The base of triangular prism is a right triangle whose hypotenuse is 65cm and one of whose legs is 16cm. Lateral edge of the prism is 10cm and makes with the base of an angle 30 degrees. Find the volume of the prism 2.the altitude of a prism is 12cm its base is a parallelogram whose sides are 8 and 5 cm and of whose angles is 60 degrees find the volume of the prism 3.the lateral area of a right prism is 162 sq.m. and the base is a rectangle 3 by 6 m find the altitude of the prism. 4.the base of the right prism is a regular hexagon 4cm on a side. If the altitude is 6cm find the total surface area of the prism. 5.the top of a trough with triangular cross section is a rectangle 40 cm by 100cm and the depth of the trough is 30cm. What is the depth of the water in a trough when it contains 24 liters of water. 6.show that the total surface of a regular octahedron 2√3e² where's represents an edge of the octahedron.
+130513 Here's the 5th one
The volume of the trough is given by (1/2)(40 * 100) (30) = 60,000 cm^3
And there are 1000cm^3 in one liter
So....when the trough contains 24L of water we have a volume of 24,000 cm^3
Now.....abritrarily assuming that the 40cm side is the width of the trough at the top and the 100cm side is the length, we can use similar triangles to show that, at any water height, the height of the water = 3/4 of the base of a cross-sectional triangle at that height
So we have
24000 = (1/2)b(3/4b)(100)
24000 = b^2(37.5) divide both sides by 37.5
24000/37.5 = b^2 take the square root of each side
sqrt(640) = b
So the height of the water when the trough contains 24000L of water = (3/4)b = (3/4)sqrt(640) cm = about 18.97 cm
BTW.....the answer would be the same if we had assumed that the 100cm side was the width of the trough and the 40 cm side was the length.......!!!!
+130513 The first one is a little confusing....you say it's a prism, yet, the lateral edge makes an angle of 30° with the base.......but.....in a prism, the lateral edge is perpendicular to the base.....thus you may have a pyramid, instead.......
If it's a prism....we can find the remaining side (leg) thusly
sqrt(65^2 - 16^2 ) = 63
And we can use Heron's formula to help us find the area of the base.... where s = (65 + 63 + 16) / 2 = 72 and a, b and c are sides of the triangle
And the base area = sqrt[s (s-a)(s-b)(s-c)] ...so we have
sqrt [ 72 * (72 - 65) * (72 - 63) * (72 -16) ] = sqrt[ 72* 7 * 9 * 56] = 504cm^2
And the volume [ if it's a prism] is given by
V = base area * height = (504cm^2) (10cm) = 5040 cm^2
If it's a pyramid....we can use the following formula to find the volume:
V = (1/3) base area * height and the height is given by 10*sin(30) = 10(1/2) = 5cm
So
V = (1/3) (504 cm^2)(5cm) = 840 cm^3
+130513 For the second one....the vloume is given by : the area of the base * height
And the area of the base is given by (5)(8)sin(60) = 40sqrt(3)/2 = 20sqrt(3) cm^2
So the volume is
(20sqrt(3) cm^2) * (12 cm) = 240*sqrt(3) cm^3 = about 415.7 cm^3
+130513 For the third one, we can use the fact that the surface area of a right prism =
2*area of the base + Perimeter of the base * height
The area of the base = (3 cm) (6 cm) = 18 cm^2
The perimeter of the base = 2(3 + 6) = 18 cm
So we have
162 cm^2 = 2(18 cm^2) + (18 cm)*h [without the units, we have]
162 = 36 + 18*h subtract 36 from each side
126 = 18*h divide both sides by 18
height = 7 cm
+130513 Here's the 4th one
If the hexagon is 4cm on a side......the total area of the hexagon is
6(1/2)(4)(4)sin(60) = 48sqrt(3)/2 = 24sqrt(3) cm^2
And this forms the base of the prism and we have two of them....so that area of the bases =
48sqrt(3) cm^2
And the area of the sides = 6 (4cm)(6cm)= 144 cm^2
So...the total surface area = [ 48sqrt(3) + 144] cm^2 = about 227.14 cm "2
+130513 Here's the 5th one
The volume of the trough is given by (1/2)(40 * 100) (30) = 60,000 cm^3
And there are 1000cm^3 in one liter
So....when the trough contains 24L of water we have a volume of 24,000 cm^3
Now.....abritrarily assuming that the 40cm side is the width of the trough at the top and the 100cm side is the length, we can use similar triangles to show that, at any water height, the height of the water = 3/4 of the base of a cross-sectional triangle at that height
So we have
24000 = (1/2)b(3/4b)(100)
24000 = b^2(37.5) divide both sides by 37.5
24000/37.5 = b^2 take the square root of each side
sqrt(640) = b
So the height of the water when the trough contains 24000L of water = (3/4)b = (3/4)sqrt(640) cm = about 18.97 cm
BTW.....the answer would be the same if we had assumed that the 100cm side was the width of the trough and the 40 cm side was the length.......!!!!
+130513 Here's the last one:
Let's post an image of the object we're talking about:
There are 8 congruent triangles each having a surface area of:
(1/2)*e^2*sin(60) =
(1/2)*e^2*sqrt(3)/2 =
(1/4)*(sqrt(3)* e^2 [sq units] ........ and since there are 8 of them, the total surface area =
8(1/4) (sqrt(3)e^2 =
2sqrt(3)*e^2 [square units]
