How do we find the area of an overlapping object?

I'm afraid I don't know exactly what you mean, but I assume you mean a shape that overlaps another shape like so:

To find the combined area of a shape and another shape that is overlapping it, you must ensure that you can find the area of both the two shapes and the the area that the two shapes overlap. In the case in the picture, we can find the area of the red and blue rectangles, along with the purple area in between them.

The formula for finding the area of a rectangle is "a = l * w" where "a" is the area, "l" is the length, and "w" is the width. Using this, we can the find the area of the red rectangle by plugging in 7 for length and 3 for width:

a = 7 * 3

or...

a = 21

Using the very same method, we can find the area of the blue rectangle using 6 as length and 4 as width:

a = 6 * 4

or...

a = 24

Finally, we can find the area of the purple rectangle (the area overlapped by the two rectangles) using 2 as length and 1 as width:

a = 2 * 1 

or...

a = 2

Now that we know this, we can add the area of the two rectangles to get:

24 + 21 = 45

However, since the shapes overlap, we need to subtract the overlapping area to get:

45 - 2 = 43

Thus, the combined area of these two rectangles is 43. I hope I helped.

-pokemonfan58

What an excellent answer Pokemonfan58!   I'd thumbs up you twice if i could.  But I can't. 

Oh. Dont forget that it is area so it is 43units²

I see that you are new.

Welcome to Web2.0calc forum.   We hope that you learn lots and have a great time here!

With answers like that, you are going to be invaluable to this forum. 

I'll second what Melody said....nice graphic, too!!!........"points" from me......Welcome aboard!!!

Here's another 'overlapping' problem.

You have two circles having radii of 15cm and 20cm and they overlap in such a way that at their common points of intersection, the tangents to the two circles are at right angles, (which would imply that the tangent to one circle would pass through the centre of the other).

Calculate their common area, (and it doesn't require calculus).

I haven't drawn it up but I think that the common area (that is the overlap) might  be

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{15}}}{{\mathtt{20}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{15}}}{{\mathtt{20}}}}\right)}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}\right)} = {\mathtt{166.041\: \!867\: \!567\: \!676\: \!442}}$$ 

I do not have time to check that it is right. I wish I did.  Maybe later.

I made an illustration of your problem Bertie.

Since the radius of the first circle from center to the common point is 15 and the radius of the second circle to the common point is 20 the distance between the two centers had to be $${\sqrt{{{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{20}}}^{{\mathtt{2}}}}} = {\mathtt{25}}$$

I'm not sure about how I can solve your question though

Since you said no calculus, I can hardly believe Melody's answer is correct .

Sorry Melody 

$${\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{15}}}{{\mathtt{20}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{20}} = {\mathtt{166.041\: \!867\: \!567\: \!676\: \!442}}$$

$$\tan^{-1} ( \frac{15}{20} )
= \tan^{-1}( \frac {1}{\frac{20}{15}} )
=\cot^{-1} ( \frac{20}{15} )
=\frac{180}{2}-\tan^{-1} ( \frac{20}{15} )\\
\Rightarrow$$

$$\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}\right){\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}\right)}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{20}} = {\mathtt{166.041\: \!867\: \!567\: \!676\: \!442}}$$

Given that you both find the same answer.

Would you care to give some explanation Bertie?

I thought you said we didn't need any calculus...

What are you onabout reinout - I didn't use any calculus!

I might've misinterpreted it then. I thought no calculus meant we could derive the area by simple logic.

I don't know that i can draw the picture. maybe I could but it would take me a while

The radii of the 2 circles join to form a kite.  The equal angles are 90 degrees.  

The long sides are 20cm and the two short sides are 15cm. Using pythagoras it is clear that the axis of symmentry is 25cm.

Consider the angle between the axis of symmetry and the 20 cm side. Let this angle be theta.

theta = atan(15/20)

So this arc is subtended from an angle 2*atan(15/20)

so the area of this sector (in the larger circle) is  [2*atan(15/20)/360]*pi*20^2

The area of the triangle is 0.5*20*20*sin[2*atan(15/20)]

So the area of the small segment of the large circle is  

{[2*atan(15/20)/360]*pi*20^2}-{0.5*20*20*sin[2*atan(15/20)]}

Then you go through the same process for the smaller circle and add the 2 bits  together.

See - I am not so stupid afterall!

Hi Melody,

the area of the two triangle together is r1*r2 = 15 * 20

Yes of course.  thanks

I definitely did not mean to offend you Melody.

I meant that it was my own mistake to think that it had to be solved by pure logic.

Off course you're not stupid, you're the 'creator' of this forum for christs sake!

I should not have put that there.  

I know that you didn't mean to offend me.

Thanks Reinout.