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No, BE is not 2.5. BE is half of BC or AB which you assert is 4.33.

For the record, I didn't solve the problem graphically either. Like you guys I solved it mathematically but, even though it was pi-day, I withheld the introduction of pi hoping that it would cancel out. It did. Let f(r) be the area of a one-eighth pie shaped portion of a circle with radius r. If needed, this is πr^2/8. The area that we are to find is the sum of three pieces which, in turn, can be expressed as differences in areas of circle sections and triangles. In total:

A = [2*f(2) - 2] (The lower left section being a quarter circle of radius two minus a triangle of area 2)
+ [f(4) - 2*f(2) - 2] (The middle section being an eighth circle of radius four minus the same triangle and quarter circle from above)
+ [8 - f(4)] (The final section being the large top left triangle minus an eighth circle of radius four)

All the circle sections cancel, leaving simply 4 as already shown.

In the second problem:

BE is 2.5. AB, BC and CD are all 5. Line AD bisects line BC so BE and CE are both 2.5. The hypotenuse of both triangles is 5.

ABE is a right triangle. If one side of a right triangle is 2.5 and the hypotenuse is 5 (given) the triangle is a 30/60 right triangle. In a 30/60 right triangle if the short side is 1 and the hypotenuse 2, then the other side is 1.732 or the square root of 3. A squared plus B squared = C squared

2.5 x 1.732 = 4.33 for AB.

Along the same lines...

This is either visually obvious...

Or... WAY OVER MY HEAD mathematically. i.e. I should have paid attention (and not been throwing spitballs) back in Caveman Math 101...

AB can't be both 5 and 4.33. We have a right triangle but not a 30/60 right triangle

Letting AD and BC both be bisected at E. Then BE is one half AB since AB equal BC. By the Pythagorean equation AE^2 = AB^2 + BE^2 = AB^2 + AB^2/4. Solving for AB^2 yields 4/5*AE^2...

Problem 3.

36 - Neat!

Hint, consider the square of the lengths AB, BC, and AC. 4 AB^2 = BC^2 = 4/5 AC^2.

I said:

AB, BC and CD are all 5.
BC is bisected by line AD and therefore BE is 2.5. So the two sides are 2.5, 4.33 and hypotenuse is 5. If you make the short side 1 and the hypotenuse 2, the longer side is the square root of three which is1.732. By definition, if the hypotenuse is double the short side it is a 30/60 right triangle.

To check:
5 squared is 25
2.5 squared is 6.25
4.33 squared is 18.75

6.25+18.75=25

Note the two places I emboldened from your earlier reply. In the first, you assert that AB is 5. In the second, you've calculated AB to be 4.33.

I don't do metric conversion.......................

Nor should you.

This is America!

Yep, should have read AE, DE and BC are all 5. I confoozed myself. 😊

FIFY - Cheers! The 4.33 is still incorrect.

The hippopotamus of the triangle be 5.

Half if the hippopotamus be 2.5...

Thus the AB etc. is 4.33.

Alert!!! The teacher got it wrong! Sure, half the hyp (AE) is 2.5, but who said BE is half the hyp (AE)? BE is half BC. If BE is also half the hyp, then BC is 5 and then NOT equal to AB which is a precondition! Get away from the assumption we are dialing with a 30/60/90 triangle and use Pythagoras.

CK, Navlav8r,

You both assume that the double ticked segment (BE) is 2.5. This segment is half of BC which equals AB. Therefore, BE must equal half of AB. Your math does not check since 2.5 is greater than half of 4.33. Your assumption is wrong.

AB is the square root of twenty or 4.47. BE is half that at the square root of five or 2.23. Adding the square of the sides yields 25 which is the square of the hypotenuse (Pythagoras). The hypotenuse is thus calculated to be 5 which verifies.

Are you trolling me?

Any right triangle:
Height square + length square = hypotenuse square

this example: 2 height equals length, and hypotenuse equals 5

height equals sq root of (25/3), length equals sq root [2x(25/3)]

19 (339) 6346^ 3387649> ( (6#3) 44/332 345.7665x0

Maybe you missed this in the OP?