Please, help to find the area of the gray triangle
- Thread starter Darya
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Hello, and welcome to FMH! 
I would begin by expressing the area \(A\) of the shaded triangle as:
[MATH]A=\frac{1}{2}x(1)=\frac{x}{2}[/MATH]
Next, let \(y\) be the length of what was the top edge of the paper that now extends below the rectangle...and by Pythagoras we may state:
[MATH]y^2=\left(2-\frac{x}{2}\right)^2-1[/MATH]
[MATH]y=\frac{\sqrt{(x-2)(x-6)}}{2}[/MATH]
And so now, using a right triangle making up half the shaded area, we may write:
[MATH]\frac{x^2}{4}+1=(2-y)^2=4-4y+y^2=4-2\sqrt{(x-2)(x-6)}+\left(2-\frac{x}{2}\right)^2-1[/MATH]
[MATH]3-x=\sqrt{(x-2)(x-6)}[/MATH]
Can you proceed?
I would begin by expressing the area \(A\) of the shaded triangle as:
[MATH]A=\frac{1}{2}x(1)=\frac{x}{2}[/MATH]
Next, let \(y\) be the length of what was the top edge of the paper that now extends below the rectangle...and by Pythagoras we may state:
[MATH]y^2=\left(2-\frac{x}{2}\right)^2-1[/MATH]
[MATH]y=\frac{\sqrt{(x-2)(x-6)}}{2}[/MATH]
And so now, using a right triangle making up half the shaded area, we may write:
[MATH]\frac{x^2}{4}+1=(2-y)^2=4-4y+y^2=4-2\sqrt{(x-2)(x-6)}+\left(2-\frac{x}{2}\right)^2-1[/MATH]
[MATH]3-x=\sqrt{(x-2)(x-6)}[/MATH]
Can you proceed?
Welcome to Free Math Help!
There are multiple ways to solve this problem. What kinds of techniques are being discussed in the place where this question was given to you?
What you've marked as [MATH]a[/MATH] in the image is a known angle whose tangent is [MATH]\frac{1}{2}[/MATH]. The angle [MATH]\theta[/MATH], then, becomes [MATH]90^{\circ} - 2tan^{-1}\left(\frac{1}{2}\right)[/MATH].
The angle [MATH]\theta[/MATH] has a tangent of [MATH]\frac{x}{1}[/MATH]. It also conveniently happens to be the case that [MATH]x[/MATH] is the area of the shaded region.
This is only one way to approach the problem, and it gets kinda jumbled when you write it out, so it may not be the way it was intended to be solved.
There are multiple ways to solve this problem. What kinds of techniques are being discussed in the place where this question was given to you?
What you've marked as [MATH]a[/MATH] in the image is a known angle whose tangent is [MATH]\frac{1}{2}[/MATH]. The angle [MATH]\theta[/MATH], then, becomes [MATH]90^{\circ} - 2tan^{-1}\left(\frac{1}{2}\right)[/MATH].
The angle [MATH]\theta[/MATH] has a tangent of [MATH]\frac{x}{1}[/MATH]. It also conveniently happens to be the case that [MATH]x[/MATH] is the area of the shaded region.
This is only one way to approach the problem, and it gets kinda jumbled when you write it out, so it may not be the way it was intended to be solved.
Hi! Of course,
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Hello, and welcome to FMH!
I would begin by expressing the area \(A\) of the shaded triangle as:
[MATH]A=\frac{1}{2}x(1)=\frac{x}{2}[/MATH]
Next, let \(y\) be the length of what was the top edge of the paper that now extends below the rectangle...and by Pythagoras we may state:
[MATH]y^2=\left(2-\frac{x}{2}\right)^2-1[/MATH]
[MATH]y=\frac{\sqrt{(x-2)(x-6)}}{2}[/MATH]
And so now, using a right triangle making up half the shaded area, we may write:
[MATH]\frac{x^2}{4}+1=(2-y)^2=4-4y+y^2=4-2\sqrt{(x-2)(x-6)}+\left(2-\frac{x}{2}\right)^2-1[/MATH]
[MATH]3-x=\sqrt{(x-2)(x-6)}[/MATH]
Can you proceed?
Hello and thank you so much for such an elaborate explanation!! You've saved me!
I also went this way at first but then realized its not the best method too. Thank you for your reply!
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