Sample Problems in Solid Mensuration Prepared by: Alcantara, Timothy John Noel Baladad, Ronnel Briones, Jefferson Mecija, Harold King Nuñez, 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 section Asc = area of the small circle Abc = area of the big circle
Asc = r 2 Asc = (4.1)2 Asc = 52.7834 ft2 Abc = r 2 Abc = (6.1)2 Abc = 116.839 116.839 ft f t2
At = Abc ± Asc At = 116.839 ft2 - 52 52.7 .783 8344 ftft2 At = 64.056 ft2
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 section Asc = area of the small circle Abc = area of the big circle
Asc = r 2 Asc = (4.1)2 Asc = 52.7834 ft2 Abc = r 2 Abc = (6.1)2 Abc = 116.839 116.839 ft f t2
At = Abc ± Asc At = 116.839 ft2 - 52 52.7 .783 8344 ftft2 At = 64.056 ft2
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.
A storage bin of circular base has 324ft of floor space. Find the radius of the floor. A = r 2 324 = (3.1416)r 2 324/3.1416 = 3.1416r 2/3.1416 ¥103.13 = ¥r 2 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 = r 2 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 bricks
V1 = abc
V2 = xyz
V1 = (144) (84) (432)
V2 = (9)(4)(2)
V1 = 5,225,472 in3
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 r 2
ARECTANGLE = (3r) (2r) = 6r 2
20 ft2 = ½ ab
20 ft2/2 = 2r 2
ATOTAL = ½ ba
ATOTAL = ½ r 2 + 6r 2 + 2r 2
ATOTAL = ½ (2r) ( 2r)
ATOTAL = ½ r 2 + 8r 2
ATOTAL =10 ft2
ATOTAL = ½ (3.14) (10 ft2) + 8 (10 ft2)
ATOTAL =
r 2
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=r 2 B= d2/4 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 = d2/4 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.
V1= BH V1= (14.12 in.)(36 in.) V1= 508.32 in.3
B = r 2 B = (3.1416) (2.12)2 B = 14.12 in.
V2= BH
W
= V2 ± V1
V2= (17.98 in.2)(36 in.)
W
= 647.28 in.3 ± 508.32 in.3
V2= 647.28 in.3
W
= 138.96 in.3
A cubic foot of water weighs about 64.2 lb. W hat 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(d2/4) D2 = 4V/h= D2 = 4(.4355)/ ft.2 D = .7446 ft. ( 12 in. / 1 ft.) D = 8.9352 in.
Wice=
25 lb.
Dice= 57.408 lb. /ft.3 Vice = 25 lb./ 57.408 lb. /ft.3 Vice= 0.4355 ft.3
L = hc C = d = 28.0708 L = 336.8496 in.2 A = d2/4 A = 62.7045 in. 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. s = 19.145 A = ¥s(s-a)(s-b)(s-c) A = ¥19.145(6.145)(4.145)(8.855) A = 65.1720 tan a = 5in/9in tan a = .55 tan a1 = 29 4¶
A = ½ bh 131.424 = ½(15)h h = 8.7616 sin a = 8.7616´/10.29´
A1 = 1/2 BH
A2=1/2 BH
A1 = 1/2(9´)(5´) A2=1/2 (12´)(9´) A1 = 22.5 ft2
A2= 22.5 ft2
tan a2 = 12in/9in a2 =
tan-1(12/9)
a2 = 53
3¶
A3=1/2 BH
sin a = .85 a3 = 58
22´
S=AT S=22.5ft2+22.5ft2+30ft2 S=172.2142ft2
A3=1/2(12´)(5´) A3= 30 ft2
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?
V=1/3 BH Cos 45 = 10/H H = Cos 45/10 H = 14.1421
V=1/3(99.9994ft2)(7 ft)
A = ½ bh
V=62/3 tons
A = (14.1424)(14.1424) A = 99.9994ft2
V=233.332ft3 {1 ton/35 ft3}
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.
V1=L1W 1H1
V2= 8V1
V1= (4ft)(3ft)(2ft)
V2=8(24ft3)
V1=24ft3
V2=194 ft3
V1/V2=H13/H23
V1/V2=W13/W23
V1/V2=L13/L23
24ft3/194ft3=23/ H23
24ft3/194ft3=33/ W 23
24ft3/194ft3=43/ L23
H23=64.667ft3
W23=218.25ft3
L23=517.333ft3
H2=4.014
W2=6.021
L2=8.028 ft
ft
ft
The volume of a certain solid is 1200 in3 and its surface area is 800 in2. W hat 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. W hat will be the cost of painting the pyramid at 2.5 cents a square foot?
H2=(61ft)2-(11ft)2
A = (1/2)(bh)
H2=3721ft2-1212
A = (1/2)(22)(60)
H=60ft
A = 660 ft2
AT= 8A 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.
V=(1/3)BH D2=(2ft)2+(4ft)2
V=(1/3)r 2H
D2=4ft2+16ft2
V=(1/3)(2.2361ft) 2(3in)
D2= (20ft2)
V=(1/3)(5.0001ft 2)(3in)
D=2.236 ft
V=(1/3)(15.0003ft 3) 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=(d2/4) V1=(1.5 in.)2/[4(120 in.)] V1= 212.0575 in.3 V2=d2/4 V2=(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
v/V = d3 /D3 v/V = (2160)3/(7920)3 Moon
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
mor e heavily built is a man whose height is 5 1 / 2 ft and weighs 160 lbs.
