Help with Quick Maths Questions! (1 Viewer)

starryblue · Oct 10, 2011

would anyone tell me how to do these quick question? i can't seem to figure them out:L

1. the surface area of a pyramid is 360cm^2. What is the length of the base if the height of the pyramid is 12cm?
2. A cone is made from a sector with radius of 10cm and has an angle of 240 degrees. find the radius of the cone.

tys for the help

lath · Oct 10, 2011

1. 10cm (this is assuming that it is a square pyramid...I kind of did trial and error so I cant really explain it to be honest)
2. ok..so if you draw a sector (part of a circle) with the radius as 10 cm and the included angle = 240 degrees..you can work out the size of the arc by working out the circumference of the whole circle and multiplying by 240/360 (since it is only a sector). This equals the circumference of the base of the cone formed by the sector. You can then work out the radius by dividing it by (2*pi). The final answer i got was 6 and two thirds centimeters.

starryblue · Oct 10, 2011

lath said:
1. 10cm (this is assuming that it is a square pyramid...I kind of did trial and error so I cant really explain it to be honest)
2. ok..so if you draw a sector (part of a circle) with the radius as 10 cm and the included angle = 240 degrees..you can work out the size of the arc by working out the circumference of the whole circle and multiplying by 240/360 (since it is only a sector). This equals the circumference of the base of the cone formed by the sector. You can then work out the radius by dividing it by (2*pi). The final answer i got was 6 and two thirds centimeters.

thanks!

lath · Oct 10, 2011

Parvee said:
1. Area of Tirangle = 0.5xHxB
360 divided by 12 = 30
30x2= 60 (x2 because 30 is only 0.5 of the base)
base = 60
i.e. (0.5x60x12=360)

It's a pyramid not a triangle.

D94 · Oct 10, 2011

Assuming a square pyramid, there are 5 sides (4 triangles, 1 square). Area of a triangle = .5 * base * slant height. (I'll denote slant height as (sh)).

There are 4 triangles, so multiply by 4 = 2b(sh). Add the base which is the base^2, to get b^2 + 2b(sh).

To get the slant height, you need the hypotenuse of a vertical cross section. Think of a right angle triangle with the base as 1/2 b and the perpendicular height as h. Essentially, it's $\sqrt{ (.5b)^{2} + (h)^{2}}$ = (sh).

So we get:

$b^{2} + 2b\sqrt{ (.5b)^{2} + (h)^{2}} = 360$

Sub in h = 12:

$b^{2} + 2b\sqrt{ (.5b)^{2} + (12)^{2}} = 360$

$b^{2} + 2b\sqrt{ .25(b)^{2} + 144} = 360$

$b^{2} + 2b\sqrt{ \frac{1}{4}(b)^{2} + 144} = 360$

$b^{2} + 2b\sqrt{ \frac{(b)^{2}+ 4*144}{4}} = 360$

$b^{2} + 2b\frac{\sqrt{(b)^{2}+ 576}}{2} = 360$

$b^{2} + b\sqrt{b^{2} + 576} = 360$

$b\sqrt{b^{2} + 576} = 360 - b^{2}$

$(b\sqrt{b^{2} + 576})^{2} = (360 - b^{2})^{2}$

$b^{2}(b^{2} + 576) = 360^{2} - 720b^{2} + b^{4}$

$b^{4} + 576b^{2} = 360^{2} - 720b^{2} + b^{4}$

$576b^{2} = 360^{2} - 720b^{2}$

$1296b^{2} = 360^{2}$

$b^{2} = 100$

$b=10$

starryblue · Oct 12, 2011

D94 said:
Assuming a square pyramid, there are 5 sides (4 triangles, 1 square). Area of a triangle = .5 * base * slant height. (I'll denote slant height as (sh)).

There are 4 triangles, so multiply by 4 = 2b(sh). Add the base which is the base^2, to get b^2 + 2b(sh).

To get the slant height, you need the hypotenuse of a vertical cross section. Think of a right angle triangle with the base as 1/2 b and the perpendicular height as h. Essentially, it's $\sqrt{ (.5b)^{2} + (h)^{2}}$ = (sh).

So we get:

$b^{2} + 2b\sqrt{ (.5b)^{2} + (h)^{2}} = 360$

Sub in h = 12:

$b^{2} + 2b\sqrt{ (.5b)^{2} + (12)^{2}} = 360$

$b^{2} + 2b\sqrt{ .25(b)^{2} + 144} = 360$

$b^{2} + 2b\sqrt{ \frac{1}{4}(b)^{2} + 144} = 360$

$b^{2} + 2b\sqrt{ \frac{(b)^{2}+ 4*144}{4}} = 360$

$b^{2} + 2b\frac{\sqrt{(b)^{2}+ 576}}{2} = 360$

$b^{2} + b\sqrt{b^{2} + 576} = 360$

$b\sqrt{b^{2} + 576} = 360 - b^{2}$

$(b\sqrt{b^{2} + 576})^{2} = (360 - b^{2})^{2}$

$b^{2}(b^{2} + 576) = 360^{2} - 720b^{2} + b^{4}$

$b^{4} + 576b^{2} = 360^{2} - 720b^{2} + b^{4}$

$576b^{2} = 360^{2} - 720b^{2}$

$1296b^{2} = 360^{2}$

$b^{2} = 100$

$b=10$

Ohh i get it...i was on the right track...=) kinda..

Help with Quick Maths Questions! (1 Viewer)

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