Triangle Questions: Get Expert Advice Now

[wailingkoh]

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Hi everyone
I need help in these 2 questions. Please advice. Tis :)

 

[Greg]

If the length of the base and the height of two triangles are equal then their area is equal.

 

[wailingkoh]

Hi there,
Thanks, how about the other question?

 

[MarkFL]

Hi there,
Thanks, how about the other question?

See if the following diagram inspires you...

View attachment 4621

 

[wailingkoh]

See if the following diagram inspires you...

Find the big tri and subtract?
But I won't be able to find the big tri

 

[MarkFL]

Find the big tri and subtract?
But I won't be able to find the big tri

Yes, but you do know the base and altitude of the big triangle, based on the diameter of the semi-circle. :D

 

[wailingkoh]

Find the big tri and subtract?
But I won't be able to find the big tri

Hmm, how to find the green part?

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Hmm, how to find the green part?

Arr...1/2 *30*18 for the big Triangle. But the green part, what do I do??

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Hmm, how to find the green part?

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Arr...1/2 *30*18 for the big Triangle. But the green part, what do I do??[/QUOTE

The small green tri has an arc to it formed by a a semi circle

 

[MarkFL]

Hmm, how to find the green part?

View attachment 4622

Take the area of the square, and subtract away the red area of the quarter circle...:D

 

[wailingkoh]

Take the area of the square, and subtract away the red area of the quarter circle...:D

Awesome! Got it and thanks for the help

 

[MarkFL]

Awesome! Got it and thanks for the help

I would let $r$ be the radius of the semi-circle, and so the area $A_T$ of the big triangle in my first diagram would be:

\(\displaystyle A_T=\frac{1}{2}(3r)(r)=\frac{3}{2}r^2\)

Now, the green area $A_S$ to be subtracted away is:

\(\displaystyle A_S=r^2-\frac{1}{4}\pi r^2=r^2\left(1-\frac{\pi}{4}\right)\)

And so the area $A$ we are asked to find is:

\(\displaystyle A=A_T-A_S=\frac{3}{2}r^2-r^2\left(1-\frac{\pi}{4}\right)=r^2\left(\frac{1}{2}+\frac{\pi}{4}\right)=\left(\frac{r}{2}\right)^2(2+\pi)\)

Now, we are given $r=9\text{ cm}$, hence:

\(\displaystyle A=\left(\frac{9\text{ cm}}{2}\right)^2(2+\pi)=\frac{81}{4}(2+\pi)\text{ cm}^2\)