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4 years ago

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What is the area of a triangle with a base length of 2 1/2 inches and a height of 2 inches

1 answer:

Sedaia [141]4 years ago

6 0

Hello I am happy to help you with this problem. So the formula to solve for a triangle is A = bxh divided by 2.
So it’s area equals base times height divided by 2. So the base is 2 1/2 which becomes 5/2 ( five over 2) times 2 which equals 5. It’s because you cancel. So 5 divided by 2 equals 5/2 which also equals 2 1/2.

I hope that helps you. If you have any questions please feel free to ask me.

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kherson [118]

Answer:

$\large\boxed{(x-2)^2+(y-1)^2=34}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

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