Problem Analysis:
The task involves solving for various variables in different geometric figures, primarily triangles. The given information includes:
- \( AB = 30 \)
- \( AC = 25 \)
- \( BC = 17 \)
We need to solve for the variables in each part of the problem. Let's break it down step by step.
---
Step 1: Solve for the Variables in Each Figure
####
Figure 1:
-
Given: A right triangle with one leg as 12 and the hypotenuse as 13.
-
To Find: The other leg (\( x \)).
Using the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 12 \), \( b = x \), and \( c = 13 \).
\[
12^2 + x^2 = 13^2
\]
\[
144 + x^2 = 169
\]
\[
x^2 = 169 - 144
\]
\[
x^2 = 25
\]
\[
x = \sqrt{25} = 5
\]
Answer for Figure 1: \( x = 5 \)
---
####
Figure 2:
-
Given: A right triangle with one leg as 8 and the hypotenuse as 17.
-
To Find: The other leg (\( y \)).
Using the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 8 \), \( b = y \), and \( c = 17 \).
\[
8^2 + y^2 = 17^2
\]
\[
64 + y^2 = 289
\]
\[
y^2 = 289 - 64
\]
\[
y^2 = 225
\]
\[
y = \sqrt{225} = 15
\]
Answer for Figure 2: \( y = 15 \)
---
####
Figure 3:
-
Given: An isosceles right triangle with one leg as 7.
-
To Find: The hypotenuse (\( z \)).
In an isosceles right triangle, the hypotenuse is \( \sqrt{2} \) times the length of each leg. Here, each leg is 7.
\[
z = 7\sqrt{2}
\]
Answer for Figure 3: \( z = 7\sqrt{2} \)
---
####
Figure 4:
-
Given: A right triangle with one leg as 12 and the other leg as 5.
-
To Find: The hypotenuse (\( w \)).
Using the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 12 \), \( b = 5 \), and \( c = w \).
\[
12^2 + 5^2 = w^2
\]
\[
144 + 25 = w^2
\]
\[
w^2 = 169
\]
\[
w = \sqrt{169} = 13
\]
Answer for Figure 4: \( w = 13 \)
---
Step 2: Solve for Variables in the Second Set of Figures
####
Figure 5:
-
Given: A square with diagonal \( d \).
-
To Find: The side length (\( s \)).
In a square, the relationship between the side length \( s \) and the diagonal \( d \) is:
\[
d = s\sqrt{2}
\]
Rearranging for \( s \):
\[
s = \frac{d}{\sqrt{2}}
\]
Answer for Figure 5: \( s = \frac{d}{\sqrt{2}} \)
---
####
Figure 6:
-
Given: A triangle with base \( b \) and height \( h \).
-
To Find: The area (\( A \)).
The formula for the area of a triangle is:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
\[
A = \frac{1}{2} \times b \times h
\]
Answer for Figure 6: \( A = \frac{1}{2}bh \)
---
####
Figure 7:
-
Given: A rectangle with two equal sides.
-
To Find: The type of quadrilateral.
Since the rectangle has two equal sides, it is a
square.
Answer for Figure 7: Square
---
####
Figure 8:
-
Given: An equilateral triangle with side length \( s \).
-
To Find: The perimeter (\( P \)).
The perimeter of an equilateral triangle is:
\[
P = 3s
\]
Answer for Figure 8: \( P = 3s \)
---
Final Answers:
1. \( x = 5 \)
2. \( y = 15 \)
3. \( z = 7\sqrt{2} \)
4. \( w = 13 \)
5. \( s = \frac{d}{\sqrt{2}} \)
6. \( A = \frac{1}{2}bh \)
7. Square
8. \( P = 3s \)
\boxed{x = 5, y = 15, z = 7\sqrt{2}, w = 13, s = \frac{d}{\sqrt{2}}, A = \frac{1}{2}bh, \text{Square}, P = 3s}
Parent Tip: Review the logic above to help your child master the concept of special right triangles 45 45 90 worksheet answers.
