AboutAzeem Hussain Expertise I can answer virtually any kind of question dealing with of Math 536 and below, my forte being in parabolic functions and analytical geometry.
I'm currently learning Linear Algebra, and cannot answer questions that deal with subject matter more advanced than that.
Experience I am neither a professor nor a teacher of this subject. I am merely a student who is gifted at mathematics and enjoys being of service to his community. I frequently tutor people in math and the results are usually great.
Publications Reflections, Riverside School Board (2005, 2006)
Education/Credentials Diploma of Secondary Studies from Chambly Academy High School, and IBO-MYP certificate as well. My lowest mark on a high school math final was 97%, peaking at 99% in 2006 and 2007 (second-highest Math 436 mark in the province). Being a Quebecer, I am fluent in English and French and can respond to questions easily in both languages.
Awards and Honors Pascal Math Competition, School Champion(2007)
Question I have a Triangular Pyramid - all edges are 3cm. How do I figure the Surface area and volume?
Answer Hi Tracy!
Every triangle is an equilateral, meaning they have angles measuring 60 degrees each. Subdivide an equilateral down the middle into two right triangles, where the common leg is the height of the equilateral triangle. By the Pythagorean theorem or basic trigonometry, you can find the height of the triangle: 3sin60 or (3√3)/2. The surface area is made up of the area of four equal triangles, so...
SA=4bh/2
SA=2bh
SA=2(3)((3√3)/2))
SA=9√3
The volume is more difficult to find because you need to find the height of the pyramid, H. It's difficult to describe how to solve for this, but I'll do my best. Visualize your pyramid and an altitude down the middle, which connects the summit to the centre of the base. Connect this spot at the centre of the base to a bottom corner of the pyramid. You now have a right angle. By using an edge of the pyramid, you can make a right triangle. This triangle's height is that of the pyramid, its hypotenuse is a pyramid edge, 3 cm, and its other leg is the distance from a corner of the base to a centre of the base. You have to solve for this.
Go back to your sketch where you found the height of each equilateral triangle. Draw a line connecting a corner to the centre of the triangle. This intersects the base halfway through. You should now have a small right triangle where the hypotenuse is a side that will help you find the base, x; half of h, (3√3)/4; and the other leg is half the edge, 3/2. Use basic trig or the Pythagorean Theorem to solve for x.
cos30=x/(3/2)
x=(3/2)/cos30
x=(3/2)/(√3/2)
x=[(3)(2)]/(2√3)
x=3/√3
x=√3
(The last step is a rationalization. Don't worry if you don't get it.)
Now to find the height, use the Pythagorean Theorem on the whole pyramid.
H=√[(3)²-(√3)²]
H=√[9-3]
H=√6
Now the volume of a pyramid is equal to one-third of its base area times its height.
V=AH/3
V=[(9√3)/4](√6)/3
V=(9√18)/12
V=9(√9√2)/12
V=9(3√2)/12
V=(27√2)/12
V=(9√2)/4
ANSWER: The pyramid's surface area is (9√3) cm² and its volume is [(9√2)/4] cm³.
This may seem more complicated than it actually is because I used radicals to retain exact values. The alternative would be using decimal notation and using a certain number of decimal places.
