Right Circular Cylinder Problems, 14

Category: Solid Geometry

"Published in Newark, California, USA"

Pass a plane through a cube of edge 6 in. so that the section formed will be a regular hexagon. Find the volume of a right circular cylinder 8 in. long, (a) whose base circumscribed this hexagon, (b) whose base is inscribed in this hexagon.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

If a cube is cut by a plane that passes through the midpoints of two adjacent sides from the upper base to the opposite lower base with the midpoints of two adjacent sides, then the intersection is a regular hexagon. By Pythagorean Theorem, the length of the sides of a regular hexagon is

 
Let's consider the section of a cube which is a regular hexagon as follows

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There are six equal equilateral triangles in a regular hexagon in which their common vertex is a center of a regular hexagon. All sides of equilateral triangles are all equal which are 3√2 in.

(a) If the base of a right circular cylinder circumscribes the regular hexagon, then the radius is equal to 3√2 in. A circle contains all the vertices of a regular hexagon. Therefore, the volume of a right circular cylinder is
 

 

 

 

(b) If the base of a right circular cylinder inscribes the regular hexagon, then the radius is tangent to all the sides of a regular hexagon. The radius of a right circular cylinder is also an apothem of a regular hexagon and an altitude of an equilateral triangle. The altitude of an equilateral triangle bisects its base. By Pythagorean Theorem, the radius of a right circular cylinder is

Therefore, the volume of a right circular cylinder is 

 

 

Triangle Inscribed in a Circle Problems, 2

Category: Plane Geometry

"Published in Newark, California, USA"

Find the area of the shaded region:

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Solution:

Consider the given figure above

Photo by Math Principles in Everyday Life

We noticed that the longest side of a triangle is also the diameter of a circle. If that's the case, the inscribed triangle is a right triangle. That is also a theorem. Let's prove that the triangle is a right triangle by Pythagorean Theorem as follows

Since both sides of the equation are equal, then the triangle is a right triangle. The base of a triangle is 12 and its altitude is 5.

Therefore, the area of the shaded region is

                              or

 

Circle Inscribed in a Trapezoid Problems

Category: Plane Geometry

"Published in Newark, California, USA"

Find the exact area of the given trapezoid in the figure:

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Solution:

Consider the given figure above

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Since the given figure is an isosceles trapezoid, then it follows that ∠A ≅ ∠B, ∠C ≅ ∠D, and AD ≅ BC. If a circle is inscribed in an isosceles trapezoid, then its radius is tangent to the sides of an isosceles trapezoid. Let's analyze and label further the given figure as follows
 

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Line segment EG that passes the center of a circle bisects the two bases of an isosceles trapezoid. Line segment OB bisects ∠B and line segment OC bisects ∠C. 

Consider rt. ∆OGB and rt. ∆OFG. If OG ≅ OF and OB ≅ OB, then it follows that BG ≅ BF. In this case, BF = 9.

Consider rt. ∆OEC and rt. ∆OFC. If OE ≅ OF and OC ≅ OC, then it follows that EC ≅ CF. In this case, CF = 4.

Hence, the length of CB = CF + FB = 9 + 4 = 13.

Photo by Math Principles in Everyday Life

The value of x is 

By Pythagorean Theorem, the altitude of an isosceles trapezoid is

 

 

 

 

Therefore, the area of an isosceles trapezoid is