Sample Problems inSolid Mensuration

Prepared by:Alcantara, Timothy John Noel

Baladad, RonnelBriones, Jefferson

Mecija, Harold KingNuñez, Jims

The section of a certain solid is bounded by two concentric circles whose radii are 6.1 ft and 4.1 ft. find the area of this section?

Let:At = area of the sectionAsc = area of the small circleAbc = area of the big circle

Asc = πr2

Asc = π(4.1)2

Asc = 52.7834 ft2

Abc = πr2

Abc = π(6.1)2

Abc = 116.839 ft2

At = Abc – Asc

At = 116.839 ft2 - 52.7834 ft2

At = 64.056 ft2

c2 = a2 + b2

(120)2 = 2a2

14,400/2 = 2a2/2√7200 = √ a2

a = 84.85 ft.

The official distance between home plates and second base in a baseball diamond (in a form of a square) is 120ft.find the distance between bases.

A storage bin of circular base has 324ft of floor space. Find the radius of the floor.

A = πr2

324 = (3.1416)r2

324/3.1416 = 3.1416r2/3.1416

√103.13 = √r2

r = 10.16 ft

The official distance between home plates and second base in a baseball diamond (in a form of a square) is 120ft.find the distance between bases.

c2 = a2 + b2

(120)2 = 2a2

14,400/2 = (2a2)/2

√7200 = √a2

a = 84.85 ft.

Find the area of the largest circle which can be cut from a square of edge 4in. what is the material wasted?

ASQUARE = s2

ASQUARE = (4)2

ASQUARE = 16 in2

ACIRCLE = r2

ACIRCLE = (3.1416) (2)2

ACIRCLE = 12.57 in2

Material wasted = ASQUARE - ACIRCLE

Material wasted = 16 in2 - 12.57 in2

Material wasted = 3.43 in2

A certain city block is in the form of a parallelogram. Two of its sides are each 421ft long the other two sides are each 227ft. in length. If the distance between the first pair of sides is 126ft, find the area of the land in the block and the length of the diagonals.

A = bh

A = (421ft)(126ft)

At = 53 046 ft2

A cube of edge A is cut by a plane and diagonally opposite edges of the cube. Find the length of the diagonal of the section formed.

x = A2 + A2

x = √2A2

x = A√2

D = √(A√2)2 + A2

D = √2A2 +A2

D = √3A2

D = A√3

What is the weight of a block of ice 24 in. by 24 in. by 24 in. if ice weights 92% as much as water, and water weights 62.5 lb per ft?

V=13 824 in3[1ft./12ft.]3

V=8ft 3

A packing box 2.2 ft by 4.9 ft by 5.5 ft is to be completely covered with tin. How many square feet of metal are needed?

T = 2(ab+bc+ca)

T = 2(2.2)(4.9) + (4.9)(5.5) + (5.5)(2.2)

T = 2(10.78 + 26.95 + 12.1)

T = 2(49.83)

T = 99.66 ft.

Building bricks are closely starked is a pile 7ft high, 36ft long, and 12ft wide. If the bricks are 2in by 4in by 9 in how many bricks are in the pile?

Let:

V1 = bldg. bricks

V2 = bricks

N = no of bricksV1 = abc

V1 = (144) (84) (432)

V1 = 5,225,472 in3

V2 = xyz

V2 = (9)(4)(2)

V2 =72 in3

N = V1/V2

N = 5,225,472 in3 / 72 in3

N = 72, 576 bricks

Find the total area of the cube from the previous problem.

T = 6A2

T = 6(17.96in)2

T = 1935.37in2

The section of a certain solid consists of a semi-circle, a rectangle, and a triangle. The altitude of a rectangle is three times the radius of the semi-circle. The altitude of the triangle is twice the same radius. The area of the triangle is 200ft2. Find the area of the section.

ATRIANGLE = ½ r2

20 ft2 = ½ ab

ARECTANGLE = (3r) (2r) = 6r2

20 ft2/2 = 2r2

ATOTAL = ½ ba

ATOTAL = ½ (2r) ( 2r)

ATOTAL =10 ft2

ATOTAL = r2

ATOTAL = ½ r2 + 6r2 + 2r2

ATOTAL = ½ r2 + 8r2

ATOTAL = ½ (3.14) (10 ft2) + 8 (10 ft2)

ATOTAL = ½ (31.4 ft2) + 80 ft2

ATOTAL = 15.7 ft2 + 80 ft2

ATOTAL = 95.71 ft2

A masonry dam of 40 ft. high has a uniform vertical cross section as shown in the figure. The dam is 80 ft. long and its material weighs 125 lbs. per as ft. Find the weights of the dam.

B= ½(b1+b2) h

B= ½(4+10)40

B= ½(20)(40)

B= ½(800)

B= 400 ft.2

V= Bh

V= (400 ft.2)(80ft.)

V= 32,000 ft.3

W/ft.3= V

125lb. /ft.3 (32,000 ft.3)

W= 4,000,000 lbs.

Find the volume of the largest cube of wood that can be cut from a log of circular cross section whose radius is 12.7 in?

c2=a2+b2

25.42 = a2 + a2

645.15 / 2 = 2a2 / 2

a = 322.58 in3

V = a3

V = 5,793 in3 [ 1ft / 12 in.]

V = 3.35 ft3

The diameter of a well is 6 ft. and the water is 7 ft. deep. How many gallons of water are there in the well reckoning 7.48 gallons to the cubic foot?

B=πr2

B= πd2/4

B= (3.1416)(62)/4

B= (3.1416) (36)/4

B= 28.27 ft.2 V= BH

V= (28.27ft.2) (7ft.)

V= 197.89 ft.3 (7.48 gal/ 1 ft.3)

V= 1,480.22 gal.

A paint manufacturer desires a cylindrical steel drum to hold 50 gal. of food paint. Food convenience handling, it is found necessary to unit the inside diameter to 2.5 ft. Find the height of the drum desired. (One gal. =23.1 cu. in.)

B = πd2/4

B = [(3.1416)(2.5)2]/4

B = 4.91 ft.2

V = 50 gal.(1ft.3 /7.48 gal.)

V/B = BH/B

H = V/B

V = 6.68 ft.3

H = V/B

H = 6.68 ft.3/4.91 ft.3

H = 1.36 ft.

Find the waste in making the largest possible cylindrical rod from a bar of iron 3 ft. long which has a square cross section whose diagonal is 6 in.

B = πr2

B = (3.1416) (2.12)2

B = 14.12 in.

V1= BH

V1= (14.12 in.)(36 in.)

V1= 508.32 in.3

V2= BH

V2= (17.98 in.2)(36 in.)

V2= 647.28 in.3

W = V2 – V1

W = 647.28 in.3 – 508.32 in.3

W = 138.96 in.3

A cubic foot of water weighs about 64.2 lb. What must be the diameter of cylinder pail 1 ft. high in order for it to hold the water from 25 lb. of ice? How many square inches of sheet tin are required to make the pail?

V = h(πd2/4)D2 = 4V/h=πD2 = 4(.4355)/ π ft.2

D = .7446 ft. ( 12 in. / 1 ft.)D = 8.9352 in.

L = hc

C = πd = 28.0708

L = 336.8496 in.2

A = πd2/4

A = 62.7045 in.

Wice= 25 lb.

Dice= 57.408 lb. /ft.3

Vice = 25 lb./ 57.408 lb. /ft.3

Vice= 0.4355 ft.3

Lt = L + A

Lt = 336.8496 in.2 + 62. 7045 in.2

Lt = 399.5541 in.2

In a pyramid AB=9in, BC=12in, and BD=5in the three faces angles at B are each 90. Calculate the three faces angles at A and the total surface of the solid.

tan a = 5in/9in

tan a = .55

tan a1 = 29○4’

tan a2 = 12in/9in

a2 = tan-1(12/9)

a2 = 53○3’

s = 19.145

A = √s(s-a)(s-b)(s-c)

A = √19.145(6.145)(4.145)(8.855)

A = 65.1720

A = ½ bh

131.424 = ½(15)h

h = 8.7616

sin a = 8.7616”/10.29”

sin a = .85

a3 = 58○22”

A1 = 1/2 BH

A1 = 1/2(9”)(5”)

A1 = 22.5 ft2

A2=1/2 BH

A2=1/2 (12”)(9”)

A2= 22.5 ft2

A3=1/2 BH

A3=1/2(12”)(5”)

A3= 30 ft2

S=AT

S=22.5ft2+22.5ft2+30ft2

S=172.2142ft2

In the corner of a cellar is a pyramid heap of coal. The base of the heap is an isosceles triangle whose hypotenuse is 20ft and the altitude of the heap is 7 ft. If there are 35 ft3 in ton of coal, how many tons are there in this heap?

Cos 45 = 10/HH = Cos 45/10H = 14.1421

A = ½ bh

A = (14.1424)(14.1424)

A = 99.9994ft2

V=1/3 BH

V=1/3(99.9994ft2)(7 ft)

V=233.332ft3 {1 ton/35 ft3}

V=62/3 tons

The inside dimension of the box are 2 ft by 3 ft. Find the dimension of packing box of the same shape which will hold 8 times as much.

V2= 8V1

V2=8(24ft3)

V2=194 ft3

V1=L1W1H1

V1= (4ft)(3ft)(2ft)

V1=24ft3

V1/V2=H13/H2

3

24ft3/194ft3=23/ H23

H23=64.667ft3

H2=4.014 ft

V1/V2=W13/W2

3

24ft3/194ft3=33/ W23

W23=218.25ft3

W2=6.021 ft

V1/V2=L13/L2

3

24ft3/194ft3=43/ L23

L23=517.333ft3

L2=8.028 ft

The volume of a certain solid is 1200 in3 and its surface area is 800 in2. What will be the surface of similar solid whose volume is 2400 in3.?

V1/V2=S1/S2

800in2/S2= [3√(112.9243in3)]/ [3√(179.2562in3)]

800in2/S2=112.9243/179.2562

S2=800in2(179.562)/112.9243

S2=1269.9212 in2

The lateral edge of a pyramid is 61 ft. Each side of its octagon base is 22 ft. What will be the cost of painting the pyramid at 2.5 cents a square foot?

H2=(61ft)2-(11ft)2

H2=3721ft2-1212

H=60ft

AΔ = (1/2)(bh)

AΔ = (1/2)(22)(60)

AΔ = 660 ft2

AT= 8AΔ

AT=5280ft2 C= AT ($0.025/ft2)

C=(5280)($0.025)

C=$132.00

Find the volume of the largest pyramid without can be cut from a rectangular parallelepiped whose edges are 2 in by 3 in by 4 in?

V=1/3 BH

V=1/3(LW)(H)

V=1/3(4ft)(2ft)(3ft)

V= 8 in3

Find the volume of the largest cone having its circular base circumscribed about a face of rectangle parallelepiped dimension 2’x3’x4’ and its vertex lying in the opposite face.

D2=(2ft)2+(4ft)2

D2=4ft2+16ft2

D2=(20ft2)

D=2.236 ft

V=(1/3)BH

V=(1/3)πr2H

V=π(1/3)(2.2361ft)2(3in)

V=π(1/3)(5.0001ft2)(3in)

V=π(1/3)(15.0003ft3)

V=47.1248/3 ft3

V=15.7083ft3

If a lead piece ¼ in thick has an inner diameter of 1 ½ in., find the number of cubic inches of lead in a piece of pipe 10 ft. long ?

10 ft. = 120 in.

V1=π(d2/4)

V1=π(1.5 in.)2/[4(120 in.)]

V1= 212.0575 in.3

V2=πd2/4

V2=π(2 in.)2/4(120 in.)

V2= 376.992 in.3

VT = V2 – V1

VT = 376.992 in.3 – 212.0575 in.3

VT = 164.9345 in.3

The diameter of the earth is 7920 miles, and that of the moon is 2160 miles. Compare their volumes.

Earth

Moon

v/V = d3/D3

v/V = (2160)3/(7920)3

v/V = 1.0078/4.4697

Which is heavier built: a man 5 1/ 2 ft tall who weights 160 lbs or one 6 ft tall who weights 200 lbs?

v/V = h3/H3

160n lbs/200lbs = (5.5ft)3/(6ft)3

160n lbs/200lbs = 166.375ft/216ft

The more heavily built is a man whose height is 5 1/ 2 ft and weighs 160 lbs.

The pyramid tank shown in the figure has a square base 11.6 ft on a side and an altitude of 14.8 ft. How many gallons of water are in it when the depth of the water is 5.725 ft?

L1/L2=X/x

14.8ft/11.6 = 5.725/x

X= 4.4827

V = 1/3 AH

V = 1/3 (4.4827)2(5.725)

V = 1/3 (115.2729)

V = 38.4243 ft 3(7.48 gal/ 1 ft3)

V = 287.4138 gal