2 m. square and its lateral edge is 2.75 m. long.

  1. 12.15 m3
  2. 10.23 m3
  3. 18.72 m3
  4. 16.23 m3
  5. 14.24 m3
The horizontal offset between corresponding upper and lower vertices equals the difference of the half-diagonals of the square bases.
$d=\frac{3.2}{\sqrt2}-\frac{1.5}{\sqrt2}=1.202$ m
With lateral edge 2.75 m, the frustum height is
$h=\sqrt{2.75^2-d^2}=2.47$ m
Base areas: $A_1=3.2^2=10.24\text{ m}^2$, $A_2=1.5^2=2.25\text{ m}^2$
Frustum volume:
$V=\frac{h}{3}(A_1+A_2+\sqrt{A_1A_2})$
$V=\frac{2.47}{3}(10.24+2.25+4.80)$
$\boxed{V\approx14.24\text{ m}^3}$

Question Bank: t2100

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Besavilla CE Pre-Board Math & Surveying

The horizontal base of a regular pyramid is an equilateral triangle 20 m. on each side. If the volume of the pyramid is 531 cu.m., what is its lateral area?

  1. 308.15 m2
  2. 356.35 m2
  3. 312.25 m2
  4. 325.80 m2
  5. 385.75 m2
Base area of the equilateral triangle:
$B=\frac{\sqrt3}{4}(20)^2=173.205\text{ m}^2$
From $V=\frac{1}{3}Bh$:
$531=\frac{1}{3}(173.205)h \Rightarrow h=9.197$ m
Base apothem of the equilateral triangle:
$a_b=\frac{20\sqrt3}{6}=5.774$ m
Slant height of a lateral face:
$l=\sqrt{h^2+a_b^2}=10.86$ m
Lateral area $=\frac{1}{2}Pl=\frac{1}{2}(60)(10.86)$
$\boxed{325.80\text{ m}^2}$

Question Bank: t2157

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / Besavilla CE Pre-Board Math & Surveying

A model of a boiler is made having an overall height of 75 mm corresponding to an overall height of the actual boiler of 6 m. If the area of metal required for the model is 12500 mm$^2$, determine the area of metal required for the actual boiler in square meters.

  1. 50 m2
  2. 60 m2
  3. 70 m2
  4. 80 m2
  5. 90 m2
Convert the actual boiler height to millimeters: $6\text{ m}=6000\text{ mm}$.
Linear scale factor $=\frac{6000}{75}=80$
Areas scale as the square of the linear scale factor.
Model area $=12500\text{ mm}^2=0.0125\text{ m}^2$
Actual area $=0.0125(80)^2$
$\boxed{80\text{ m}^2}$