Please help I will reward brainly

Match each set of vertices with the type of triangle they form.

Andrew [12]

Answer: The calculations are done below.

Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

$AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.$

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

$AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.$

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

$AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.$

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

$AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.$

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

$AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.$

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

4 0

3 years ago

Read 2 more answers

Tia’s tent is in the form of a triangular prism as shown below if Tia plans to waterproof the tent excluding the pace what is th

Marat540 [252]

just multiple it all together

7 0

3 years ago

6. Find the slope of the given points (3,4) and (7,-4).

Grace [21]

Answer:

-2, C

Step-by-step explanation:

Use the formula: (y2-y1)/(x2-x1)

Substitute the numbers:

(4-(-4)) / (3-7)

=8 / -4

= -2

The slope is -2, or C

7 0

3 years ago

Read 2 more answers

The five circles making up this archery target have diameters of length $2,4,6,8,$ and $10$. What is the total red area?

Sliva [168]

Answer:

$A=15\pi\ units^2$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

It is given that the diameter of 5 circles making up the archery is 2,4,6,8, and 10.

To determine the total red area, we use the formula for area of the circle

$A=\pi r^{2}$

step 1

Find the Area of the 1st red circle

$r=2/2=1\ unit$ ---> the radius is half the diameter

$A_1=\pi (1)^{2}=\pi\ units^2$

step 2

Find the Area of the 2nd white circle

$r=4/2=2\ units$ ---> the radius is half the diameter

$A_2=\pi (2)^{2}=4\pi\ units^2$

step 3

Find the Area of the 3rd red circle

$r=6/2=3\ units$ ---> the radius is half the diameter

$A_3=\pi (3)^{2}=9\pi\ units^2$

step 4

Find the Area of the 4th white circle

$r=8/2=4\ units$ ---> the radius is half the diameter

$A_4=\pi (4)^{2}=16\pi\ units^2$

step 5

Find the Area of the 5th red circle

$r=10/2=5\ units$ ---> the radius is half the diameter

$A_5=\pi (5)^{2}=25\pi\ units^2$

The total red area is given by

$A=A_5-A_4+A_3-A_2+A_1$

substitute

$A=25\pi-16\pi+9\pi-4\pi+\pi$

$A=15\pi\ units^2$

7 0

3 years ago

Add 1} + (-5) using the number line.

gizmo_the_mogwai [7]

Answer:

1/2

Step-by-step explanation:

1 1/3+(-5/6)

1 1/3=4/3

4/3+(-5/6)

4/3-5/6

8/6-5/6

3/6

simplify

1/2

6 0

3 years ago