This is a hardest 5th Grade Math Problem From China? (To Identify Talented Students)

Can You Solve A 5th Grade Math Problem?

(Identify your talent)

I tried many times this problem.but i was 

Not able to found this solution.

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Finally ,i solved this question after long time .really this is great question.

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.keep continue..

Question. ABCD is a parallelogram. In the diagram (see below), the areas of yellow regions are 8, 10, 72, and 79. Find the area of red triangle. The diagram is not to scale.

Math prerequisites:
1. Basic arithmetic.
2. Area of parallelogram and area of triangle

Solution:

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.Answer To A 5th Grade Math Problem From China

It turns out the problem can be solved with a simple calculation. The area of the red triangle is:

79 + 10 – 72 – 8 = 9

Wow! But why does this calculation work? Even though I did not solve the problem, I was very interested in learning the method in order to verify the solution and to improve my problem solving skills.
Now we start the solution for this question.
First we calculate

The area of a triangle is (base × height)/2, and the area of a parallelogram is (base × height).

A triangle whose base equals one side of the parallelogram, and whose height reaches the opposite side of the parallelogram, has exactly half the area of a parallelogram. The same is true for a pair of triangles, if the pair of triangles span one side and if their heights reach the opposite side.

Let’s first label the unknown areas with letters a, b, c, d, e, and f. And let’s label the area we want with the letter x.

First consider the two triangles below whose sides together span the top side (length) of the parallelogram. As both of these triangles touch the bottom side of the parallelogram, each triangle has the same vertical height as the parallelogram does between its bottom and top sides.

Therefore, these two triangles have a combined area equal to 1/2 the area of the parallelogram. Their area is:

Upper triangles = (x + a) + (72 + b + 8) = (area of parallelogram)/2

Now consider the triangle with a base of the left side (width) of the parallelogram that touches the right width of the parallelogram. As this triangle touches two opposite sides of the parallelogram, the triangle has the same horizontal distance the parallelogram does between its left and right sides.

And ,

Therefore, this triangle also has an area equal 1/2 the area of the parallelogram, and its area equals:

Left triangle = a + 79 + b + 10 = (area of parallelogram)/2

As both equations equal half the area of the parallelogram, we can set these areas equal to each other.

(x + a) + (72 + b + 8) = a + 79 + b + 10

We can cancel the terms a and b on both sides and then solve for x.

x + 72 + 8 = 79 + 10

x= 79 + 10 – 72 – 8 = 9

And like magic we have found the solution!

Finally his answer is 9 for this question.

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