Mathematics, 21.01.2020 20:31 softballlover487

2. the coordinates of the vertices of abc are a (5, 2), b (2, 4), and c (7, 5). determine whether abc is a right triangle.

3. the coordinates of the vertices of quadrilateral qrst are q (5, 1), r (8, 7), s (14, 10) and t (10, 2). iron man states that quadrilateral qrst is a parallelogram. prove or disprove iron man’s statement. explain your method (fill out the brainstorming area) and show your work.

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Answer from: hePandaKing3750

Part 2) Triangle ABC is a right triangle (see the explanation)

Part 3) Quadrilateral QRST is not a parallelogram (see the explanation)

Step-by-step explanation:

Part 2) we have

A (5, 2), B (2, 4), and C (7, 5)

Plot the figure to better understand the problem

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the length side AB

A (5, 2), B (2, 4)

substitute the values in the formula

$d=\sqrt{(4-2)^{2}+(2-5)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$d_A_B=\sqrt{13}\ units$

step 2

Find the length side BC

B (2, 4), C (7, 5)

substitute the values in the formula

$d=\sqrt{(5-4)^{2}+(7-2)^{2}}$

$d=\sqrt{(1)^{2}+(5)^{2}}$

$d_B_C=\sqrt{26}\ units$

step 3

Find the length side AC

A (5, 2), C (7, 5)

substitute the values in the formula

$d=\sqrt{(5-2)^{2}+(7-5)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$d_A_C=\sqrt{13}\ units$

step 4

Verify if the triangle ABC is a right triangle

we know that a right triangle must satisfy the Pythagorean Theorem

BC^2=AB^2+AC^2

Remember that the hypotenuse is the greater side

substitute the values

$(\sqrt{26})^2=(\sqrt{13})^2+(\sqrt{13})^2$

26=13+13

26=26 ----> is true

therefore

Triangle ABC is a right triangle

Part 3) we have

Q (5, 1), R (8, 7), S (14, 10) and T (10, 2)

we know that

The opposite sides of a parallelogram are parallel and congruent

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step 1

Find the length side QR

Q (5, 1), R (8, 7)

substitute in the formula

$d=\sqrt{(7-1)^{2}+(8-5)^{2}}$

$d=\sqrt{(6)^{2}+(3)^{2}}$

$d_Q_R=\sqrt{45}\ units$

Step 2

Find the length side RS

R (8, 7), S (14, 10)

substitute in the formula

$d=\sqrt{(10-7)^{2}+(14-8)^{2}}$

$d=\sqrt{(3)^{2}+(6)^{2}}$

$d_R_S=\sqrt{45}\ units$

Step 3

Find the length side ST

S (14, 10), T (10, 2)

substitute in the formula

$d=\sqrt{(2-10)^{2}+(10-14)^{2}}$

$d=\sqrt{(-8)^{2}+(-4)^{2}}$

$d_S_T=\sqrt{80}\ units$

Step 4

Find the length side QT

Q (5, 1), T (10, 2)

substitute in the formula

$d=\sqrt{(2-1)^{2}+(10-5)^{2}}$

$d=\sqrt{(1)^{2}+(5)^{2}}$

$d_Q_T=\sqrt{26}\ units$

Step 5

Compare the length of the opposite sides

QR and ST

$\sqrt{45}\ units \neq \sqrt{80}\ units$

$d_Q_R \neq d_S_T$

RS and QT

$\sqrt{45}\ units \neq \sqrt{26}\ units$

$d_R_S \neq d_Q_T$

Opposite sides are not congruent

therefore

Quadrilateral QRST is not a parallelogram

2. the coordinates of the vertices of abc are a (5, 2), b (2, 4), and c (7, 5). determine whether 

Answer from: Quest

$32,660

step-by-step explanation:

Answer from: Quest

answer: $\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

$\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

which is the required equation for finding the value of x,

$\frac{x}{sin40^{\circ}}= \frac{12}{sin60^{\circ}}$

$x= 12\times \frac{sin40^{\circ}}{sin60^{\circ}}$

$x= 12\times 0.74222719896$

$x= 8.90672638762\approx 8.967$

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