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3 years ago

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What kind of triangle must have an altitude that is outside the triangle?

2 answers:

Tanya [424]3 years ago

7 0

Answer:

Brainliest?

'''For an obtuse-angled triangle, the altitude is outside the triangle. For such triangles, the base is extended and then a perpendicular is drawn from the opposite vertex to the base. For an obtuse triangle, the altitude is shown in the triangle below.'''

Gekata [30.6K]3 years ago

6 0

Answer:

Obtuse

Step-by-step explanation:

When examining the shape of an obtuse triangle, due to it's wide nature, you will notice at least one of the side lengths will extend outward, which will create an altitude.

As outlined in this picture:

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Can anyone help me find the coordinates (?,?) given that the shape is a parallelogram? (15 points)

xxTIMURxx [149]

The missing coordinates of the parallelogram is (m + h, n).

Solution:

Diagonals of the parallelogram bisect each other.

Solve using mid-point formula:

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Here $x_1=m, y_1=n, x_2=h, y_2=0$

$$=\left( \frac{m+h}{2}, \frac{n+0}{2}\right )$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n}{2}\right )$

<u>To find the missing coordinate:</u>

Let the missing coordinates by x and y.

Here $x_1=0, y_1=0, x_2=x, y_2=y$

$$\text{Midpoint}=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{x}{2}, \frac{y}{2}\right )$

Now equate the x-coordinate.

$$ \frac{m+h}{2}=\frac{x}{2}$

Multiply by 2 on both sides of the equation, we get

m + h = x

x = m + h

Now equate the y-coordinate.

$$\frac{n}{2} = \frac{y}{2}$

Multiply by 2 on both sides of the equation, we get

n = y

y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

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