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 Calculus I, 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)
Expert: Azeem Hussain Date: 6/23/2008 Subject: Length and Width of a Rectangle
Question QUESTION: The length of a rectangle is 2 inches less than twice the width. The diagonal is 5 inches.
Find the length and width of the rectangle.
Let the width equal (x) and let the length equal (x-2). These two sides are perpendicular to one another and they make a right triangle with the 5-inch diagonal. By using the Pythagorean Theorem and a quadratic equation, it is possible to solve for (x).
c^2=a^2+b^2
(5)^2=(x)^2+(x-2)^2
25=x^2+x^2-4x+4
0=2x^2-4x-21
x=-2.39116... x=4.39116...
In this situation, it is not possible for (x) to be negative, so its value must be 4.39. The width must equal 4.39, so the length, (x-2), will be 2.39.
Thanks for asking,
Azeem
---------- FOLLOW-UP ----------
QUESTION: I believe you made a mistake
I agree with the formula c^2=a^2+b^2
but you made a mistake in b
b does not equal (x-2)^2
b = [(x+x)-2]^2
so the answer to the problem would be diffrent
please review and advise
thanks
Peter
Answer Hey Peter,
"2 inches less than TWICE the width." Here's the proper math:
c^2=a^2+b^2
(5)^2=(x)^2+(2x-2)^2
25=x^2+4x^2-8x+4
0=5x^2-8x-21
0=(5x+7)(x-3)
x=-7/5 x=3
x=-1.4 x=3
The width is 3, and the length is 4 [2(3)-2=4]. You know have two 3, 4, 5 Pythagorean triangles.
