Since the lengths of the sides are consecutive integers and the shortest side is , the three sides are , , and .

We then use the Pythagorean Theorem:

To find the distance between points  and , split the square into two 45-45-90 triangles and find the hypotenuse. The side ratio of the 45-45-90 triangle is , so if the sides have a length of 5, the hypotenuse must be .

This is solving for the hypotenuse of a triangle. Using the Pythagorean Theorem, which says that  

Susie walks 30 meters north, then 80 meters west, then 30 meters north again. Thus, she walks 60 meters north and 80 meters west. These two directions are 90 degrees away from one another.

At this point, construct a right triangle with one leg that measures 60 meters and a second leg that is 80 meters.

You can save time by using the 3:4:5 common triangle. 60 and 80 are  and , respectively, making the hypotenuse equal to .

We can solve for the length of the missing hypotenuse by applying the Pythagorean theorem:

Substitute the following known values into the formula and solve for the missing hypotenuse: side .

Susie will walk 100 meters to reach her house.

First, define the sides of the triangle. Because the side lengths are consecutive odd numbers, if we define the shortest side will be as , the next side will be defined as , and the longest side will be defined as . We can then find the perimeter of a triangle using the following formula:

Substitute in the known values and variables.

Subtract 6 from both sides of the equation.

Divide both sides of the equation by 3. 

Solve.

This is not the answer; we need to find the length of the longest side, or . 

Substitute in the calculated value for  and solve.

The longest side of the triangle is 21 centimeters long.

 cannot be the side lengths of a right triangle.  does not equal . Also, special right triangle  and  rules can eliminate all the other choices.

The altitude of a right triangle from the vertex of its right triangle to its hypotenuse divides it into two similar triangles.

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

.

The ratio of the smaller sides of these similar triangles is 

The ratio of the smaller side of  to that of  is 

 or  : 1,

so this is also the ratio of the perimeter of  to that of .

Because the perimeter is equal to the sum of the lengths of the three sides of a triangle, we can add the three expressions for the lengths and set them equal to 40.

Perimeter:

Simplify the x terms.

Simplify the constants.

Subtract 8 from both sides.

Divide by 4

One side is 8.

The second side is

.

The third side is

.

Thus, the sides of the triangle are 8, 15, and 17.

The question asks us for the area of the triangle, which is given by the formula (1/2)bh. We are told it is a right triangle, so we can use one of the legs as the base, and the other leg as the height, since the legs will intersect at right angles. The legs of the right triangle must be the smallest sides (the longest must be the hypotenuse), which in this case are 8 and 15. So, let's assume that 8 is the base and 15 is the height. 

The area of a triangle is (1/2)bh. We can substitute 8 and 15 for b and h.

.

The answer is 60 units squared.

Because PA and PB are tangent to circle O, angles PAO and PBO must be right angles; therefore, triangles PAO and PBO are both right triangles.

Since AO and OB are both radii of circle O, they are congruent. Furthermore, because PA and PB are external tangents originating from the same point, they must also be congruent.

So, in triangles PAO and PBO, we have two sides that are congruent, and we have a congruent angle (all right angles are congruent) between them. Therefore, by the Side-Angle-Side (SAS) Theorem of congruency, triangles PAO and PBO are congruent.

Notice that quadrilateral PAOB can be broken up into triangles PAO and PBO. Since those triangles are congruent, each must comprise one half of the area of quadrilateral PAOB. As a result, if we find the area of one of the triangles, we can double it in order to find the area of the quadrilateral.

Let's determine the area of triangle PAO. We have already established that it is a right triangle. We are told that PO, which is the hypotenuse of the triangle, is equal to 17. We are also told that the diameter of circle O is 16, which means that every radius of the circle is 8, because a radius is half the size of a diameter. Since segment AO is a radius, its length must be 8.

So, triangle PAO is a right triangle with a hypotenuse of 17 and a leg of 8. We can use the Pythagorean Theorem in order to find the other leg. According to the Pythagorean Theorem, if a and b are the lengths of the legs of a right triangle, and c is the length of the hypotenuse, then:

a² + b² = c²

Let us let b represent the length of PA.

8² + b² = 17²

64 + b² = 289

Subtract 64 from both sides.

b² = 225

Take the square root of both sides.

b = 15

This means that the length of PA is 15.

Now let's apply the formula for the area of a right triangle. Because the legs of a right triangle are perpendicular, one can be considered the base, and the other can be considered the height of the triangle.

area of triangle PAO = (¹/₂)bh

= (¹/₂)(8)(15) = 60

Ultimately, we must find the area of quadrilateral PAOB; however, we previously determined that triangles PAO and PBO each comprise half of the quadrilateral. Thus, if we double the area of PAO, we would get the area of quadrilateral PAOB.

Area of PAOB = 2(area of PAO)

= 2(60) = 120 square units

The answer is 120.

The answer is 12. In this circumstance, the area of the triangle cannot be smaller than its hypotenuse length, and cannot be bigger than its hypotenuse squared (that would be the area of a square).