Area List

Find x, such that the triangle has an area of 4.

(-5, 1), (0, 2), (-2, x)

Can someone please help me solve for x using a graphing calculator?
Thanks!

I hope you’ll understand what i’m going` to write (i’m from Romania, in the last year of high school)
First of all, I would note the determinant of the matrix with D and the area with A , to simplify our work.
The formula of the area is : A=1/2 * | D |
the determinant looks like that :
-5 1 1
0 2 1
-2 x 1
now, we’ll solve the determinant with the triangle’s formula =>
D= -10 + 0 -2 + 4 + 5x + 0 = -12 + 4 + 5x = 5x – 8
We know that A = 4 => 1/2 * | D | = 4 => | D | = 8 => 5x – 8 = 8 =>
x = 16/5
hope it will help you in some way

"The sum of the area of two triangles is 120cm^2. The triangles overlap to create another triangle, which is shaded. If 2/3 of the smaller triangle is unshaded, and 8/9 of the larger triangle is unshaded, what is the area of the shaded overlapping triangle?"

I tried to set up a system relating the two, but had two variables in one, and three in the other. Substitution only proved x = x for me. :p

Could someone get me started on the right path?

The part that is overlapping (shaded) is 1/3 of the smaller triangle, and 1/9 of the larger triangle.
S = area of the smaller triangle
L = area of the larger triangle

(1/3)S = (1/9)L
S = (1/3)L

You know that S + L = 120

By substitution, (1/3)L + L = 120
(4/3)L = 120
L = 90
S = 30

(1/3) of 30 = 10
(1/9) of 90 = 10

The overlapping area is 10cm²

Let be the triangle with vertices at P=(3,−2,1), Q=(−5,2,−5) and R=(3,−3,2).

The area of is:

The angle ∠QPR is _____ degrees
The angle ∠PQR is _____ degrees
The angle ∠PRQ is _____ degrees

14 degrees

23 degrees

97 degrees

The base of the triangle is 24cm and the area is 228cm^2. I have to find out the height
how???
Pleasee helpp x

A = 1/2BH
228 = 1/2*24*H
H = 228/12 = 19 CM.

An isosceles triangle is placed inside a circle of radius r with its vetices on the boundary of the circle. How should this be accomplished if the area of the triangle is to be maximized?

This problem relates to using derivatives… I have no idea what to do or how to set it up.. can someone please help? Thank you so much!

Draw a triangle in a circle with a height greater than the radius. The height is r+x.
The 1/2 of the base is sqrt(r^2 – x^2).
The area is A = (1/2) b h = (r+x)*sqrt(r^2 – x^2)

The max area occurs when dA/dx =0

Compute the derivative and solve for the value of x as a fraction of r or simply let r=1.

My sister is coming to visit me in Chapel Hill in a few days. She has sort of a thing for Mexicans and she would really love to visit any salsa dancing club or any bars/clubs that latinos generally hang out. I’m new to the area so I don’t know any locals to ask. I looked online but I couldn’t find anything. Thanks for your help!

Red Room Tapas Lounge in downtown Raleigh has a "Latin Night". But I think it’s more Spanish (as in actual Spain) than Mexican.

A right triangle has a hypotenuse of length 10 and a leg of length 7. What is the area of the triangle?

Please tell me how to solve this problem?

Hello!

A=1/2bh

We need to know the base and height.

If we’re given the base and height, we can multiply it and divide by 2 (:

For your question:

We need to find the base for your question.

We use Pythagorean Theorem:

a^2+b^2=c^2

7^2+b^2=10^2
49+b^2=100
b^2=51
b=√51
b=7.141428429

The base would be around 7

So now we apply the formula:

A=1/2(7)(7)
A=24.5

Hope this helps!

Sincerely,
Mr.Math

Message me if you need more help.

The instruction is : use a determinant and the vertices of the triangle to find the area of the triangle.

Vertices: (2,-2), (8,5), (6,-10)

The answer is 26. Does anyone know how to get the answers?
the answer to this problem is 26!!!! How can you get 26 as the answer?!!!!!

Use this formula(it uses the third-order determinant)

If the 3 points are
(x1,y1)
(x2,y2)
(x3,y3)
Then the area A is:

A =
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|

^_^

A seamstress is designing a triangular scarf so that the length of the base of the triangle, in inches, is 7 less than twice the height, h. Express the area of the scarf as a function of the height.

I know the base would be 2h-7
and the height would be h.

But I have no idea where to go from there or how to express the area as a function of the height.

A = 1/2 * b * h

b = 2h – 7

A = (2h – 7) * h / 2

A = (2h^2 – 7h) / 2

That’s all it is.

You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a,0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

The Area of the triangle is A = (1/2)ab

We know that a and b are related by a^2 + b^2 = 20^2
Use this to eliminate a from the area equation since
a^2 = 400-b^2 or a = sqrt(400-b^2)

So A = (1/2)* [sqrt(400-b^2)]*b
Now we want the Area A to be a max. So take the derivative dA/db and set equal to zero:
A ‘(b) = (1/2)*{ sqrt(400-b^2) + b*(1/2)(-2b)/sqrt(400-b^2) }
using the product and chain rule.
Set this = 0 and solve for b: you get
b = 20/sqrt(2)
then a = sqrt(400-b^2) = sqrt(400/2) = 20/sqrt(2)

Therefore, a = b for the maximum area.

(I’m assuming you’re in first year single variable calculus; in multivariable calculus, this can be done even quicker by other methods)