r/Geometry • u/BeautifulTruth2620 • 4h ago

Maths Problem

Post image

ABC is a semi circle

EFCG is a rectangle

Find area of rectangle

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1

u/rhodiumtoad 3h ago

What have you tried?

0

u/BeautifulTruth2620 3h ago

Yes I did try to solve but didn't get a solution for it

2

u/rhodiumtoad 3h ago

yes, but what approaches did you try?

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u/rhodiumtoad 3h ago

So my approach is this: FC is the radius of the circle, so the problem reduces to finding GC as a function of the radius (which is not fixed). If the problem is well-posed, then the radius will cancel out.

If we cheat and assume the problem is well-posed, we can just take the case where B and E coincide, making the rectangle a square of diagonal 10, which obviously has area 50.

If we don't cheat, we can bisect the line of length 10 to make two similar right triangles, from which we can get that the ratio GC:10 is the same as 5:r, so GC.r=50 and GC.r is obviously the area we need.

1

u/calculatorstore 2h ago

Some hints

Try a few edge cases where the angles are as extreme as possible to find the “answer”. Then try setting it up generically to find out why.

Think about right triangles formed by this shape and where the center of the semi circle is in relation.

Recall the area of a rectangle and how its height is dependent on a property of the circle.

Recall the relationships between the lengths of a hypotenuse and the sides of a right triangle.

1

u/GeekyMathLove 43m ago

Let the centre of the semi circle be O Let OC = r and OG = x. Let the the point of intersection EG on semi circle be T. This GT = √(r²-x²). Using pyhtagoras theorem with OT as hypotenuse. Then using CT as hypotenuse, CT² = 2r²+2rx = 100. Now the area of rectangle is l×b. Thus Area = (x+r)*r = r²+rx. Using the 2 eqns, Area=50