To solve the problem of finding the area of the composite shapes, we will break each shape into simpler geometric figures and calculate their areas step by step. Let's go through each question:
---
Problem 1:
The shape is a square with a triangle removed from it. We need to find the area of the white space.
#### Step 1: Calculate the area of the square.
The side length of the square is given as 12 meters.
\[
\text{Area of the square} = \text{side}^2 = 12 \times 12 = 144 \, \text{m}^2
\]
#### Step 2: Calculate the area of the triangle.
The triangle has a base of 12 meters and a height of 8 meters.
\[
\text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 8 = 48 \, \text{m}^2
\]
#### Step 3: Subtract the area of the triangle from the area of the square.
\[
\text{Area of the white space} = \text{Area of the square} - \text{Area of the triangle} = 144 - 48 = 96 \, \text{m}^2
\]
#### Final Answer for Problem 1:
\[
\boxed{96}
\]
---
Problem 2:
The shape is a combination of two rectangles. We need to find the total area.
#### Step 1: Identify the dimensions of the two rectangles.
- The larger rectangle has dimensions 15 cm (height) and 10 cm (width).
- The smaller rectangle has dimensions 5 cm (height) and 12 cm (width).
#### Step 2: Calculate the area of the larger rectangle.
\[
\text{Area of the larger rectangle} = \text{length} \times \text{width} = 15 \times 10 = 150 \, \text{cm}^2
\]
#### Step 3: Calculate the area of the smaller rectangle.
\[
\text{Area of the smaller rectangle} = \text{length} \times \text{width} = 5 \times 12 = 60 \, \text{cm}^2
\]
#### Step 4: Add the areas of the two rectangles.
\[
\text{Total area} = \text{Area of the larger rectangle} + \text{Area of the smaller rectangle} = 150 + 60 = 210 \, \text{cm}^2
\]
#### Final Answer for Problem 2:
\[
\boxed{210}
\]
---
Problem 3:
The shape is a combination of two rectangles. We need to find the total area.
#### Step 1: Identify the dimensions of the two rectangles.
- The larger rectangle has dimensions 9 meters (height) and 11 meters (width).
- The smaller rectangle has dimensions 3 meters (height) and 2 meters (width).
#### Step 2: Calculate the area of the larger rectangle.
\[
\text{Area of the larger rectangle} = \text{length} \times \text{width} = 9 \times 11 = 99 \, \text{m}^2
\]
#### Step 3: Calculate the area of the smaller rectangle.
\[
\text{Area of the smaller rectangle} = \text{length} \times \text{width} = 3 \times 2 = 6 \, \text{m}^2
\]
#### Step 4: Add the areas of the two rectangles.
\[
\text{Total area} = \text{Area of the larger rectangle} + \text{Area of the smaller rectangle} = 99 + 6 = 105 \, \text{m}^2
\]
#### Final Answer for Problem 3:
\[
\boxed{105}
\]
---
Final Answers:
1. \(\boxed{96}\)
2. \(\boxed{210}\)
3. \(\boxed{105}\)
Parent Tip: Review the logic above to help your child master the concept of area and perimeter worksheet 6th grade.