Quiz Question

One of a continuing series

Just when you thought we were finished with the math questions. This is from the Internet, so no  fair running to Google for an answer.

The red right triangle is circumscribed by the large circle. The two sides of the triangle are diameters of the smaller circles. Prove the blue area is equal to the red area.

Post your answer as a comment below.

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Quiz Question

One of a continuing series

Keeping with a run of math questions… This problem is on the Internet. You have to provide an answer without going to the Internet.

The large arc is centered at O, The small arc is centered at D. Prove the two shaded areas are equal.

Post  your answer as a comment below.

Update and solution

Mike and Steve have provided correct solutions. See the comments. Steve worked out the math, and Mike stated the path to resolution rather cryptically. Both invoked π, which is not necessary. Try this approach.

The triangle is a right, equilateral triangle. The hypotenuse is √2 times the base and is also the diameter of the small semicircle. You will have no problem from that point concluding the small semicircle’s area is ½ the area of the large semicircle. The area A of the small semicircle is equal to the area of the triangle + the circle segment subtended by the triangle’s hypotenuse. The area of the triangle is A – the area of the segment. The area of the lune outside the large semicircle is A – the area of the segment. Therefore the two areas are the same.

Quiz Question

One of a continuing series

Back to math questions for a change. Full disclosure: I don’t make up all of these. This is from an Internet site. No fair going to the Internet to get the answer.

The triangle is equilateral. Prove the shaded area is equal to the inner circle. Post your answer as a comment below.

Update and solution

Mike is the first and only to provide the correct solution. A reasoning goes like this.

It is easy to demonstrate (exercise left to the reader) that the inner circle is ¼ the area of the outer circle. Then the region between the inner and outer circles is ¾ the area of the outer circle. The blue-shaded regions total 1/3 of this difference or ¼ the area of the outer circle. The inner circle is equal to the blue-shaded area.

Quiz Question

One of a continuing series

 

Taking the easy road this week. I pulled this week’s Quiz Question from an Internet site, so don’t go searching math puzzles on Google. Copied and pasted from the site:

Use the numerals 1, 9, 9 and 6 exactly in that order to make the following numbers: 28, 32, 35, 38, 72, 73, 76, 77, 100 and 1000

You can use the mathematical symbols +, -, ×, /, √, ^ (exponent symbol) and brackets.

Example: 63 = 1×9+9×6

Post your answer in the comments section below. The winner will be whoever posts the greatest number of correct solutions.

Quiz Question

One of a continuing series

This is from somebody else. It showed up on my Facebook feed just in  time, when I needed inspiration for a new Quiz Question. It’s easy. Give yourself about 15 seconds. The problem was posed as:

There are three boxes and three statements. There is a car in only one of the boxes. Only one statement is true. Which statement is true, and in which box is the car?

Post your answer as a comment below.

Quiz Question

One of a continuing series

Got this one from the Internet, so no fair going to Google for the answer.

ABCDEF × 3 = BCDEFA

Substitute a digit for each letter to provide the correct equation. Post your answer as a comment below. The solution will be provided next week (or sooner).

Update

No solution given yet. I have not taken the time to solve this, but here are some hints.

Note that A < 4 and A ≠ 0. A ≠ 0 is not stated in the problem, but I’m taking it as assumed. If A > 3, then multiplying by three would produce overflow and a number with more digits.

BCDEFA is divisible by 3, which means ABCDEF is divisible by 3, since both have the same digital root.

BCDEFA is divisible by 9.

That should get people going, so I’m going to give more time to come up with an answer.

Quiz Question

One of a continuing series

Easy one for a change, so give yourself 10 seconds to work it. It’s a single water hose with the ends uncoupled. Where are the ends?

Post your answer in the comment section below.

Update and solution

The solution is straight-forward. See the revised picture below.

Draw circles (ellipses) around A and B. Each has three hoses crossing into (or out of) the ellipse. Therefore, there must be a hose end within each of the two ellipses. Since there are only two ends (one hose), the ends must be under A and B. You don’t need to examine C and D, but if you do you will observe an even number of crossings.

Quiz Question

One of a continuing series

mathematics-gemoetrydivideidentical

Readers have been getting off easy recently. I’m going back to geometry questions, so give your brain a work out.

I found this on the Internet, but you shouldn’t go looking for the solution without first coming up with a solution. With a single line, does not need to be straight, divide the shape shown above into two identical parts. Post your answer as a comment below.

Actually, send me a copy of your solution by email, and I will post it.

Update

No solution. I have not solved it. Mike proposed a solution. See his comment below. Unable to post a graphic, he indicated the shape of the solution as follows:

XX
XXX
X

See the figure below:

Shape A is the original, turned upright. Shape B is Mike’s proposed solution in graphical form. My apologies if I misinterpreted Mike’s rendition.

What is apparent to me is that shape B cannot be fitted twice into shape A. I’m calling the Quiz Question still  unanswered.