This is the 8th problem of the week of MTAP Reviewer.
The diagonal of a square is 6cm. Find the area of the square that is twice its size.
We can divide a square into several ways, one of which is using its diagonal.
Click on the image to enlarge
Now, using the diagonal of the square as side of the larger square, we can create a larger square as shown above. As seen, the diagonal of the square cuts the square into half. This half region makes a fourth of the larger square (Can you see why?).
Since the length of the side of the larger square is the diagonal of the smaller square 6, the area of the larger square is 6× 6 = 36.
This is the eighth problem in our Problem of the Week.
Two semi-circles are placed inside a square as shown below. What is the area of the shaded part?
To get the area of the shaded part, we have to get the area of the square and then subtract the area of the two semicircles. Continue reading
My apologies for not sustaining the Problem of the Week section of this site. Anyway, I’ll do my best to make Problem of the Week a really weekly post. This week, we are going to tackle two problems since they are quite easy. These problem are classified under the third Grade and were taken from Grade 3 Sample Problem Set 1, so these are actually very easy.
My uncle’s age is 7 more than twice my eldest brother’s age. My eldest brother is 19 years old. How old is my uncle?
The eldest brother’s age is 19. The uncle’s age is 7 more than twice the brother’s age. Now, twice 19 is equal to 38. Since the uncle is 7 more than twice, we add 7 to 38. So, the uncle is 38 + 7 = 45 years old. Continue reading
The number 25a where a is a digit is divisible by 6. What is the largest possible value of a?
A number is divisible by 6 if it is both divisible by 2 and divisible by 3.
First, a number if divisible by 2 if it is even. Now, for 25a to be even, a must be even. Since a is even, it could be one of the following numbers: 0, 2, 4, 6 or 8.
Second, a number is divisible by 3 if the sum of its digit is divisible by 3. So, we substitute the possible numbers above, add the digits, and see if they are divisible by 3.
For a = 0
2 + 5 + 0 = 7 Continue reading
The last problem in the previous post asked about the least wrapper that can be used to wrap the box below. As we can see, if we wrap each side with a rectangular wrapper, it is equivalent to finding the areas of all the faces or finding the surface area of the prism. The original problem is as follows.
The Minimum Gift Wrapper Problem
Rina placed a gift for her friend in a box whose dimensions are shown below. She wants to cover the box. What is the least area of the wrapper that she could use? Continue reading