Admission to a car show cost $5 for each ticket. After selling only 20 tickets, they decided to lower the price to $3 each. If t
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A right triangle is a triangle that has a right angle of 90 degrees. In this exercise, we have two of this type of triangles. To solve this problem, we will apply some properties of triangles, so:
1. Which step can be used to prove that triangle EFG is also a right triangle?
To prove this, we need to use Pythagorean Theorem. So, for triangle EGF, if the formula:
is true, then EFG is also a right triangle. The steps are:
- Square a
- Square b
- Square c
- If the sum of the square of a and the sum of the square of b equals the square of c, then this is a right triangle.
2. Prove that the sum of a and c is greater than b.
We need to prove that:
From trigonometry, we know that:
By substituting these values in the previous inequality:
Since is an angle greater than 0 and less than 90 degrees,
are numbers between 0 (greater than) and 1. Therefore, this inequality is always true. So, the statement has been proved.
3. Prove that the sum of a and b is greater than c.
We need to prove that:
Applying the same reasoning as in 2:
This inequality is always true. So, the statement has been proved.
4. Prove that triangles are congruent by SSS property and hence angle EGF is equal to angle KML.
SSS means that we know the three sides of the triangle and want to find the missing angles. For ΔEGF, we know the three sides, and for KML, since this is a right triangle KL = c, therefore the corresponding sides of these two triangles are equal. Accordingly, these two triangles are congruent by SSS and angle EGF is equal to angle KML.
5. Prove that the ratio of EF and KL is greater than 1 and hence the triangles are similar by AA postulate.
The ratio of EF and KL is:
The statement is true if and only if EF > KL. Then, according to AA postulate, two triangles are similar if they have two corresponding angles congruent. So:
- EGF is congruent to KML and equals 90 degrees
- GEF is congruent to MKL
- EFG is conguent to KLM
Accordingly, the triangles are similar.