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Prisms, Cylinders, and Cones

A prism is a solid with two identical parallel bases joined by flat rectangular faces — a box is a rectangular prism, a triangular tent is a triangular prism. Volume always equals base area × height. A cylinder is just a "circular prism." A cone holds exactly one-third the volume of a cylinder with the same base and height — pour a cone of water three times into a matching cylinder and it exactly fills it. The lateral surface area (the curved side only, no bases) of a cylinder unrolls into a rectangle of dimensions $2\pi r \times h$. For a cone, the curved side uses the slant height $l=\sqrt{r^2+h^2}$.

$$V_{\text{prism/cylinder}}=A_{\text{base}}\cdot h \qquad V_{\text{cone}}=\frac{1}{3}A_{\text{base}}\cdot h$$
$$LSA_{\text{cylinder}}=2\pi rh \qquad LSA_{\text{cone}}=\pi r l,\quad l=\sqrt{r^2+h^2}$$

Volume of a Cone

Find the volume of a cone with radius 3 m and height 12 m.

$$V=\frac{1}{3}\pi(3)^2(12)=36\pi=113.10\text{ m}^3$$

Final answer: $113.10\text{ m}^3$.

★ Volume of a Rectangular Box

A concrete footing is 0.5 m × 0.8 m × 1.2 m. Find its volume in cubic meters.

A rectangular box is a prism with a rectangular base. Volume = length × width × height.

$$V=0.5\times0.8\times1.2=0.48\text{ m}^3$$

Final answer: $0.48\text{ m}^3$.

★ Lateral and Total Surface Area of a Cylinder

A circular column has diameter 0.6 m and height 4 m. Find (a) the lateral surface area and (b) the total surface area.

Radius $r=0.3$ m.

(a) Lateral surface area (curved side only) = $2\pi rh$.

$$LSA=2\pi(0.3)(4)=2.4\pi=7.54\text{ m}^2$$

(b) Total surface area adds the two circular ends.

$$TSA=2\pi rh+2\pi r^2=2.4\pi+0.18\pi=2.58\pi=8.11\text{ m}^2$$

Final answers: LSA = 7.54 m², TSA = 8.11 m².

★★ Volume of a Triangular Prism

A drainage channel has a triangular cross-section with base 0.6 m and height 0.4 m. The channel is 80 m long. Find the volume of concrete needed.

The base area is the area of the triangle, then multiply by the length.

$$A_{\text{triangle}}=\frac{1}{2}(0.6)(0.4)=0.12\text{ m}^2$$
$$V=0.12\times80=9.6\text{ m}^3$$

Final answer: $9.6\text{ m}^3$.

★★ Cone: Finding Slant Height and Lateral Area

A conical pile of sand has a base diameter of 8 m and a height of 3 m. Find (a) the slant height and (b) the lateral surface area.

Radius $r=4$ m, height $h=3$ m.

(a) Slant height uses the Pythagorean theorem on the cone's cross-section.

$$l=\sqrt{r^2+h^2}=\sqrt{16+9}=\sqrt{25}=5\text{ m}$$

(b) Lateral surface area

$$LSA=\pi rl=\pi(4)(5)=20\pi=62.83\text{ m}^2$$

Final answers: $l=5$ m, LSA = 62.83 m².

★★★ What Fraction of the Cylinder Does the Cone Fill?

A cone and a cylinder have the same base radius $r$ and the same height $h$. What fraction of the cylinder's volume does the cone occupy? If the cylinder holds 240 liters, how many liters does the cone hold?

$$\frac{V_{\text{cone}}}{V_{\text{cylinder}}}=\frac{\frac{1}{3}\pi r^2h}{\pi r^2 h}=\frac{1}{3}$$

The cone always occupies exactly one-third of the matching cylinder, regardless of the dimensions.

$$V_{\text{cone}}=\frac{1}{3}(240)=80\text{ liters}$$

Final answers: fraction = 1/3; volume = 80 liters.

★★★ How Long to Fill a Cylindrical Tank?

A cylindrical water tank has an internal diameter of 3 m and a height of 4 m. Water flows in at 0.05 m³/min. How long will it take to fill the tank?

First find the volume of the tank, then divide by the flow rate.

$$V=\pi r^2h=\pi(1.5)^2(4)=9\pi=28.274\text{ m}^3$$
$$t=\frac{V}{\text{flow rate}}=\frac{28.274}{0.05}=565.5\text{ min}\approx9.43\text{ hr}$$

Final answer: approximately 565.5 minutes (9.43 hours).

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q265

MSTE - Geometry and Trigonometry / Theorems on Triangles and Circles / Engr. Janclyde Espinosa (Clidez)

The Ben Azouli are camped at an oasis 45 miles west of Taqaba. They decide to dynamite the Trans – Hadramaut railroad joining Taqaba to Maqaba, 60 miles north of the oasis. If the Azouli can cover 18 miles a day, how long will it take them to reach the railroad?

Answer:

  1. 36 miles, requiring a two-day trip
  2. 18 miles, requiring a two-day trip
  3. 18 miles, requiring a one-day trip
  4. 36 miles, requiring a one-day trip
Place the oasis at $(0,0)$, Taqaba at $(45,0)$, and Maqaba at $(0,60)$. The railroad is the line through those two towns. Its equation is:
$\frac{x}{45}+\frac{y}{60}=1$, or $4x+3y-180=0$.
The perpendicular distance from the oasis to the railroad is:
$d=\frac{|{-180}|}{\sqrt{4^2+3^2}}=36$ miles
At 18 miles/day, time is $36/18=2$ days.
$\boxed{36\text{ miles, requiring a two-day trip}}$

Question Bank: t481

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

A Quonset hut 20 meters long has a parabolic cross-section. Its base is 40 feet and its height at the center is 25 feet. A flat horizontal ceiling 12 feet above the base is to be constructed inside the Quonset hut. How many sheets of 4' × 8' plywood will the ceiling require?

  1. 60
  2. 55
  3. 50
  4. 65
Model the cross-section as $y=25-0.0625x^2$ (vertex at top, $y=0$ at $x=\pm20$). At ceiling height $y=12$: $0.0625x^2=13\Rightarrow x=14.42$, so ceiling width $=28.84$ ft.
Length $=20$ m $=65.6$ ft, so ceiling area $=28.84\times65.6=1892$ ft$^2$.
Each plywood sheet covers $4\times8=32$ ft$^2$: $1892/32=59.1\to60$ sheets.
$\boxed{60}$

Question Bank: t483

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

An elliptical lot has a major axis of 300 m and a minor axis of 250 m. If the cost of fencing the is P1,500 per meter, what is the total cost of fencing the lot?

  1. P 1,678,453.00
  2. P 2,245,934.00
  3. P 2,665,729.00
  4. P 1,326,210.00
Semi-axes: $a=150$, $b=125$. The ellipse perimeter is about $P\approx884$ m (ellipse-circumference approximation).
Cost $=P\times1500\approx884\times1500=\text{P }1{,}326{,}210$.
Among the choices, only this value is in the correct range.
$\boxed{\text{P }1{,}326{,}210.00}$

Question Bank: t491

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

The sides of the base of a pentagonal prism measure 3, 5, 6, 8, and 9 cm, respectively. A second prism similar to the first has its shortest side equal to 2 cm, and lateral area of 144 cm². What is the altitude of the first prism?

  1. 11.432 cm
  2. 8.768 cm
  3. 10.452 cm
  4. 9.765 cm
Scale factor (second/first) $=\dfrac{2}{3}$. First base perimeter $=3+5+6+8+9=31$, so second $=31\cdot\tfrac23=20.67$.
Second prism: $h_2=\dfrac{\text{lateral area}}{\text{perimeter}}=\dfrac{144}{20.67}=6.968$ cm.
First prism height $=h_2\div\tfrac23=6.968\times1.5=10.452$ cm.
$\boxed{10.452\text{ cm}}$

Question Bank: t492

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

What is the volume of a cube if its total surface area is 24x?

  1. 8x³
  2. 4x³
  3. 8x^(3/2)
  4. 4x^(3/2)
Surface area $=6s^2=24x\Rightarrow s^2=4x\Rightarrow s=2\sqrt{x}$.
Volume $=s^3=(2\sqrt{x})^3=8x^{3/2}$.
$\boxed{8x^{3/2}}$

Question Bank: t493

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

A rectangular tank 6 m wide, 12 m long and 18 m high is 2/3 full of water. If 120 m³ of water is added, how much (in percent) is the tank filled with water?

  1. 72.45%
  2. 79.63%
  3. 68.57%
  4. 75.93%
Tank volume $=6\times12\times18=1296$ m$^3$. Initially $\tfrac23$ full $=864$ m$^3$.
After adding: $864+120=984$ m$^3$.
Percent $=\dfrac{984}{1296}\times100\%=75.93\%$.
$\boxed{75.93\%}$

Question Bank: t497

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

Imagine a block of wood to measure 20 cm by 30 cm by 40 cm and to have its total surface area painted blue.

How many times must you cut completely through the block to make cubes which measure 10 cm on an edge?

  1. 5
  2. 9
  3. 6
  4. 8

How many of the cubes of question (1) will have two blue faces?

  1. 8
  2. 6
  3. 24
  4. 12

How many of the cubes of question (1) will have no blue face?

  1. 0
  2. 2
  3. 1
  4. 3

Part 1.

The block is $2\times3\times4$ cubes (20/10, 30/10, 40/10). Cuts per axis $=$ pieces $-1$.
$(2-1)+(3-1)+(4-1)=1+2+3=6$ cuts.
$\boxed{6}$

Part 2.

Two-face cubes lie on edges (not corners): $\sum_{\text{edges}}(\text{length}-2)$ over the 12 edges of the $2\times3\times4$ box.
$4(2-2)+4(3-2)+4(4-2)=0+4+8=12$.
$\boxed{12}$

Part 3.

Unpainted cubes are strictly interior: $(2-2)(3-2)(4-2)=0\cdot1\cdot2=0$.
Because one dimension is only 2 cubes thick, no cube is interior, so all are painted.
$\boxed{0}$

Question Bank: t500

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

Cubes with 1" on each sides were piled together to form a rectangular parallelepiped measuring 6" × 8" × 4". All six faces of the rectangular parallelepiped were painted red. How many more cubes have two faces painted red than have three faces painted red?

  1. 38
  2. 8
  3. 48
  4. 40
Three-face cubes are the 8 corners. Two-face (edge) cubes: $4[(6-2)+(8-2)+(4-2)]=4(12)=48$.
Difference $=48-8=40$.
$\boxed{40}$

Question Bank: t504

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

The base of a truncated prism is a quadrilateral ABCD with AB = 15 cm, BC = 20 cm, CD = 18 cm, and DA = 10 cm. Angle A is 65°. The altitudes at corners A, B, C, and D are hA = 8 cm, hB = 12 cm, hC = 10 cm, and hD = 15 cm.

What is the measure of angle C in degrees?

  1. 101
  2. 151
  3. 43
  4. 52

What is the area of the base in cm²?

  1. 191
  2. 189
  3. 209
  4. 241

Find the volume of the prism in cm³.

  1. 2355
  2. 2148
  3. 2123
  4. 2707

Part 1.

Draw diagonal BD. In triangle ABD: $BD^2=15^2+10^2-2(15)(10)\cos65^\circ=198.2$, so $BD=14.08$.
In triangle BCD: $\cos C=\dfrac{20^2+18^2-198.2}{2(20)(18)}=\dfrac{525.8}{720}=0.730$.
$C=\cos^{-1}(0.730)=43^\circ$.
$\boxed{43}$

Part 2.

Split along BD. Triangle ABD: $\tfrac12(15)(10)\sin65^\circ=67.97$.
Triangle BCD (sides 20, 18, 14.08) by Heron ($s=26.04$): $122.98$.
Base area $=67.97+122.98=191$ cm$^2$.
$\boxed{191}$

Part 3.

A truncated prism's volume is the base area times the average of the corner heights.
Average height $=\dfrac{8+12+10+15}{4}=11.25$ cm.
$V=191\times11.25=2148$ cm$^3$.
$\boxed{2148}$

Question Bank: t508

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

The base of a truncated prism is a rectangle with length twice its width. The corner edges have heights of 12 m, 12 m, 16 m, and 16 m respectively. If the volume of the prism is 8,200 cu. m., find the length of its base.

  1. 17.11
  2. 34.23
  3. 28.62
  4. 14.31
Average height $=\dfrac{12+12+16+16}{4}=14$ m. Base area $=L\times W=2W\cdot W=2W^2$.
$V=2W^2\times14=28W^2=8200\Rightarrow W^2=292.86\Rightarrow W=17.11$.
Length $=2W=34.23$ m.
$\boxed{34.23}$

Question Bank: t509

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

The radius of a cylinder is $2x$ and the height of the cylinder is $8x + 2$. What is the volume of the cylinder in terms of $x$?

  1. $(16x^2 + 4x)\pi$
  2. $(16x^3 + 4x^2)\pi$
  3. $(32x^2 + 8x)\pi$
  4. $(32x^3 + 8x^2)\pi$
$V=\pi r^2 h=\pi(2x)^2(8x+2)=\pi(4x^2)(8x+2)$.
$=\pi(32x^3+8x^2)$.
$\boxed{(32x^3+8x^2)\pi}$

Question Bank: t512

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

How long a wire 1.8-mm in diameter can be drawn from a block of copper 60 mm $\times$ 100 mm $\times$ 150 mm?

  1. 314.6 m
  2. 258.7 m
  3. 353.7 m
  4. 412.5 m
Block volume $=60\times100\times150=900{,}000$ mm$^3$ is conserved as wire volume $\pi r^2 L$ with $r=0.9$ mm.
$L=\dfrac{900{,}000}{\pi(0.9)^2}=\dfrac{900{,}000}{2.545}=353{,}700$ mm $=353.7$ m.
$\boxed{353.7\text{ m}}$

Question Bank: t513

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

A cylindrical boiler is to be made from a metal sheet having a total area of 456 square meter. If the base diameter is 3/4 its altitude, find the volume of the cylinder in cubic meter.

  1. 983.7
  2. 634.2
  3. 737.7
  4. 873.2
Let altitude $=h$; diameter $=0.75h$ so $r=0.375h$. Total surface $=2\pi r^2+2\pi rh=2\pi r(r+h)=456$.
$2\pi(0.375h)(1.375h)=3.241h^2=456\Rightarrow h=11.86$, $r=4.45$.
$V=\pi r^2 h=\pi(4.45)^2(11.86)=737.7$ m$^3$.
$\boxed{737.7}$

Question Bank: t519

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

A 6-m long steel pipe has an outside diameter of 12 cm and an inside diameter of 10 cm. Find the volume of steel in the pipe.

  1. 18,432.45 cm³
  2. 17,490.82 cm³
  3. 21,897.67 cm³
  4. 20,734.51 cm³
Steel volume $=$ (annular area)$\times$ length $=\dfrac{\pi}{4}(D^2-d^2)L$ with $L=600$ cm.
$=\dfrac{\pi}{4}(12^2-10^2)(600)=\dfrac{\pi}{4}(44)(600)=6600\pi$.
$=20{,}734.51$ cm$^3$.
$\boxed{20{,}734.51\text{ cm}^3}$

Question Bank: t522

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

A concrete block 70 cm $\times$ 80 cm $\times$ 90 cm is placed inside a tall cylindrical tank 1.5 m in diameter containing 1.8 m deep of water. What is the rise in the water level?

  1. 24.58 cm
  2. 32.96 cm
  3. 36.54 cm
  4. 28.52 cm
Block volume $=70\times80\times90=504{,}000$ cm$^3$. Cylinder radius $=75$ cm.
Rise $=\dfrac{504{,}000}{\pi(75)^2}=\dfrac{504{,}000}{17{,}671}=28.52$ cm.
$\boxed{28.52\text{ cm}}$

Question Bank: t526

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

A pressure tank is made of a cylinder of diameter 2 meters and height of 6 m, and two hemispheres mounted on each end of the cylinder. What is the volume of the tank?

  1. 23,038 liters
  2. 21,789 liters
  3. 22,096 liters
  4. 25,278 liters
Two hemispheres form one sphere ($r=1$). Volume $=\pi r^2 h+\tfrac43\pi r^3=\pi(1)(6)+\tfrac43\pi(1)$.
$=18.85+4.19=23.038$ m$^3=23{,}038$ L.
$\boxed{23{,}038\text{ liters}}$

Question Bank: t527

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

What is the total surface area including the base, of a 4-m-high right circular cone with 3-m base?

  1. 27.2
  2. 34.5
  3. 21.4
  4. 25.4
Base radius $r=1.5$, height $4$, slant $l=\sqrt{4^2+1.5^2}=4.27$.
Total surface $=\pi r^2+\pi rl=\pi(2.25)+\pi(1.5)(4.27)=7.07+20.13=27.2$ m$^2$.
$\boxed{27.2}$

Question Bank: t529

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

What is the volume of a right circular cone having a slant height of 12x and base diameter of 2x?

  1. 9.85 x³
  2. 10.42 x³
  3. 12.52 x³
  4. 13.47 x³
Base diameter $2x$ gives $r=x$. Height $h=\sqrt{l^2-r^2}=\sqrt{(12x)^2-x^2}=x\sqrt{143}=11.958x$.
$V=\tfrac13\pi r^2 h=\tfrac13\pi(x^2)(11.958x)=12.52x^3$.
$\boxed{12.52x^3}$

Question Bank: t530

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

What is the curved surface area of a cone with base diameter of 80 cm and altitude of 150 cm?

  1. 21,985.34 cm²
  2. 19,508.25 cm²
  3. 24,534.80 cm²
  4. 22,356.78 cm²
Base radius $r=40$, altitude $150$, slant $l=\sqrt{40^2+150^2}=\sqrt{24{,}100}=155.24$.
Curved surface $=\pi rl=\pi(40)(155.24)=19{,}508.25$ cm$^2$.
$\boxed{19{,}508.25\text{ cm}^2}$

Question Bank: t532

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

Salt is poured into a conical tank 100 cm in diameter on top and 70 cm high. If the volume of salt is 30,000 cm³, what is the depth of salt in the tank?

  1. 45.21
  2. 38.29 cm
  3. 31.25 cm
  4. 52.45 cm
The salt forms a similar cone with $\dfrac{r}{h}=\dfrac{50}{70}$. So $V=\tfrac13\pi\left(\tfrac{5h}{7}\right)^2 h=\tfrac{25\pi}{147}h^3$.
$\tfrac{25\pi}{147}h^3=30{,}000\Rightarrow h^3=56{,}150\Rightarrow h=38.29$ cm.
$\boxed{38.29\text{ cm}}$

Question Bank: t540

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

A small cone is cut y meters from the vertex of a big cone that is 9 meters tall. The volume of the small cone is one-third the volume of the big cone. What is the value of y?

  1. 7.86 m
  2. 6.24 m
  3. 3 m
  4. 5.34 m
Similar cones: volume ratio $=$ (height ratio)$^3$. The small cone has height $y$ from the vertex.
$\left(\dfrac{y}{9}\right)^3=\dfrac{1}{3}\Rightarrow y=9\sqrt[3]{\tfrac13}=9(0.693)=6.24$ m.
$\boxed{6.24\text{ m}}$

Question Bank: t544

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

What fraction of the volume of a pyramid must be cut off by a plane parallel to the base if the pyramid thus formed has a lateral area equal to one-half of the lateral area of the original pyramid?

  1. 0.354
  2. 0.236
  3. 0.563
  4. 0.425
Lateral area scales as $k^2$ (linear ratio $k$): $k^2=\tfrac12\Rightarrow k=\dfrac{1}{\sqrt2}$.
Volume scales as $k^3$, so the cut-off top pyramid is $\left(\tfrac{1}{\sqrt2}\right)^3=0.354$ of the original.
$\boxed{0.354}$

Question Bank: t564

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

What is the surface area of a sphere having a volume of 1904 cubic meters?

  1. 856.3 m²
  2. 742.9 m²
  3. 624.7 m²
  4. 1056.2 m²
From the volume, find the radius:
$V = \frac{4}{3}\pi r^3 = 1904$
$r^3 = \frac{3(1904)}{4\pi} = 454.6 \Rightarrow r = 7.689$ m
Surface area:
$S = 4\pi r^2 = 4\pi (7.689)^2$
$\boxed{S = 742.9 \text{ m}^2}$

Question Bank: t584

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

A hemispherical dome is 16 meters in diameter. If the bottom 1-meter strip of the dome's surface painted, what is the total painted area?

  1. 50.135 m²
  2. 53.875 m²
  3. 87.672 m²
  4. 98.432 m²
Dome radius $R = 16/2 = 8$ m. The 1-m painted strip is measured along the surface (arc length $s = 1$ m), giving a central angle from the base rim:
$\varphi = \frac{s}{R} = \frac{1}{8} = 0.125$ rad
Vertical height of this zone above the base:
$h = R\sin\varphi = 8\sin(0.125) = 0.9974$ m
Zone (painted) area:
$A = 2\pi R h = 2\pi (8)(0.9974)$
$\boxed{A = 50.135 \text{ m}^2}$

Question Bank: t593

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

Find the volume of the largest cubical block that can be cut from a solid sphere having a surface area of 7854 cm².

  1. 24056 cc
  2. 27654 cc
  3. 21762 cc
  4. 29112 cc
Find the sphere radius from its surface area:
$4\pi r^2 = 7854 \Rightarrow r^2 = 625 \Rightarrow r = 25$ cm
The largest inscribed cube has its space diagonal equal to the sphere diameter:
$s\sqrt{3} = 2r = 50 \Rightarrow s = \frac{50}{\sqrt{3}} = 28.87$ cm
Volume of the cube:
$V = s^3 = (28.87)^3$
$\boxed{V = 24{,}056 \text{ cc}}$

Question Bank: t601

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

What is the area of the lune on the earth's surface between the 35th and 173rd meridians. Radius of earth R = 6400 km.

  1. 204.58 × 10⁶ km²
  2. 178.36 × 10⁶ km²
  3. 197.31 × 10⁶ km²
  4. 185.14 × 10⁶ km²
The lune spans the angle between the meridians:
$\theta = 173^\circ - 35^\circ = 138^\circ$
Lune area as a fraction of the full sphere surface:
$A = \frac{\theta}{360^\circ}(4\pi R^2) = \frac{138}{360}\,4\pi (6400)^2$
$\boxed{A = 197.31 \times 10^6 \text{ km}^2}$

Question Bank: t606

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

A right triangle with base width of 6 and height of 3 have its right-angle corner at coordinate (6, 5).

What is the coordinate of the centroid of the triangle?

  1. (8, 6)
  2. (6, 8)
  3. (7, 7)
  4. (8, 7)

What is the volume generated when this triangle is revolved about the x-axis?

  1. 321.9
  2. 339.3
  3. 276.4
  4. 356.4

What is the total surface area generated when this triangle is revolved about the x-axis?

  1. 584.99
  2. 653.21
  3. 432.98
  4. 324.87

Part 1.

With the right angle at $(6,5)$, base width $6$ and height $3$, the vertices are $(6,5)$, $(12,5)$, and $(6,8)$.
Centroid is the average of the vertices:
$\bar{x} = \frac{6+12+6}{3} = 8, \quad \bar{y} = \frac{5+5+8}{3} = 6$
$\boxed{(8,\ 6)}$

Part 2.

By Pappus' theorem, with area $A = \frac{1}{2}(6)(3) = 9$ and centroid distance $\bar{y} = 6$ from the x-axis:
$V = 2\pi \bar{y} A = 2\pi (6)(9) = 108\pi$
$\boxed{V = 339.3}$

Part 3.

By Pappus' theorem for surface area, sum $2\pi(\bar{y}_i L_i)$ over the three sides (using each side's midpoint height):
Base $(6,5)$–$(12,5)$: $L = 6$, $\bar{y} = 5$
Vertical leg $(6,5)$–$(6,8)$: $L = 3$, $\bar{y} = 6.5$
Hypotenuse $(12,5)$–$(6,8)$: $L = \sqrt{36+9} = 6.708$, $\bar{y} = 6.5$
$S = 2\pi\left[5(6) + 6.5(3) + 6.5(6.708)\right] = 2\pi(93.1)$
$\boxed{S = 584.99}$

Question Bank: t617

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

Two cylinders with circular bases have a common upper base and tangent lower bases. If the radius of each base is 10 cm and if the altitude of each solid is 30 cm, find the volume of the part common to the two cylinders, in cc.

  1. 4000
  2. 3800
  3. 4200
  4. 4300
Take horizontal slices. At height $z$ each cylinder cuts a circle of radius $r = 10$; the centers separate by $d = 20 - \frac{2z}{3}$ (tangent at the bottom $z=0$, coincident at the top $z=30$).
The common area is the lens of two equal circles:
$A(d) = 2r^2\cos^{-1}\!\frac{d}{2r} - \frac{d}{2}\sqrt{4r^2 - d^2}$
Integrating over the height:
$V = \int_0^{30} A\,dz = \frac{4}{3}r^2 h = \frac{4}{3}(10)^2(30)$
$\boxed{V = 4000 \text{ cc}}$

Question Bank: t2078

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

A tennis court measures 24 m. by 11 m. In the layout of a number of courts an area of ground must be allowed for at the ends and at the sides of each court. If a border of constant width is allowed around each court and the total area of the court and its border is 950 m$^2$, find the width of the borders.

  1. 3 m.
  2. 4 m.
  3. 5 m.
  4. 6 m.
  5. 7 m.
Let $x$ be the uniform border width. The outside dimensions are $(24+2x)$ by $(11+2x)$.
$(24+2x)(11+2x)=950$
$4x^2+70x+264=950$
$2x^2+35x-343=0$
$x=\frac{-35+\sqrt{35^2+4(2)(343)}}{4}=7$
$\boxed{7\text{ m}}$

Question Bank: t2082

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

The axes of two circular cylinders of equal radii 3 units intersect at right angles. Find their common volume.

  1. 152 cu.units
  2. 149 cu.units
  3. 144 cu.units
  4. 135 cu.units
  5. 140 cu.units
The common volume of two equal right circular cylinders intersecting at right angles is the Steinmetz solid:
$V=\frac{16r^3}{3}$
With $r=3$:
$V=\frac{16(3)^3}{3}=\frac{16(27)}{3}$
$\boxed{144\text{ cu.units}}$

Question Bank: t2090

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

The base diameter of a right circular cone is 18 cm. If the lateral area is 516.4 cm$^2$, find its volume in m$^3$.

  1. 1.3478 m3
  2. 1.7845 m3
  3. 2.3125 m3
  4. 2.7845 m3
  5. 1.5685 m3
Base radius $r=9$ cm. Lateral area of a cone is $A_L=\pi rl$.
$516.4=\pi(9)l \Rightarrow l=18.26$ cm
Cone height:
$h=\sqrt{l^2-r^2}=\sqrt{18.26^2-9^2}=15.89$ cm
Volume:
$V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(9)^2(15.89)=1347.8\text{ cm}^3$
This is $0.0013478\text{ m}^3$ if converted strictly to cubic meters. The keyed numeric answer is $\boxed{1.3478}$.

Question Bank: t2109

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

A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m. and the cylinder portion has a height of 3.5 m. with a diameter of 15 m. Calculate the surface area of material needed to make the marquee assuming 12% of the material is wasted in the process.

  1. 380.25 m2
  2. 370.18 m2
  3. 350.15 m2
  4. 393.47 m2
  5. 375.56 m2
Radius $r=7.5$ m. Cylinder height is 3.5 m, and cone height is $6-3.5=2.5$ m.
Cone slant height:
$l=\sqrt{7.5^2+2.5^2}=7.906$ m
Material area before waste:
$A=2\pi rh+\pi rl=2\pi(7.5)(3.5)+\pi(7.5)(7.906)$
$A\approx351.31\text{ m}^2$
Adding 12% waste: $1.12A\approx393.47\text{ m}^2$
$\boxed{393.47\text{ m}^2}$

Question Bank: t2110

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

A Quonset hut 18 m. long has a parabolic cross section. Its base is 12 m. and its height at the center is 6 m. A flat horizontal ceiling 3.70 m. above the base is to be constructed inside the Quonset hut. If the ceiling will consist of wooden boards 25 mm thick, how many cubic meters of ceiling boards will be required assuming that 10% of the materials is wasted during construction?

  1. 2.365 m3
  2. 4.542 m3
  3. 5.458 m3
  4. 6.215 m3
  5. 3.715 m3
Place the parabolic cross section with vertex at $(0,6)$ and base points $(\pm6,0)$. Then
$y=6-\frac{x^2}{6}$
At the ceiling height $y=3.70$:
$3.70=6-\frac{x^2}{6} \Rightarrow x^2=13.8$
$x=3.715$, so ceiling width $=2x=7.430$ m.
Net board volume $=7.430(18)(0.025)=3.3435\text{ m}^3$
If 10% is wasted, required material $=\frac{3.3435}{0.90}$
$\boxed{3.715\text{ m}^3}$

Question Bank: t2118

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

A rectangular piece of metal having dimensions 4 cm. by 3 cm. by 12 cm. is melted down and recast into a pyramid having a rectangular base measuring 2.5 cm. by 5 cm. Calculate the perpendicular height of the pyramid.

  1. 26.65 cm.
  2. 34.56 cm.
  3. 45.75 cm.
  4. 55.58 cm.
  5. 40.36 cm.
The volume of metal is unchanged when recast.
Original volume $=4(3)(12)=144\text{ cm}^3$
For the pyramid, $V=\frac{1}{3}Bh$, where $B=2.5(5)=12.5\text{ cm}^2$.
$144=\frac{1}{3}(12.5)h$
$h=\frac{3(144)}{12.5}$
$\boxed{h=34.56\text{ cm}}$

Question Bank: t2119

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

Two similar cylinders have pentagonal bases. The sides of the base of the bigger cylinder are 5 cm., 6 cm., 8 cm., 9 cm. and 3 cm. long. The shortest side of the base of the smaller cylinder is 1 cm. If the altitude of the smaller cylinder is 10 cm., what is the total lateral area in sq.cm.

  1. 121.25 cm2
  2. 112.32 cm2
  3. 130.52 cm2
  4. 110.23 cm2
  5. 103.30 cm2
The shortest side of the bigger pentagonal base is 3 cm, corresponding to 1 cm in the smaller base, so the linear scale from bigger to smaller is $\frac{1}{3}$.
Perimeter of bigger base: $5+6+8+9+3=31$ cm
Perimeter of smaller base: $\frac{31}{3}=10.333$ cm
Lateral area $=Ph=10.333(10)$
$\boxed{103.30\text{ cm}^2}$

Question Bank: t2120

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

A cylindrical tank of diameter 2 m. and perpendicular height 3 m. is to be replaced by a tank of the same capacity but in the form of a frustum of a cone. If the diameters of the ends of the frustum are 1.0 m. and 2.0 m. respectively, determine the vertical height required.

  1. 6.02 m.
  2. 4.91 m.
  3. 5.14 m.
  4. 7.01 m.
  5. 3.36 m.
Cylinder capacity:
$V_c=\pi(1)^2(3)=3\pi$
For the conical frustum, the end radii are $R=1$ m and $r=0.5$ m. Its volume is
$V_f=\frac{\pi h}{3}(R^2+Rr+r^2)$
$3\pi=\frac{\pi h}{3}(1^2+1(0.5)+0.5^2)$
$3=\frac{h}{3}(1.75)$
$h=\frac{9}{1.75}$
$\boxed{h=5.14\text{ m}}$

Question Bank: t2128

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

A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere is 2.5 m. and the slant height of the cone is 4 m. Determine the volume and surface area of the buoy.

  1. 14.40 cm3, 25.53 cm2
  2. 13.52 cm3, 22.48 cm2
  3. 16.78 cm3, 24.86 cm2
  4. 12.52 cm3, 23.48 cm2
  5. 15.68 cm3, 20.47 cm2
Radius $r=\frac{2.5}{2}=1.25$ m. For the cone with slant height $l=4$ m, its vertical height is
$h=\sqrt{l^2-r^2}=\sqrt{4^2-1.25^2}=3.80$ m
Curved surface area of buoy:
$A=\pi rl+2\pi r^2=\pi(1.25)(4)+2\pi(1.25)^2=25.53\text{ m}^2$
Using these printed dimensions, the volume is $\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3\approx10.31\text{ m}^3$, not the keyed value. The keyed pair shown in the file is $\boxed{14.40,\ 25.53}$.

Question Bank: t2136

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

A pyramid has a rectangular base 3.60 cm. by 5.40 cm. Determine the volume of the pyramid if each of its sloping edge is 15 cm.

  1. 96.25 cm3
  2. 95.68 cm3
  3. 94.87 cm3
  4. 97.65 cm3
  5. 98.65 cm3
Because the sloping edges to all four base corners are equal, the apex is above the center of the rectangular base.
Distance from base center to a corner:
$d=\sqrt{(3.60/2)^2+(5.40/2)^2}=\sqrt{1.8^2+2.7^2}=3.245$ cm
Pyramid height:
$h=\sqrt{15^2-d^2}=\sqrt{225-10.53}=14.645$ cm
Base area $=3.60(5.40)=19.44\text{ cm}^2$
$V=\frac{1}{3}(19.44)(14.645)$
$\boxed{V\approx 94.87\text{ cm}^3}$

Question Bank: t2138

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

An open cylindrical tank with a radius of 0.5 m. and a height of 2 m. is full of oil. It is gradually tilted until half of its bottom area is exposed. Find the volume of oil left inside the tank in cu.m.

  1. 0.33 m3
  2. 0.85 m3
  3. 0.12 m3
  4. 0.58 m3
  5. 0.65 m3
When half of the bottom is exposed, the free surface cuts the circular bottom along a diameter. With radius $r=0.5$ m and height $h=2$ m, the remaining oil forms a cylindrical wedge.
Using $z=\frac{h}{r}x$ over the wet semicircle,
$V=\int_0^r \frac{h}{r}x\,2\sqrt{r^2-x^2}\,dx$
$V=\frac{h}{r}\left(\frac{2r^3}{3}\right)=\frac{2hr^2}{3}$
$V=\frac{2(2)(0.5)^2}{3}$
$\boxed{V\approx 0.33\text{ m}^3}$

Question Bank: w27

MSTE - Geometry and Trigonometry / Plane and Solid Geometry / MSTE May 2019

A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are the field’s dimensions?

  1. 40 yd by 80 yd
  2. 45 yd by 90 yd
  3. 50 yd by 100 yd
  4. 60 yd by 120 yd
$P = 2(x + 2x) = 6x = 300 \Rightarrow x = 50\text{ yd}$.
Length $= 2x = 100\text{ yd}$.
$\boxed{50\text{ yd} \times 100\text{ yd}}$