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Paladinen [302]

2 years ago

8

Find the values of x and y that make these triangles congruent by the HL Theorem

1 answer:

Misha Larkins [42]2 years ago

7 0

<span>The correct answer is A. X=5, Y=6. Both triangles are right triangles. Each triangle has corresponding sides that are congruent which makes the triangles congruent. The answer A has the values that make each side of both triangles equal to the other triangle.</span>

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Which statements are true regarding triangle LMN? Check all that apply.

dimaraw [331]

Answer:

$NM = x$

$LM = x\sqrt{2}$

$tan (45) = 1$

Step-by-step explanation:

Step 1: Pythagoras Theorem

Pythagoras theorem relates the three sides of the triangle in such a way that the sum of the square of base and perpendicular is equal to hypotenuse, such as:

$LM^{2} =LN^{2} +NM^{2}$

Step 2: Trigonometric Functions

Only for a right angle triangle following three trigonometric relations are valid

$sin (\theta) = \frac{opposite}{hypotenuse}$

$cos (\theta) = \frac{adjacent}{hypotenuse}$

$tan (\theta)=\frac{sin (\theta)}{cos (\theta)} = \frac{opposite}{adjacent}$

Step 3: Verifying all the possible answers

A: Since, LN = x and using $tan (45) =1$

we can calculate

$tan (\theta)= \frac{opposite}{adjacent}$

$tan (45)= \frac{NM}{x} =1$

therefore, NM = x (true)

B: As NM = x therefore it can not be equal to $x\sqrt{2\\}$ .

C: Using Pythagoras Theorem

$LM^{2} =LN^{2} +NM^{2}$

$LM^{2} =x^{2} +x^{2}$

$LM^{2} =2x^{2}$

$LM = \sqrt{2x^{2}} = x\sqrt{2}$

It can also be proved using trigonometric relation

$cos (45) = \frac{x}{LM}$

$LM = \frac{x}{cos (45)}$

As, $\frac{1}{cos (45)}= \sqrt{2}$

Therefore

$LM = x\sqrt{2}$

D and E:

Using same approach similar to part A

Since, LN = x and NM = x

we can calculate

$tan (\theta)= \frac{opposite}{adjacent}$

$tan (45)= \frac{x}{x} =1$

Therefore, $tan (45) = 1$ and not equal to $\frac{\sqrt{2} }{2}$

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