Geometry help please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
The figure below shows a circular prop with two triangles labeled ABC and AOC made on it. The center of the prop is labeled O. https://www.connexus.com/content/media/532465-1182011-110442-AM-321962674.jpg What is the area of the cloth required tomake the circular prop?
I can't figure it out and I have to get a good grade to pass the class, anyone good at geometry?
do you know the answer? is it 39.25?
Heres the choices; # 17.83 ft2 # 23.60 ft2 # 29.63 ft2 # 39.27 ft2
39.27
ok so the angle at the centre must be double the angle at the top so: angle ABC is 45 so angle AOC must be 90, hence the triangle AOC is right angled and as yu can tell the hypotenuse is 5 Now the radius of the circle is OA and OC which means the lengths OA and OC are the same.... Therefore if yu let OA=x=OC then x^2 + x^2 = 5^2 2x^2 = 25 solve for x, to get the radius then use the area of a circle equation to get the area of the circular cloth
answer is 39.27 ft2
Thanks!!!!!! could you help me with one more?
sure ill try
An aquarium sells cubic fish tanks of different sizes. The length of the small size fish tank is 4 feet. The dimensions of the jumbo size fish tank are double the dimensions of the small size fish tank. Which expression can be used to find the ratio of the volume of the small size fish tank to the volume of the jumbo size fish tank?
post a separate question
https://www.connexus.com/content/media/532465-1182011-113631-AM-2108352531.jpg
volume = x^3 now if yu double the length volume = (2x)^3 = 8x^3 so V of small tank is 64 and V of big tank must be 512 ratio = 1:8
these are the choices https://www.connexus.com/content/media/532465-1182011-113631-AM-2108352531.jpg is it d?
dat link just says 4/8 ^
the answer is 1/8 , well dats the answer that i got
4/8^2 8/4^2 4/8^3 8/4^3
yep so its d, 8/4^3
ok
one more? lol im horrible at this
haha dats fine, sure ill try to help
Alex designed the model of a cylindrical water tank to be installed in a park. The model has a radius of 5 cm and a height of 14 cm. The dimensions of the actual water tanks that will be installed in the park are 8 times larger than the model. The total surface area of the actual water tanks, rounded to the nearest whole number, is _________ cm². Use 22/7 for pi.
so SA of model is 2*pi*r*h + pi*r*r SA of actual model is pi*8r*8r + 2*pi*8r*8h = 64 (2*pi*r*h + pi*r*r) = 64*(SA of model) hence SA of model is 778.5 and SA of actual model is 778.5*64 = 49826.5 cm2
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