Chapter Test
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Substitute the given values for a and b into the Pythagorean Formula and verify if the value of c is 5. Recall the definition of the converse of a conditional statement.
Conditional Statement: Yes, see solution.
Converse: No, see solution.
We are given the following conditional statement. If a right triangle has a leg that is 3 in. long and a leg that is4 in. long, then the hypotenuse is5 in. long. First, let's determine whether the given statement is true. Then, we will analyze its converse.
In order to determine if the statement is true, let's use the Pythagorean Theorem.
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Pythagorean Theorem |
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In any right triangle, the sum of the squares of the lengths of the legs a and b is equal to the square of the length of the hypotenuse c. [0.1cm] a^2+b^2=c^2 |
We can substitute a with 3 and b with 4 and then solve the equation for c.
a= 3, b= 4
Calculate power
Add terms
Rearrange equation
sqrt(LHS)=sqrt(RHS)
The length of the hypotenuse of this right triangle is 5 inches, as the given statement says. Therefore, it is true.
In the if-then form of a conditional statement, the hypothesis comes after if
and the conclusion comes after then.
Using this information, we can identify the hypothesis and the conclusion of the given conditional statement.
& Statement
Hypothesis: & a right triangle has a leg that is
& 3 in. long and a leg that is4 in. long
Conclusion: & the hypotenuse is5 in. long
The converse of a conditional statement exchanges the hypothesis and the conclusion of the conditional statement.
& Converse
Hypothesis: & the hypotenuse is5 in. long
Conclusion: & a right triangle has a leg that is
& 3 in. long and a leg that is4 in. long
By adding if
before the hypothesis and then
before the conclusion, we can write the converse of the given conditional statement.
If the hypotenuse is5 in. long,
then the right triangle has a leg that is
3 in. long and a leg that is4 in. long.
Is this sentence always true? It is given that c equals 5, so we can substitute this value into the Pythagorean Formula.
a^2+b^2= 5^2 ⇔ a^2+b^2=25
Let's try to find values of a and b, other than 3 and 4, which make this equation true.
| a | b | Equation |
|---|---|---|
| 1 | sqrt(24) | 1^2+( sqrt(24))^2=25 |
| sqrt(5) | sqrt(20) | ( sqrt(5))^2+( sqrt(20))^2=25 |
| sqrt(11) | sqrt(14) | ( sqrt(11))^2+( sqrt(14))^2=25 |
As we can see, there are values of a and b other than 3 and 4 that make the equation true. The conclusion of the converse is not always met, which allows us to conclude that the converse is not always true.