All categories

3 years ago

What additional information is needed to prove triangle LMN is congruent to triangle KMN using the HL theorem?

Mathematics

2 answers:

Alexxx [7]3 years ago

6 0

<h3>Answer</h3>

leg MN of both triangle is equal.

<h3>Step-by-step explanation</h3>

HL stands for "Hypotenuse, Leg" (the longest side of a right-angled triangle is called the "hypotenuse", the other two sides are called 'legs', or 'base' and 'height'.

It means we have two right-angled triangles with

the same length of hypotenuse
the same length for one of the other two legs

Since ∠ LMN and ∠KMN are right angle , the hypotenuse LN and and one leg LN of one right-angled triangle LMN are equal to the corresponding hypotenuse KN and leg MN of another right-angled triangle MKN, hence the two triangles are congruent.

IgorC [24]3 years ago

3 0

<h2>Answer:</h2>

∠LMN is a right angle

<h2>Step-by-step explanation:</h2>

If we want to prove that two right triangles are congruent by knowing that the corresponding hypotenuses and one leg are congruent, we begin as follows:

Since two legs are congruent and we know this by the hash marks, then the triangle ΔLKN is isosceles.
By definition LN ≅ NK

If ∠LMN is a right angle, then MN is the altitude of triangle ΔLKN
Also MN is the bisector of LK, so KM ≅ ML
So we have two right triangles ΔLMN and ΔKM having the same lengths of corresponding sides

In conclusion, ΔLMN ≅ ΔKMN

You might be interested in

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

2 years ago

A German Shepard puppy weighs 9 pounds and is gaining 5 pounds each month. A Great Dane puppy weighs 11 pounds and is gaining 4

bezimeni [28]

Answer:

2 months

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find an equation of variation where y varies directly as x and y = 7 when x = 0.5 .

Alisiya [41]

Answer:

y = 14x

Step-by-step explanation:

Use the direct variation equation, y = kx

Plug in 7 as y and 0.5 as x, and solve for k:

y = kx

7 = k(0.5)

14 = k

Plug this into the equation:

y = kx

y = 14x

So, the equation of variation is y = 14x

4 0

2 years ago

What is the best measure of center for the following data set?

Rom4ik [11]

Answer:

b) median: there are 7 numbers listed; 20 is the middle number among them

range: subtract lowest from highest number 25-2=23

mean: add all the numbers together =129

mode: place numbers in order (smallest to greatest), then see which number appears the most 2, 18, 18, 20, 22, 24, 25=18

Step-by-step explanation:

5 0

2 years ago

Is negative 5 a rational number

katen-ka-za [31]

Yes and rational numbers are numbers that can be written as a ratio of two integers

8 0

2 years ago