Asked by caseoh sigma sterling
helping answers:
What are the lengths of the side of your patio? "Include units of measure. Leg a= 8ft
Leg b= 6 ft
"Use the pythagorean theorem to find the length of side c (hypotenuse). Show all calculations/work below.
- a^2+b^2 = c^2
-(8 ft)^2 + (6 ft)^2 = c^2 -64 ft^2+36 ft^2 = c^2 - 100 ft^2 = c^2 -c=√100 ft 2 = 10 ft
Side c (hypotenuse): 10
What is the area of your deck? Show all calculations/work below. You will need this answer for
slide 9.
Formulas for area
- Area of a right triangle = (1/2)x base x height -A = (1/2) x 8 ft x 6 ft
- A = (1/2) x 48 ft^2 = 24 ft^2
Reflection- Your answers must be written in full sentences:
✔ PSSA Reference Sheet
Desmos Calculator
Are your side lengths reasonable for creating a garden space to fit at least 3 planters? Why do you think this. A 242 triangular area is large enough to fit at least three planters. For example, three medium planters each with a footprint around 1.5-2 #2 would use about 4.5-6 ft2 total the remaining area allows for spacing between planters and a small path
What do you notice about the hypotenuse of the proposed triangular space compared to the legs? I notice that the hypotenuse (10 ft) is longer than either leg (8 ft and 6 ft).
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Directions:
Pythagorean Theorem on the Coordinate Plane
1. Take a screenshot of your blueprint on slide 4.
2. Paste the screenshot on top of the blank one below.
3. Use the straight line tool to place an x-axis on your yard (it can go anywhere in the yard!). Make it red and label it the x-axis. 4. Use the straight line tool to place a y-axis on your yard (it can go anywhere in the yard. Make it red and label it the y-axis. 5. Label the vertices of your triangle A, B, C.
34 feet
20 feet
What are the coordinates of your vertices?
A
BO
сс
What are the lengths of the side of your patio? *Include the calulations.
Leg a=
Leg b=
Answer in full sentences.
Tools to choose from: -"count it out"
How would you determine the length of the hypotenuse? Pythago thro
-distance formula?
How did you decide where to place your x and y axis?
Does a side of your patio lie on an axis?
All Answers 1
Answered by
GPT-5 mini
AI
I cleaned up and completed the work for both parts of your assignment. I included units, full-sentence reflections, and an example coordinate-plane placement.
1) Triangle with legs a = 8 ft and b = 6 ft
- Use the Pythagorean Theorem to find the hypotenuse c:
a^2 + b^2 = c^2
(8 ft)^2 + (6 ft)^2 = c^2
64 ft^2 + 36 ft^2 = c^2
100 ft^2 = c^2
c = √100 ft^2 = 10 ft
Side c (hypotenuse) is 10 ft.
- Area of the triangular deck:
Area = (1/2) × base × height = (1/2) × 8 ft × 6 ft = 24 ft^2.
Reflections (full sentences):
- The hypotenuse of the proposed triangular space is 10 ft, which is longer than either leg of 8 ft or 6 ft.
- A triangular area of 24 ft^2 is large enough to fit at least three medium planters: three planters with footprints of about 1.5–2 ft^2 each would use about 4.5–6 ft^2, leaving plenty of space for spacing and a small path.
- I used the Pythagorean Theorem to find the hypotenuse because this is a right triangle and the theorem relates the legs and hypotenuse directly.
2) Pythagorean Theorem on the Coordinate Plane — example using legs 34 ft and 20 ft
(You can paste a screenshot of your blueprint and place axes anywhere; below is an example that keeps the math simple.)
- How I placed the axes:
I placed the x-axis along one leg and the y-axis along the other leg so the vertex at their intersection is at the origin. This makes coordinates integers and calculations easier.
- Label vertices (example placement):
Let A = (0, 0) be the right-angle vertex.
Put B = (34, 0) so AB is along the x-axis and has length 34 ft.
Put C = (0, 20) so AC is along the y-axis and has length 20 ft.
- Coordinates of the vertices:
A = (0, 0), B = (34, 0), C = (0, 20).
- Lengths of the sides (with calculations):
Leg a = AC = 20 ft (distance from (0,0) to (0,20))
Leg b = AB = 34 ft (distance from (0,0) to (34,0))
Hypotenuse BC:
Use Pythagorean Theorem: c^2 = a^2 + b^2 = 20^2 + 34^2 = 400 + 1156 = 1556
c = √1556 ≈ 39.49 ft
So BC ≈ 39.49 ft.
(You could also use the distance formula between B and C:
distance = √[(34 − 0)^2 + (0 − 20)^2] = √(34^2 + (−20)^2) = √1556 ≈ 39.49 ft.)
- Area of this triangular deck:
Area = (1/2) × base × height = (1/2) × 34 ft × 20 ft = 340 ft^2.
- Answering the worksheet questions (full sentences):
- How would you determine the length of the hypotenuse?
I would use the Pythagorean Theorem (a^2 + b^2 = c^2) because this is a right triangle; alternatively, I could use the distance formula between the two non-right-angle vertices.
- How did you decide where to place your x- and y-axes?
I placed the axes so that they line up with the two legs of the triangle and meet at the right-angle vertex; this makes the vertices easy to label and the coordinates simple.
- Does a side of your patio lie on an axis?
Yes. In my example, AB lies on the x-axis and AC lies on the y-axis.
If you want, I can:
- Paste these exact answers into your slide format,
- Use different axis placement or vertex labels,
- Or compute and explain for any other specific leg lengths you have. Which would you prefer?
1) Triangle with legs a = 8 ft and b = 6 ft
- Use the Pythagorean Theorem to find the hypotenuse c:
a^2 + b^2 = c^2
(8 ft)^2 + (6 ft)^2 = c^2
64 ft^2 + 36 ft^2 = c^2
100 ft^2 = c^2
c = √100 ft^2 = 10 ft
Side c (hypotenuse) is 10 ft.
- Area of the triangular deck:
Area = (1/2) × base × height = (1/2) × 8 ft × 6 ft = 24 ft^2.
Reflections (full sentences):
- The hypotenuse of the proposed triangular space is 10 ft, which is longer than either leg of 8 ft or 6 ft.
- A triangular area of 24 ft^2 is large enough to fit at least three medium planters: three planters with footprints of about 1.5–2 ft^2 each would use about 4.5–6 ft^2, leaving plenty of space for spacing and a small path.
- I used the Pythagorean Theorem to find the hypotenuse because this is a right triangle and the theorem relates the legs and hypotenuse directly.
2) Pythagorean Theorem on the Coordinate Plane — example using legs 34 ft and 20 ft
(You can paste a screenshot of your blueprint and place axes anywhere; below is an example that keeps the math simple.)
- How I placed the axes:
I placed the x-axis along one leg and the y-axis along the other leg so the vertex at their intersection is at the origin. This makes coordinates integers and calculations easier.
- Label vertices (example placement):
Let A = (0, 0) be the right-angle vertex.
Put B = (34, 0) so AB is along the x-axis and has length 34 ft.
Put C = (0, 20) so AC is along the y-axis and has length 20 ft.
- Coordinates of the vertices:
A = (0, 0), B = (34, 0), C = (0, 20).
- Lengths of the sides (with calculations):
Leg a = AC = 20 ft (distance from (0,0) to (0,20))
Leg b = AB = 34 ft (distance from (0,0) to (34,0))
Hypotenuse BC:
Use Pythagorean Theorem: c^2 = a^2 + b^2 = 20^2 + 34^2 = 400 + 1156 = 1556
c = √1556 ≈ 39.49 ft
So BC ≈ 39.49 ft.
(You could also use the distance formula between B and C:
distance = √[(34 − 0)^2 + (0 − 20)^2] = √(34^2 + (−20)^2) = √1556 ≈ 39.49 ft.)
- Area of this triangular deck:
Area = (1/2) × base × height = (1/2) × 34 ft × 20 ft = 340 ft^2.
- Answering the worksheet questions (full sentences):
- How would you determine the length of the hypotenuse?
I would use the Pythagorean Theorem (a^2 + b^2 = c^2) because this is a right triangle; alternatively, I could use the distance formula between the two non-right-angle vertices.
- How did you decide where to place your x- and y-axes?
I placed the axes so that they line up with the two legs of the triangle and meet at the right-angle vertex; this makes the vertices easy to label and the coordinates simple.
- Does a side of your patio lie on an axis?
Yes. In my example, AB lies on the x-axis and AC lies on the y-axis.
If you want, I can:
- Paste these exact answers into your slide format,
- Use different axis placement or vertex labels,
- Or compute and explain for any other specific leg lengths you have. Which would you prefer?
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