Question

Solve for the value of a/b.​

Answer 1

Step-by-step explanation:

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Solve for the value of a/b.

step by step solution:

First let’s review a principle. If two circles are tangent, then their centers and the point of tangency will be collinear. This is because each circle’s radius is perpendicular to the tangent line.

• We can solve the problem by constructing a line segment between the semicircle’s center, the tangent point, and the inscribed circle. As the inscribed circle is tangent to adjacent sides of the rectangle, its center is equidistant from the two sides of the rectangle. So we can extend the line segment to the corner of the rectangle, which bisects the rectangle’s corner into two equal 45 degree angles.

• We have thus constructed two isosceles right triangles: one for the small circle with legs equal to b and hypotenuse b√2, and one for the semicircle with legs equal to a and hypotenuse a√2.

• But the hypotenuse of the larger triangle is also equal to the sum of the hypotenuse of the small triangle, plus the radii b and a. Thus we can equate the two ways to express the length of the hypotenuse, which leads to the answer:

a√2 = a + b + b√2

a√2 – a = b + b√2

a(√2 – 1) = b(√2 + 1)

a/b = (√2 + 1)/(√2 – 1) ≈ 5.828

Answer 2

First let’s review a principle. If two circles are tangent, then their centers and the point of tangency will be collinear. This is because each circle’s radius is perpendicular to the tangent line.

First let’s review a principle. If two circles are tangent, then their centers and the point of tangency will be collinear. This is because each circle’s radius is perpendicular to the tangent line.Solve for the value of a/b.​We have thus constructed two isosceles right triangles: one for the small circle with legs equal to b and hypotenuse b√2, and one for the semicircle with legs equal to a and hypotenuse a√2.

But the hypotenuse of the larger triangle is also equal to the sum of the hypotenuse of the small triangle, plus the radii b and a. Thus we can equate the two ways to express the length of the hypotenuse, which leads to the answer:

a√2 = a + b + b√2
a√2 – a = b + b√2
a(√2 – 1) = b(√2 + 1)
a/b = (√2 + 1)/(√2 – 1) ≈ 5.828We can solve the problem by constructing a line segment between the semicircle’s center, the tangent point, and the inscribed circle. As the inscribed circle is tangent to adjacent sides of the rectangle, its center is equidistant from the two sides of the rectangle. So we can extend the line segment to the corner of the rectangle, which bisects the rectangle’s corner into two equal 45 degree angles.We can solve the problem by constructing a line segment between the semicircle’s center, the tangent point, and the inscribed circle. As the inscribed circle is tangent to adjacent sides of the rectangle, its center is equidistant from the two sides of the rectangle. So we can extend the line segment to the corner of the rectangle, which bisects the rectangle’s corner into two equal 45 degree angles.We have thus constructed two isosceles right triangles: one for the small circle with legs equal to b and hypotenuse b√2, and one for the semicircle with legs equal to a and hypotenuse a√2.

But the hypotenuse of the larger triangle is also equal to the sum of the hypotenuse of the small triangle, plus the radii b and a. Thus we can equate the two ways to express the length of the hypotenuse, which leads to the answer:

a√2 = a + b + b√2
a√2 – a = b + b√2
a(√2 – 1) = b(√2 + 1)
a/b = (√2 + 1)/(√2 – 1) ≈ 5.828We have thus constructed two isosceles right triangles: one for the small circle with legs equal to b and hypotenuse b√2, and one for the semicircle with legs equal to a and hypotenuse a√2.

But the hypotenuse of the larger triangle is also equal to the sum of the hypotenuse of the small triangle, plus the radii b and a. Thus we can equate the two ways to express the length of the hypotenuse, which leads to the answer:

a√2 = a + b + b√2
a√2 – a = b + b√2
a(√2 – 1) = b(√2 + 1)
a/b = (√2 + 1)/(√2 – 1) ≈ 5.828First let’s review a principle. If two circles are tangent, then their centers and the point of tangency will be collinear. This is because each circle’s radius is perpendicular to the tangent line.