Semi circle and triangle problem

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MathsFormula

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MathsFormula said:
http://study.com/cimages/multimages/16/illustrationsemicircle.png


http://upload.wikimedia.org/wikiped.../195px-IsoscelesTriangleProofTextbook.svg.png



Semi circle and triangle problem


A semicircle and a isosceles triangle have the same base BC and the same area. Find the angle X in the triangle. The base of the triangle is BC and the angle X is at the corner B and C. Shown above are links to pictures similar to the text book.
Click to expand...

Hint: Area of the semicircle = π/8 * B2 and the area of the triangle = 1/2 * B * [1/2 * B * tan(x)]

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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I'm having trouble editing the thread above so I'll post my attempt at a solution here:

Lets call the length of the semi circle base and the base of the triangle Y.

So Area of the semi circle is (1/2)πY2

Call the sides of the triangle T so area of triangle is (1/2)TTSinX

Then
(1/2)πY2 = (1/2)TTSinX


Cant do anymore. Please help. book answer is 57.5 degrees






 
Last edited:
MathsFormula said:
I'm having trouble editing the thread above so I'll post my attempt at a solution here:

Lets call the length of the semi circle base and the base of the triangle Y.

So Area of the semi circle is (1/2)πY2

Incorrect

Area SC = π/8 * Y2

Call the sides of the triangle T so area of triangle is (1/2)TTSinX

use Base = Y and height = Y/2 * tan(X) →

Area of triangle = 1/4 * tan(X) * Y2

Then
(1/2)πY2 = (1/2)TTSinX


Cant do anymore. Please help. book answer is 57.5 degrees
Click to expand...
.
 
Hello,
Thanks for your ideas.

I've not come across your formulae in my studies yet.


I only know:

Area of circle = πr2 so for a semi circle this would mean


πr2/2

Also for area of a triangle with angles I've only come across the formula A = (1/2)ab sinC where C is the angle and 'a' and 'b' are the sides leading from that angle.








 
I get that Area=(pi r2)2 and if r = 1 then A =pi/2 But how can you just assume r = 1



Also how can area of a triangle be hr


Totally confused
 
MathsFormula said:
I get that Area=(pi r2)2 and if r = 1 then A =pi/2 But how can you just assume r = 1



Also how can area of a triangle be hr


Totally confused
Click to expand...
First, about assuming r=1: Changing the radius would not change the angles [which is what you are after] but would just change the size.

About the area of the triangle, have you seen/do you know the formula
Area of triangle is (1/2) Base * Height
In your formula
A = (1/2) a b sin(C)
the a sin(C) would, for example, be the Height and the b would be the Base.

What is the Base in terms of the radius? Knowing that you should be able to get the height, as Denis pointed out, by just equating the formulas for the area of the triangle and semi-circle which will lead to the equation Denis gives if the radius were 1.

From that point, again as Denis pointed out, just use the law of sines to calculate the needed angles.
 
FINALLY GOT IT. Non stop attempts with help from all replies. Thank you. Here's my solution:


Let the semi-circle base be called ‘y’
Let the triangle height be called ‘h’
Let the base angle of the isosceles triangle be called ‘X’

Consider the semi-circle.
Area of a circle is PIr2
So area of the semi-circle is PI(y/2)2 …….. (i)

Consider the triangle.
Area of a triangle is (1/2)yh ……….. (ii)

(i) = (ii)
So PI(y/2)2 = (1/2)yh
h = (y*PI)/(2) ………. (iii)

TanX = h/y
h = y*TanX ………… (iv)

(iii) = (iv)

(y*PI)/(2) = y*TanX

PI/2 = TanX

X = 57.5 degrees
 
Denis said:
Do you see now that any value could be assigned
to the radius? The y's cancel out...
Click to expand...

That's something I wouldn't know because I'm not familiar with maths the way you are. I can't see that far ahead in the solution pathway to know what gets canceled. I have to go step by step
 
Thanks Denis for the suggestions but your thinking is beyond my level of capability. I'm High School maths level. Tan = opposite/adjacent is as far as I can go.

I'm pleased with myself that I figured out the solution (with nudging from the forum posters with their clues). Thanks all
 
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