Lesson HOW TO solve problems on squares

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How to solve problems on squares


In this lesson typical problems on squares are presented. These problems relate to the side measures of squares.

Problem 1
Find the perimeter of the square, if its side is 9 cm long.

Solution
The perimeter of a square is equal to the sum of the measures of its four sides:
4*9 cm = 36 cm.

Answer. The perimeter of the square is 36 cm.


Problem 2
Find the side measure of the square, if its perimeter is 52 cm.

Solution
The side of a square is one fourth of its perimeter.
So, the side length of our square is 52/4 cm = 13 cm.

Answer. The side of the square is 13 cm long.


Problem 3
Find the diagonal of the square, if its side is in 7 cm long.

Solution
Consider the right triangle formed by the two adjacent sides of the square and its diagonal.
By the Pythagoren Theorem, the hypotenuse of this triangle is

sqrt%287%5E2+%2B+7%5E2%29+=+sqrt%282%2A49%29+=+7sqrt%282%29 =~ 1.414*7 = 9.899.

This is the length of the square diagonal.

Answer. The square diagonal is 7sqrt%282%29 =~ 9.899 cm long.


Problem 4
Find the side of the square, if it is in 2 cm shorter than the diagonal.

Solution
Let x be the measure in centimeters of the side of the square.
Let us consider the right triangle formed by the two adjacent sides of the square and its diagonal.
This triangle has the legs x cm long and the hypotenuse x+2 cm long.
Apply the Pythagorean Theorem to this triangle. You get the quadratic equation

x%5E2+%2B+x%5E2+=+%28x%2B2%29%5E2.

To solve this equation, open the brackets first:

x%5E2+%2B+x%5E2+=+x%5E2+%2B+4x+%2B+4.

Then cancel the terms x%5E2 in both sides, move all the terms from the right side to the left with the opposite signs, and combine the like terms:

x%5E2+-+4x+-+4+=+0.

This is the quadratic equation in the standard form. Solve it using the quadratic formula

.

So, the two roots are x%5B1%5D+=+2+%2B+2sqrt%282%29 and x%5B2%5D+=+2+-+2sqrt%282%29.
Only first root suits to the problem condition. The second root gives the negative measure of the square side, which doesn't make sense.

So, the square sides are 2+%2B+2sqrt%282%29 cm long.
Then the diagonal length is equal to 4+%2B+2sqrt%282%29 cm in accordance with the condition.

Let us check that the Pythagorean Theorem is valid in this case.
The sum of squares of the legs of the right triangle is

.

From the other side, the square of the hypotenuse is

%284+%2B+2sqrt%282%29%29%5E2+=+16+%2B+8sqrt%282%29+%2B+8+=+24+%2B+8sqrt%282%29.

The check shows that the solution is correct.

Answer. The square sides are 2+%2B+2sqrt%282%29 cm =~ 4.828 cm long.

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