Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

\(\displaystyle a^2+b^2=c^2\)

In this equation:

\(\displaystyle a \textup{ and }b=\textup{legs}\)

\(\displaystyle c=\textup{hypotenuse}\)

We can count the number of units on the coordinate plane that were used to create the legs of our drawn triangle. Afterwards, we can use the Pythagorean Theorem to solve for the length of the hypotenuse, or the original diagonal line. 

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

\(\displaystyle 7^2+5^2=c^2\)

\(\displaystyle 49+25=c^2\)

\(\displaystyle 74=c^2\)

\(\displaystyle \sqrt{74}=\sqrt{c^2}\)

\(\displaystyle 8.6=c\)

Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

\(\displaystyle a^2+b^2=c^2\)

In this equation:

\(\displaystyle a \textup{ and }b=\textup{legs}\)

\(\displaystyle c=\textup{hypotenuse}\)

We can count the number of units on the coordinate plane that were used to create the legs of our drawn triangle. Afterwards, we can use the Pythagorean Theorem to solve for the length of the hypotenuse, or the original diagonal line. 

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

\(\displaystyle 4^2+10^2=c^2\)

\(\displaystyle 16+100=c^2\)

\(\displaystyle 116=c^2\)

\(\displaystyle \sqrt{116}=\sqrt{c^2}\)

\(\displaystyle 10.8=c\)

Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

\(\displaystyle a^2+b^2=c^2\)

In this equation:

\(\displaystyle a \textup{ and }b=\textup{legs}\)

\(\displaystyle c=\textup{hypotenuse}\)

We can count the number of units on the coordinate plane that were used to create the legs of our drawn triangle. Afterwards, we can use the Pythagorean Theorem to solve for the length of the hypotenuse, or the original diagonal line. 

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

\(\displaystyle 18^2+9^2=c^2\)

\(\displaystyle 324+81=c^2\)

\(\displaystyle 405=c^2\)

\(\displaystyle \sqrt{405}=\sqrt{c^2}\)

\(\displaystyle 20.1=c\)

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 \(\displaystyle 3\cdot 20\) and \(\displaystyle 4\cdot 20\), respectively, making the hypotenuse equal to \(\displaystyle 5\cdot 20=100\).

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

\(\displaystyle a^{2}+b^{2}=c^{2}\)

Substitute the following known values into the formula and solve for the missing hypotenuse: side \(\displaystyle c\).

\(\displaystyle a= 60,\ b= 80,\ c= ?\)

\(\displaystyle (60)^{2}+(80)^{2}=c^{2}\)

 \(\displaystyle 3600+6400=c^{2}\)

\(\displaystyle 10,000=c^{2}\)

\(\displaystyle c=100\)

Susie will walk 100 meters to reach her house.

First, convert the dimensions to cubic centimeters by multiplying by \(\displaystyle 100\): the cone has height \(\displaystyle 300cm\), and its base has radius \(\displaystyle 200cm\).

\(\displaystyle 3m*\frac{100cm}{1m}=300cm\ \ \2m*\frac{100cm}{1m}=200cm\)

Its volume is found by using the formula and the converted height and radius.

\(\displaystyle V=\frac{1}{3} \pi r^{2}h\)

\(\displaystyle V=\frac{1}{3} \pi (200cm)^{2} ( 300cm )\approx 12,566,371 \; \textrm{cm}^{3}\)

Now multiply this by \(\displaystyle 0.45\frac{g}{cm^3}\) to get the mass.

\(\displaystyle m=V*\rho\)

\(\displaystyle 12,566,371cm^3*0.45\frac{g}{cm^3} \approx 5,654,867g\)

Finally, convert the answer to kilograms.

\(\displaystyle 5654867g *\frac{1kg}{1,000g} \approx 5,655kg\)

The volume of a cone is:

\(\displaystyle Volume=\frac{1}{3}\pi r^2h\)

where \(\displaystyle r\) is the radius of the circular base, and \(\displaystyle h\) is the height (the perpendicular distance from the base to the vertex).

As the circular base area is \(\displaystyle \pi r^2\), so we can rewrite the volume formula as follows:

\(\displaystyle Volume =\frac{A\times h}{3}\)

where \(\displaystyle A\) is the circular base area and known in this problem. So we can write:

\(\displaystyle Volume =\frac{A\times h}{3}=\frac{4\times 4}{3}=\frac{16}{3}\approx 5.333 m^3\)

We know that density is defined as mass per unit volume or:

\(\displaystyle \rho =\frac{m}{v}\)

Where \(\displaystyle \rho\) is the density; \(\displaystyle m\) is the mass and \(\displaystyle v\) is the volume. So we get:

\(\displaystyle m=\rho v=1000\times 5.333=5333 Kg\)

The volume of a cone is:

\(\displaystyle Volume=\frac{1}3{}\pi r^2h\)

where \(\displaystyle r\) is the radius of the circular base, and \(\displaystyle h\) is the height (the perpendicular distance from the base to the vertex).

\(\displaystyle Volume=\frac{1}3{}\pi r^2h =\frac{1}3{}\pi t^2\times 2t=\frac{2}{3}\pi t^3\)

The volume of a cone is given by:

\(\displaystyle Volume=\frac{1}{3}\pi r^2h\)

where \(\displaystyle r\)is the radius of the circular base, and \(\displaystyle h\) is the height; the perpendicular distance from the base to the vertex. Substitute the known values in the formula:

\(\displaystyle Volume=\frac{1}{3}\pi r^2h\Rightarrow 8\pi=\frac{1}{3}\pi\times (2t)^2\times 3t\)

\(\displaystyle \Rightarrow 8\pi=4\pi t^3\Rightarrow t^3=2\Rightarrow t=\sqrt[3]{2}\)

First, divide the diameter in half to find the radius.

\(\displaystyle \frac{d}{2}=r\)

\(\displaystyle \frac{12}{2}=6\:m\)

Now, use the formula to find the volume of the cone.

\(\displaystyle \text{Volume}=\frac{1}{3}\pi r^2 h\)

\(\displaystyle \text{Volume}=\frac{1}{3}\pi \times 6^2\times4=\frac{1}{3}\pi \times 36\times4=\frac{144}{3}\pi=48\pi\:m^3\)

The volume of a cylinder is found by multiplying the area of one end of the cylinder (base) by its height or:

 \(\displaystyle Volume=\pi r^2h\)

where \(\displaystyle r\) is the radius of the circular end of the cylinder and \(\displaystyle h\) is the height of the cylinder. So we can write:

\(\displaystyle Volume=\pi r^2h=\pi \times 3^2\times 3=84.82in^3\)

The surface area of the cylinder is given by:

\(\displaystyle A=2\pi r^2+2\pi rh\)

where \(\displaystyle A\) is the surface area of the cylinder, \(\displaystyle r\) is the radius of the cylinder and \(\displaystyle h\) is the height of the cylinder. So we can write:

\(\displaystyle A=2\pi r^2+2\pi rh\)

\(\displaystyle A=2\pi (3)^2+2\pi \times 3\times 3\)

\(\displaystyle A=18\pi+18\pi\)

\(\displaystyle A=36\pi\)

\(\displaystyle A=113.10\)