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3 years ago

15

Find the value of the variables in each figure. Explain your reasoning.

1 answer:

sergiy2304 [10]3 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

The figure has one pair of parallel sides and is a trapezoid.

Each lower base angle is supplementary to the upper base angle on the same side, thus

15x + 30 + 10x = 180 , that is

25x + 30 = 180 ( subtract 30 from both sides )

25x = 150 ( divide both sides by 25 )

x = 6

Also

3y + 18 + 90 = 180 , that is

3y + 108 = 180 ( subtract 108 from both sides )

3y = 72 ( divide both sides by 3 )

y = 24

-------------------------------------------------------------------

2x, 90 and x lie on a straight line and sum to 180 , thus

2x + 90 + x = 180 , that is

3x + 90 = 180 ( subtract 90 from both sides )

3x = 90 ( divide both sides by 3 )

x = 30

------------

2y and x are alternate angles and congruent , thus

2y = x = 30 ( divide both sides by 2 )

y = 15

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

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Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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