Question - Calculating Missing Dimensions for Cones in a Worksheet
mathothercone dimensions calculationmissing dimensions worksheetcalculate cone volumegeometry formulas
Solution:
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:\[ r = 20/2 \]\[ r = 10 \] units3. 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 unitsSo, 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 units2. Area of the base for row 3: $$ 30.25\pi $$ square units3. Diameter for row 3: 11 units4. Volume for row 4: $$ 36\pi $$ cubic units
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com