Learning Task 1

Directions: Solve the following problems. Show your complete solutions and

answers on a piece of paper.

1. Manny painted a 4 sided figure inside a circle whose radius measures 30

cm. Each side of the 4 sided figure he painted measures 13 cm. What is the

figure that Manny painted? What is the area of the circle that is not painted

2. The figure below is composed of a rectangle and a half circle. Find the

area of the figure. (Figure is not drawn to scale)

2cm

3 cm

2 cm

9 cm

3. Calculate the shaded region. Show only your solution and answer

8.7 cm

10 cm

## Learning Task 1

- 2,657 cm²
- 69.28 cm²
- 9.26 cm²

### Solutions:

1. **Given: **circle with r = 30 cm

4 – sided figure with s = 13 cm

**First, find the area of the circle.**

A = πr²

A = (3.14)(30 cm)²

A = (3.14)(900 cm²)

**A = 2,826 cm²**

**Therefore, the area of the circle is 2,826 cm².**

**Next, find the area of the 4 – sided figure.**

A = s²

A = (13 cm)²

**A = 169 cm²**

**Subtract the area of the 4 – sided figure from the area of the circle.**

Area of the Circle – Area of the 4 – sided figure

2,826 cm² – 169 cm² = **2,657 cm²**

**Therefore, the area of the circle that was not painted is 2, 657 cm².**

2. **Given:** rectangle with dimensions 9 cm x 7 cm

semi – circle with diameter 9 cm – (2 cm + 3 cm)

**Solve first for the area of the rectangle.**

A = l x w

A = 9 cm x 7 cm

**A = 63 cm²**

**Therefore, the area of the rectangle is 63 cm².**

**Next, solve for the area of the semi – circle.**

A = [tex]\frac{1}{2}[/tex]πr²

A = [tex]\frac{1}{2}[/tex](3.14)[__9cm – (2cm + 3 cm)__]²

2

A = [tex]\frac{1}{2}[/tex](3.14)__[(9cm – 5 cm)]__²

2

A = [tex]\frac{1}{2}[/tex] (3.14)__(4cm)__²

2

A = [tex]\frac{1}{2}[/tex](3.14)(2cm)²

A = [tex]\frac{1}{2}[/tex](3.14)(4 cm²)

A = [tex]\frac{1}{2}[/tex](12.56 cm²)

A = [tex]\frac{12.56}{2}[/tex] cm²

**A = 6.28 cm²**

**Therefore, the area of the semi – circle is 6.28 cm².**

**Add the areas of the two figures.**

Area of the rectangle + Area of the semi – circle

63 cm² + 6.28 cm² = **69.28 cm²**

**Therefore, the area of the given figure is 69.28 cm².**

3. **Given: **scalene triangle with sides 8.7 cm and 10 cm

square with dimensions 2 cm x 2 cm

**First, find the area of the triangle **

A = [tex]\frac{1}{2}[/tex](8.7 cm)(10 cm)(sin 60°) assuming that the triangle is equiangular

A = [tex]\frac{1}{2}[/tex](8.7 cm)(10 cm)(0.3048106211)

A = [tex]\frac{1}{2}[/tex](87 cm²)(0.3048106211)

A = [tex]\frac{1}{2}[/tex](26.52 cm²)

**A = 13.26 cm²**

**Next, find the area of the square.**

A = s²

A = (2 cm)²

**A = 4 cm²**

**Subtract the area of the square from the area of the scalene triangle.**

Area of the scalene triangle – Area of the square

13.26 cm² – 4 cm² = **9.26 cm²**

**Therefore, the area of the shaded region is 9.26 cm².**

**What are the area formulas: https://brainly.ph/question/1836692**

** #Let’sStudy**