# Thread: Finding the area of a window

1. ## Finding the area of a window

Hello I have an assignment to do, and i need help in the last question of this assignment

I am allowed to get help with this assignment but I need to prove that i can do the questions because after submitting the assignment and marking, the teacher would ask random students to come up to the board and do the questions
Here I will attach the question

I just cant get to connect the perimeter to the area
Also fyi we never studied intergration and differentiation(Or calculus), some suggested me to use Calculus but since it wasnt taught to us yet, i shouldnt use it
I am guessing I have to use an Algebric Equation here?

2. ## Re: Finding the area of a window

Have you found the length of the slant side in terms of a and b?

3. ## Re: Finding the area of a window

Originally Posted by Debsta
Have you found the length of the slant side in terms of a and b?
I am honestly not sure how to do this one at all.the perimeter written there had me confused for a while now. I tried using the formula of area, using 1/2 bh and area of rectangle lb equal to 750, but one of my friends said that wont work since the value of "750" is the "Perimeter"

4. ## Re: Finding the area of a window

The triangle is isosceles (of course).
Assuming b is rectangle's width, then triangle's height = 2b.
Let c = triangle's equal sides.
Then c = sqrt(4b^2 + a^2/4) [1]

So we have a + 2b + 2c = 750
Substitute [1] to get: a + 2b + 2(4b^2 + a^2/4) = 750
Leads to the quadratic 12b^2 + 3000b - 4ab + 1500a - 750^2 = 0
Solving for b:
b = (a - 750 + sqrt(a^2 - 6000a + 2250000)) / 6

I used brute strength to get maximum area 33786.8 where:
a = 231, b = 73.13, c = 186.37

Is the solution given? If so, are my calculation ok?
And perhaps Debsta can confirm...or tell me I'm wrong!!

5. ## Re: Finding the area of a window

Originally Posted by DenisB
The triangle is isosceles (of course).
Assuming b is rectangle's width, then triangle's height = 2b.
Let c = triangle's equal sides.
Then c = sqrt(4b^2 + a^2/4) [1]

So we have a + 2b + 2c = 750
Substitute [1] to get: a + 2b + 2(4b^2 + a^2/4) = 750
Leads to the quadratic 12b^2 + 3000b - 4ab + 1500a - 750^2 = 0
Solving for b:
b = (a - 750 + sqrt(a^2 - 6000a + 2250000)) / 6

I used brute strength to get maximum area 33786.8 where:
a = 231, b = 73.13, c = 186.37

Is the solution given? If so, are my calculation ok?
And perhaps Debsta can confirm...or tell me I'm wrong!!
Hello!
Unfortunately she didnt give us the solution but a classmate of mine did the same method too upto the "maximum area" it seems
He isnt sure if its correct or not tho but he did exactly like you did
Wasnt sure what "max area" meant

6. ## Re: Finding the area of a window

I'd say the "width" of the window is a not b.

7. ## Re: Finding the area of a window

Let the height of the triangle be 2a. This leads to the slant being $\displaystyle \frac{\sqrt5}{2}a$

So Perimeter = $\displaystyle 2* \frac{\sqrt5}{2}a + a + 2b = 750$

This gives $\displaystyle b=\frac{750 - (\sqrt5+1)a}{2}$ … **

Now Area of window = area of rectangle + area of triangle = $\displaystyle a*b + \frac{1}{2}a*2a = ab + a^2$

Now sub in ** for b and you'll have the Area in terms of a only. It will be a quadratic expression in a.

Graph (as it suggests in the question), it will be an "upside-down" parabola and find the maximum area ie the highest point on the graph.

(I got max area of approx. 56884 when a=303.38. Check that by doing it yourself - I did it very quickly.)

You also have to find b, so sub a=303.38 into **.

8. ## Re: Finding the area of a window

Originally Posted by Debsta
Let the height of the triangle be 2a. This leads to the slant being $\displaystyle \frac{\sqrt5}{2}a$
Disagree. Should be slant = sqrt[(2a)^2 + (a/2)^2] = aSQRT(17) / 2

Also, area of 56,884 is evidently impossible:
if instead the 750 was the perimeter of a square,
area = (750/4)^2 = 35556

9. ## Re: Finding the area of a window

Originally Posted by Debsta
I'd say the "width" of the window is a not b.
Why? If his classmate used b, then the class must have been told b was the width.

10. ## Re: Finding the area of a window

Originally Posted by DenisB
Why? If his classmate used b, then the class must have been told b was the width.
No I think his classmate just made that assumption. The problem is so much more "do-able" if the width is a. We can't assume the diagram is drawn to scale.

11. ## Re: Finding the area of a window

Originally Posted by DenisB
Why? If his classmate used b, then the class must have been told b was the width.
No I think his classmate just made that assumption. The problem is so much more "do-able" if the width is a. We can't assume the diagram is drawn to scale.

12. ## Re: Finding the area of a window

Originally Posted by DenisB
Disagree. Should be slant = sqrt[(2a)^2 + (a/2)^2] = aSQRT(17) / 2

Also, area of 56,884 is evidently impossible:
if instead the 750 was the perimeter of a square,
area = (750/4)^2 = 35556
Yes you are correct. I made a dumb mistake when simplifying Pythagoras' rule.
I'll leave it to eobardrush to fix it up from there.

13. ## Re: Finding the area of a window

Originally Posted by Debsta
Let the height of the triangle be 2a. This leads to the slant being $\displaystyle \frac{\sqrt17}{2}a$

So Perimeter = $\displaystyle 2* \frac{\sqrt17}{2}a + a + 2b = 750$

This gives $\displaystyle b=\frac{750 - (\sqrt17+1)a}{2}$ … **

Now Area of window = area of rectangle + area of triangle = $\displaystyle a*b + \frac{1}{2}a*2a = ab + a^2$

Now sub in ** for b and you'll have the Area in terms of a only. It will be a quadratic expression in a.

Graph (as it suggests in the question), it will be an "upside-down" parabola and find the maximum area ie the highest point on the graph.

(I got max area of approx. 56884 when a=303.38. Check that by doing it yourself - I did it very quickly.) This is obviously not correct now.

You also have to find b, so sub a=303.38 into **.
Thanks Denis for pointing out my mistake. As they say "Two heads are better than one!"

14. ## Re: Finding the area of a window

Stinking(!) out loud: Debsta, do you see anything wrong with making the rectangle a square?
That's since a square produces max. area...haven't tried it...too lazy!

15. ## Re: Finding the area of a window

Originally Posted by DenisB
Stinking(!) out loud: Debsta, do you see anything wrong with making the rectangle a square?
That's since a square produces max. area...haven't tried it...too lazy!
No that wouldn't necessarily work, because the window is not just the rectangle.

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