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I was recently going through a mensuration sum from a tenth grade board exam book. This one particular question stumped me, and I spent the entire evening thinking of this, but to no avail. The question goes like this:

ABC is a right angled triangle at B. AB, AC and BC are diameters of 3 semicircles. Given, AC = 14 cm, find the total area of the semicircles.

Pardon me for my awful drawing skills in paint. To clarify, the "curved" triangles on top of AC, BC and AB are semicircles (Sorry, couldn't do it in Paint!). Here's what I tried. By Pythagoras theorem,

$x^2 + y^2 = 14^2$.

This is one of the equations for the quadratic. The question is how do I proceed next? Remember that it is a scalene triangle, with all sides unequal, and the hypotenuse given 14 cm. I thought of another equation, but I guess it's incorrect:

$x^2(1-\frac{\pi}{4})$ + $y^2(1-\frac{\pi}{4})$ = $14^2 - 49\pi$

I arrived at the above equation keeping in mind that squares on sides of a right angled triangle add up to the area of the square of the hypotenuse side. I then subtracted the area of the semicircles from their areas of the square respectively. Here's the image:

enter image description here

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  • $\begingroup$ The $x^2$ and the $y^2$ (combined) together will be absorbed in the total area calculateion. They need not be evaluated separatedly. $\endgroup$ – Mick Jul 22 '15 at 13:48
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A semicircle of diameter $d$ has area $\pi d^2/8$. Thus the three semicircles together have area $\pi x^2/8+\pi y^2/8+\pi h^2/8=\pi(x^2+y^2+h^2)/8=\pi(h^2+h^2)/8=\pi h^2/4=7^2\pi\,\text{cm}^2\approx154\,\text{cm}^2$, where $h=14\,\text{cm}$ is the length of the hypotenuse.

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  • $\begingroup$ I guess 49*3.14 is almost equal to 154? $\endgroup$ – Mission Coding Jul 22 '15 at 16:03
  • $\begingroup$ @MissionCoding: Sorry, don't know what I typed into the calculator there :-) Corrected. $\endgroup$ – joriki Jul 22 '15 at 16:05
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Since $x^2 + y^2 = h^2$ (by the Pythagorean theorem), and the area of a semicircle is $\pi d^2/8$, you can write

$A = \dfrac{\pi}{8}(x^2 + y^2 + h^2) = \dfrac{\pi}{8}(2h^2) = \dfrac{\pi h^2}{4} = 49 \pi \approx 153.9\:cm^2$

Or you could argue that the semicircles are similar to one another, so we have directly $A = 2 \times \dfrac{\pi}{8}h^2$

These similarity arguments lead to a truly elegant proof of the Pythagorean Theorem, originally due to George Polya, I believe (see the second proof at the following link):

http://dankalman.net/AUhome/pythag/index.html

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As you said, squares on sides of a right angled triangle add up to the area of the square of the hypotenuse side. Lets call the sides of the triangle x,y and h for the hipothenuse.

We have: h^2=x^2+y^2

The total area of the three squares is:

A(squares)= h^2 + x^2 + y^2 = h^2 + h^2 = 2*h^2

In the other side, you know that the area of a semicircle of diameter h es equal to:

A(semicircle)= (pih^2/4)/2= pih^2/8

Thus, the relation between the area of a square of side h and the area of a semicircle of diameter h is:

A(semicircle)/A(square)= (pi*h^2/8)/h^2 = pi/8

Now we can find the total area of the three semicircles by multiplying the total area of the squares by pi/8:

A(total)=(pi/8)*A(squares)=(pi/8)*2*h^2= pi*h^2/4= pi*14^2/4 ≈ 153,94

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