ometry hw – 10.1 pythagorean theorem and its converse
name:
date:
period:
find each missing side length in simplest radical form and as a decimal approximation rounded to the nearest tenth. use the ≈ symbol for approximations (answers that you have rounded).
recall the pythagorean triples:
pythagorean triples list
(3, 4, 5) (5, 12, 13) (7, 24, 25)
(8, 15, 17) (9, 40, 41) (11, 60, 61)
(12, 35, 37) (13, 84, 85) (15, 112, 113)
(16, 63, 65) (17, 144, 145) (19, 180, 181)
(20, 21, 29) (20, 99, 101) (21, 220, 221)
2. find a.
right triangle with legs 5, a and hypotenuse 13
a =
3. find b
right triangle with legs 6, b and hypotenuse 8
b =
4. find d.
rectangle with length 8, width 6, diagonal d
d =
5. find s.
right triangle with legs 24, 10 and hypotenuse s
s =
6. find c.
right triangle with leg 6, other leg (same as 6? maybe isoceles right triangle) and hypotenuse c
c =
7. find x
square with side x, diagonal √18
x =
8. find s.
right triangle with leg 1.5, hypotenuse 3.9 and leg s
s =
9. find x.
cone with slant height 41, radius 9 and height x
x =

Question

ometry hw – 10.1 pythagorean theorem and its converse
name:
date:
period:
find each missing side length in simplest radical form and as a decimal approximation rounded to the nearest tenth. use the ≈ symbol for approximations (answers that you have rounded).
recall the pythagorean triples:
pythagorean triples list
(3, 4, 5)  (5, 12, 13)  (7, 24, 25)
(8, 15, 17)  (9, 40, 41)  (11, 60, 61)
(12, 35, 37)  (13, 84, 85)  (15, 112, 113)
(16, 63, 65)  (17, 144, 145)  (19, 180, 181)
(20, 21, 29)  (20, 99, 101)  (21, 220, 221)
2. find a.
right triangle with legs 5, a and hypotenuse 13
a =
3. find b
right triangle with legs 6, b and hypotenuse 8
b =
4. find d.
rectangle with length 8, width 6, diagonal d
d =
5. find s.
right triangle with legs 24, 10 and hypotenuse s
s =
6. find c.
right triangle with leg 6, other leg (same as 6? maybe isoceles right triangle) and hypotenuse c
c =
7. find x
square with side x, diagonal √18
x =
8. find s.
right triangle with leg 1.5, hypotenuse 3.9 and leg s
s =
9. find x.
cone with slant height 41, radius 9 and height x
x =

Accepted Answer

### Problem 2: Find $ a $
# Explanation:
## Step 1: Apply Pythagorean Theorem
The triangle is right - angled, so by the Pythagorean theorem $ a^{2}+5^{2}=13^{2} $.
$ a^{2}+25 = 169 $
## Step 2: Solve for $ a^{2} $
Subtract 25 from both sides: $ a^{2}=169 - 25=144 $
## Step 3: Solve for $ a $
Take the square root of both sides: $ a=\sqrt{144} = 12 $ (since length can't be negative)
# Answer:
$ a = 12 $

### Problem 3: Find $ b $
# Explanation:
## Step 1: Apply Pythagorean Theorem
For the right - angled triangle, $ 6^{2}+b^{2}=8^{2} $
$ 36 + b^{2}=64 $
## Step 2: Solve for $ b^{2} $
Subtract 36 from both sides: $ b^{2}=64 - 36 = 28 $
## Step 3: Simplify the square root
$ b=\sqrt{28}=\sqrt{4\times7}=2\sqrt{7}\approx2\times2.6 = 5.2 $ (rounded to the nearest tenth)
# Answer:
In simplest radical form $ b = 2\sqrt{7} $, as a decimal approximation $ b\approx5.2 $

### Problem 4: Find $ d $
# Explanation:
## Step 1: Recognize the right triangle
The diagonal of a rectangle forms a right triangle with the length and width. The length is 8 and the width is 6. By the Pythagorean theorem $ d^{2}=8^{2}+6^{2} $
## Step 2: Calculate $ d^{2} $
$ d^{2}=64 + 36=100 $
## Step 3: Solve for $ d $
Take the square root: $ d=\sqrt{100}=10 $
# Answer:
$ d = 10 $

### Problem 5: Find $ s $
# Explanation:
## Step 1: Apply Pythagorean Theorem
The right - angled triangle has legs 24 and 10, and hypotenuse $ s $. So $ s^{2}=24^{2}+10^{2} $
## Step 2: Calculate $ s^{2} $
$ s^{2}=576+100 = 676 $
## Step 3: Solve for $ s $
Take the square root: $ s=\sqrt{676}=26 $
# Answer:
$ s = 26 $

### Problem 6: Find $ c $
(Assuming the other leg is also 6, since the triangle is isosceles right - angled? Wait, if it's a right - angled triangle with both legs 6)
# Explanation:
## Step 1: Apply Pythagorean Theorem
$ c^{2}=6^{2}+6^{2} $
$ c^{2}=36 + 36=72 $
## Step 2: Simplify the square root
$ c=\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\approx6\times1.4 = 8.4 $ (rounded to the nearest tenth)
# Answer:
In simplest radical form $ c = 6\sqrt{2} $, as a decimal approximation $ c\approx8.4 $

### Problem 7: Find $ x $
# Explanation:
## Step 1: Recognize the square
The figure is a square, so the diagonal of a square with side $ x $ is given by $ \text{diagonal}=x\sqrt{2} $. We know the diagonal is $ \sqrt{18} $
## Step 2: Set up the equation
$ x\sqrt{2}=\sqrt{18} $
## Step 3: Solve for $ x $
$ x=\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3 $
# Answer:
$ x = 3 $

### Problem 8: Find $ s $
# Explanation:
## Step 1: Apply Pythagorean Theorem
The right - angled triangle has one leg 1.5 and hypotenuse 3.9. Let the other leg be $ s $. So $ s^{2}+1.5^{2}=3.9^{2} $
## Step 2: Calculate $ s^{2} $
$ s^{2}=3.9^{2}-1.5^{2}=(3.9 + 1.5)(3.9 - 1.5) $ (using $ a^{2}-b^{2}=(a + b)(a - b) $)
$ s^{2}=(5.4)(2.4)=12.96 $
## Step 3: Solve for $ s $
Take the square root: $ s=\sqrt{12.96}=3.6 $
# Answer:
$ s = 3.6 $

### Problem 9: Find $ x $
(Assuming the cone has a slant height of 41 and radius of 9, and we need to find the height $ x $)
# Explanation:
## Step 1: Apply Pythagorean Theorem for cone
For a cone, the slant height $ l $, radius $ r $ and height $ x $ are related by $ l^{2}=r^{2}+x^{2} $. Here $ l = 41 $, $ r = 9 $
So $ x^{2}+9^{2}=41^{2} $
## Step 2: Calculate $ x^{2} $
$ x^{2}=41^{2}-9^{2}=(41 + 9)(41 - 9) $ (using $ a^{2}-b^{2}=(a + b)(a - b) $)
$ x^{2}=(50)(32)=1600 $
## Step 3: Solve for $ x $
Take the square root: $ x=\sqrt{1600}=40 $
# Answer:
$ x = 40 $

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QUESTION IMAGE

Question

Explanation:

Problem 2: Find \( a \)

Step 1: Apply Pythagorean Theorem

Step 2: Solve for \( a^{2} \)

Step 3: Solve for \( a \)

Step 1: Apply Pythagorean Theorem

Step 2: Solve for \( b^{2} \)

Step 3: Simplify the square root

Step 1: Recognize the right triangle

Step 2: Calculate \( d^{2} \)

Step 3: Solve for \( d \)

Answer:

Problem 3: Find \( b \)