SOLUTION: Find, correct to 1 decimal place, the percentage areas for these situations. a) The largest square inside a circle. b) The largest circle inside a square. c) The largest square

Question 1128669: Find, correct to 1 decimal place, the percentage areas for these situations.
a) The largest square inside a circle.
b) The largest circle inside a square.
c) The largest square inside a right isosceles triangle.
d) The largest circle inside a right isosceles triangle.
Answer by KMST(5345) (Show Source): You can put this solution on YOUR website!
A good strategy to calculate the percentages required is to
assign a length of to a side or a radius,
because "scaling up or down" the drawing for any of the cases
would not change the ratio of areas of the geometrical figures.

a) The largest square inside a circle

If the radius of that circle is ,
the area of the circle is ;
each of the isosceles right triangles forming the square
has legs measuring and area = ,
and the area of the square is .
The area of the square as a percentage of the area of the square as a fraction/percentage of the area of the circle is

b) The largest circle inside a square

If the radius of that circle is ,
the area of the circle is ;
the length of the side of the square is ,
and the area of the square is .
The area of the circle as a fraction/percentage of the area of the square is

c) The largest square inside a right isosceles triangle

There are 4 small congruent triangles inside the large right isosceles triangle,
with of those small triangles forming the square,
so the square is of the triangle.

d) The largest circle inside a right isosceles triangle

If the radius of that circle is ,
and the length of each leg of the right isosceles triangle is ,
the area of the circle is ,
and the area of the triangle is .
The right isosceles triangle is made of three smaller triangles,
each with height ,
and each having for a base one side of the right isosceles triangle.
Two of those triangles have a base of ,
and the area of each one of those is .
The third small triangle has for a base,
and its area is .
The areas of those three small triangles adds up to the area of the right isosceles triangle, so
.
multiplying both sides of the equal sign times 2,
or .
Dividing both sides of the equal sign by ,
.
So, the area of the right isosceles triangle is
.
The area of the circle as a fraction/percentage of the area of the right isosceles triangle is

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