It can help to draw your own figure with the right angles better represented. You will notice that right triangles PQR and PRS share side RS. PQ is the hypotenuse of one of them; RS is the hypotenuse of the other.

We can write two equations that make use of the Pythagorean theorem:

PQ² = PR² +RQ²

RS² = QS² +RQ²

Then we can swap sides of the first equation and subtract the second equation. This gives ...

(PR² +RQ²) -(RS²) = (PQ²) -(QS² +RQ²)

Adding RQ², we can write this as ...

PR² -RS² +2RQ² = PQ² -QS²

The given relationship PR > RS means that PR² -RS² will be positive, so the left side of this equation must be positive.

Since PQ² - QS² is a positive number, PQ > QS.

6 0

4 years ago

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the point P = (x, 1/2) lies on the unit curcle shown below. what is the value of x in the simplest form

Mrac [35]

Answer:

x=sqrt(3)/2 or -sqrt(3)/2

Step-by-step explanation:

(x,1/2) is on the unit circle at pi/6 and 5pi/6.

so x is either sqrt(3)/2 or -sqrt(3)/2

5 0

4 years ago

Examine the diagram and information to answer the question.

Morgarella [4.7K]

Answer:

23.4 units

Step-by-step explanation:

Given the point of the vertices of triangle ABC as A(−5,1), B(2,1), and C(6,4).

The distance (d) between two points X (x₁ , y₁) and Y(x₂, y₂) is given as:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . We can use this to calculate the length of the sides of the triangle. Therefore:

$AB=\sqrt{(2-(-5))^2+(1-1)^2}=\sqrt{7^2}=7\\ AC=\sqrt{(6-(-5))^2+(4-1)^2}=\sqrt{11^2+3^2}=11.4\\\\BC=\sqrt{(6-(2))^2+(4-1)^2}=\sqrt{4^2+3^2}=5\\$

The perimeter of the triangle ABC = AB + AC + BC = 7 + 11.4 + 5 = 23.4 units

4 0

3 years ago