Example Question - geometry formulas
Here are examples of questions we've helped users solve.
Calculating Missing Dimensions for Cones in a Worksheet
The image shows a worksheet with a cone and a table where we need to fill in the missing dimensions. The volume of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Let's fill in the missing dimensions for each row: 1. For the first row with missing height: We have the Diameter (8 units), Radius (4 units), Area of the Base \( (16\pi \) square units), and Volume of Cone \( (16\pi \) cubic units). The volume is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] Now, we solve for \( h \): \[ 16\pi = \frac{1}{3} \pi \times 4^2 \times h \] \[ 16\pi = \frac{1}{3} \pi \times 16 \times h \] \[ 16\pi = \frac{16}{3} \pi \times h \] \[ h = 16\pi / (\frac{16}{3} \pi) \] \[ h = 3 \] So, the missing height is 3 units. 2. For the second row with missing radius: We have the Diameter (20 units), so the Radius is half of the diameter: \[ r = 20/2 \] \[ r = 10 \] units 3. For the third row with missing diameter and area of the base: We have the Radius \( (5 \frac{1}{2} \) units) and the Volume of Cone \( (200\pi \) cubic units), but no height. First, let’s find the height. We can rearrange the volume formula to solve for \( h \): \[ 200\pi = \frac{1}{3} \pi \times (5 \frac{1}{2})^2 \times h \] \[ 200\pi = \frac{1}{3} \pi \times \frac{121}{4} \times h \] \[ 200\pi = \frac{121}{12} \pi \times h \] \[ h = \frac{200\pi}{\frac{121}{12}\pi} \] \[ h = \frac{200 \times 12}{121} \] \[ h = \frac{2400}{121} \] \[ h = 19.83 \] (approx.) Now, let’s find the Diameter and the area of the base: Since the Radius is \( 5 \frac{1}{2} \), the Diameter is: \[ d = 2r \] \[ d = 2 \times 5 \frac{1}{2} \] \[ d = 2 \times \frac{11}{2} \] \[ d = 11 \] The area of the base is: \[ A = \pi r^2 \] \[ A = \pi \times (5 \frac{1}{2})^2 \] \[ A = \pi \times \frac{121}{4} \] \[ A = 30.25\pi \] square units So, the missing diameter is 11 units, and the area of the base is \( 30.25\pi \) square units. 4. For the fourth row with missing volume: We have the Radius (3 units), the Area of the Base \( (9\pi \) square units), and the Height (\( 12 \) units). The volume is calculated as follows: \[ V = \frac{1}{3} \pi r^2 h \] \[ V = \frac{1}{3} \pi \times 3^2 \times 12 \] \[ V = \frac{1}{3} \pi \times 9 \times 12 \] \[ V = 36\pi \] So, the missing volume is \( 36\pi \) cubic units. Fill the table with these missing dimensions: 1. Height: 3 units 2. Area of the base for row 3: \( 30.25\pi \) square units 3. Diameter for row 3: 11 units 4. Volume for row 4: \( 36\pi \) cubic units
Parallelogram Base and Height Calculation
The image shows a diagram of a parallelogram with a base labeled as 15 units and a diagonal labeled as 17 units. There is also an altitude drawn from the top vertex opposite the base downward to the base, forming a right triangle with the base and the diagonal, with the altitude labeled as 10 units. The question asks for the base and height of the parallelogram. The base is clearly labeled as 15 units. The height is the length of the altitude, which is given as 10 units. Therefore: - base = 15 units - height = 10 units
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com