Three Tangent Circles, 2

Category: Plane Geometry

"Published in Vacaville, California, USA"

Three tangent circles of radius 10 cm are drawn. All centers lie on the line AB. The tangent AC to the right-hand circle is drawn, intersecting the middle circle at D and E. Find the length of the segment DE.

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Solution:

Consider the given figure above

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Since AC is tangent to the third circle at C, then the radius of third circle is perpendicular to AC. Hence, the three lines becomes a right triangle as follows

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By Pythagorean Theorem, the length of AC is

At the center of second circle, draw a line which is perpendicular to DE. Extend that line at the opposite ends of second circle as follows

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If SQ is perpendicular to DE and passes thru the center of the circle P, then SQ bisects DE. Since RP is perpendicular to AC and CO is perpendicular to AC, then RP is parallel to CO. Because of this ∆ARP and ∆ACO are similar.

If ∆ARP and ∆ACO are similar, then we can solve for RP by ratio and proportion as follows

If a theorem says "When two chords intersect inside a circle, the product of the segments of one chord equals the product of the segments of the other chord.", then the working equation for the second circle is

Therefore, the length of DE is

 

Triangle and Trapezoid Problems, 2

Category: Plane Geometry

"Published in Vacaville, California, USA"

Find the area of the shaded region

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Solution:

Consider the given figure above

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The small triangle is an isosceles triangle because the two adjacent sides are congruent. Same thing with the big triangle. If the bases of two triangles are parallel with the same or common vertex, then two triangles are similar. 

Since the two bases of the shaded region is parallel and the opposite sides are congruent, then the figure of the shaded region is an isosceles trapezoid. 

The length of the lower base as well as the altitude of an isosceles trapezoid are unknown, then we need to label further the figure first as follows

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The altitude of big triangle bisects the bases of the small and big triangles with their same or common vertex.

By using ratio and proportion since two triangles are similar, the length of the lower base of an isosceles trapezoid is

By Pythagorean Theorem, the altitude of the small triangle is

By using ratio and proportion since two triangles are similar, the altitude of an isosceles trapezoid is 

Therefore, the area of the shaded region which is an isosceles trapezoid is

                               or

 

Triangle and Trapezoid Problems

Category: Plane Geometry

"Published in Newark, California, USA"

Find the area of the given figure

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Solution:

Consider the given figure above 
 

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The given figure consists of equilateral triangle and a right trapezoid. The altitude of a right trapezoid is also a side of an equilateral triangle. Since the altitude of an equilateral triangle is not given, then we have to solve it first as follows

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By Pythagorean Theorem, the altitude of an equilateral triangle is

 

 

 

 

The area of an equilateral triangle is

The area of a right trapezoid is

 

 

Therefore, the area of a given figure is
 

 

 

                             or