A quarter is 1/5 of an inch thick and 5 quarters weigh one ounce. Stephanie built a toothpick bridge that could safely hold 3 po
1 answer:
Answer:
Maximum height of quarter tower = 9.6 inches.
Step-by-step explanation:
The thickness of a quarter = inches
Weight of a quarter = 1 ounce
The maximum weight that could safely hold = 3 pounds.
<u>The conversion factor for ounce to pound is 1 ounce = 0.0625 pound</u>
The number of quarter used in the tower = no of ounces equivalent to 3 pounds.
No of ounces equivalent to 3 pounds =
No of ounces equivalent to 3 pounds = 48
So the height of quarter tower = 48 \times inches
= 9.6 inches.
You might be interested in
The area of the shaded region /ACDEH/ is 325.64cm²
Step 1 - Collect all the facts
First, let's examine all that we know.
- We know that the octagon is regular which means all sides are equal.
- since all sides are equal, then all sides are equal 10cm.
- if all sides are equal then all angles within it are equal.
- since the total angle in an octagon is 1080°, the sum of each angle within the octagon is 135°.
Please note that the shaded region comprises a rectangle /ADEH/ and a scalene triangle /ACD/.
So to get the area of the entire region, we have to solve for the area of the Scalene Triangle /ACD/ and add that to the area of the rectangle /ADEH/
Step 2 - Solving for /ACD/
The formula for the area of a Scalene Triangle is given as:
A =
This formula assumes that we have all the sides. But we don't yet.
However, we know the side /CD/ is 10cm. Recall that side /CD/ is one of the sides of the octagon ABCDEFGH.
This is not enough. To get sides /AC/ and /AD/ of Δ ACD, we have to turn to another triangle - Triangle ABC. Fortunately, ΔABC is an Isosceles triangle.
Step 3 - Solving for side AC.
Since all the angles in the octagon are equal, ∠ABC = 135°.
Recall that the total angle in a triangle is 180°. Since Δ ABC is an Isosceles triangle, sides /AB/ and /BC/ are equal.
Recall that the Base angles of an isosceles triangle is always equal. That is ∠BCA and ∠BAC are equal. To get that we say:
180° - 135° = 45° [This is the sum total of ∠BCA and ∠BAC. Each angle therefore equals
45°/2 = 22.5°
Now that we know all the angles of Δ ABC and two sides /AB/ and /BC/, let's try to solve for /AC/ which is one of the sides of Δ ACD.
According to the Sine rule,
=
=
Since we know side /BC/, let's go with the first two parts of the equation.
That gives us
Cross multiplying the above, we get
/AC/ =
Side /AC/ = 18.48cm.
Returning to our Scalene Triangle, we now have /AC/ and /CD/.
To get /AD/ we can also use the Sine rule since we can now derive the angles in Δ ABC.
From the Octagon the total angle inside /HAB/ is 135°. We know that ∠HAB comprises ∠CAB which is 22.5°, ∠HAD which is 90°. Therefore, ∠DAC = 135° - (22.5+90)
∠DAC = 22.5°
Using the same deductive principle, we can obtain all the other angles within Δ ACD, with ∠CDA = 45° and ∠112.5°.
Now that we have two sides of ΔACD and all its angles, let's solve for side /AD/ using the Sine rule.
=
Cross multiplying we have:
/AD/ =
Therefore, /AD/ = 24.15cm.
Step 4 - Solving for Area of ΔACD
Now that we have all the sides of ΔACD, let's solve for its area.
Recall that the area of a Scalene Triangle using Heron's formula is given as
A =
Where S is the semi-perimeter given as
S= (/AC/ + /CD/ + /DA/)/2
We are using this formula because we don't have the height for ΔACD but we have all the sides.
Step 5 - Solving for Semi Perimeter
S = (18.48 + 10 + 24.15)/2
S = 26.32
Therefore, Area =
A =
A = Square cm.
A of ΔACD = 85.49cm²
Recall that the shape consists of the rectangle /ADEH/.
The A of a rectangle is L x B
A of /ADEH/ = 240.15cm²
Step 6 - Solving for total Area of the shaded region of the Octagon
The total area of the Shaded region /ACDEH/, therefore, is 240.15 + 85.49
= 325.64cm²
See the link below for more about Octagons:
brainly.com/question/4515567
I hope this helps, let me know if it doesn't make sense
