Question

What is the area of the composite shape. Use 3.14 for n and round to the nearest hundredth. Show all your work
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Answer 1

area of triangle

• = 1/2bh
• = 1/2*3*2
• = 3 units

area of rectangle

• = bh
• = 2*3
• = 6 units

area of regular pentagon

• = 1/2*p*a
• = 1/2*5*3*2
• = 15 units

area of sector

• = θ/360° x πr^2
• = 66°/360° x 3.14*3*3
• = 5.181 units

thus, the total area would be

• = 3 + 6 + 15 + 5.181
• = 29.181 units

Answer 2

Answer:

• Area of triangle:  3 square units
• Area of rectangle:  6 square units
• Area of regular polygon:  15 square units
• Area of sector:  5.18 square units
• Total:  29.18 square units

Step-by-step explanation:

Area of the triangle

The area of any triangle is half the product of its base and height:

$\boxed{\begin{array}{l}\underline{\textsf{Area of a triangle}}\\\\A=\dfrac{1}{2}bh\\\\\textsf{where:}\\\phantom{ww}\bullet\; \textsf{ A is the area.}\\ \phantom{ww}\bullet\;\textsf{ b is the base.}\\ \phantom{ww}\bullet\;\textsf{ h is the height.}\end{array}}$

In this case:

• b = 3
• h = 2

Substitute the values of b and h into the area equation and solve for A:

$A=\dfrac{1}{2} \cdot 3 \cdot 2\\\\\\A=\dfrac{1}{2}\cdot 6\\\\\\A=3\; \sf square\;units$

Therefore, the area of the triangle is:

$\Large\boxed{\boxed{\textsf{Area of triangle:}\quad 3\;\sf units^2}}$

$\dotfill$

Area of the rectangle

The area of any rectangle is the product of its width and length:

$\boxed{\begin{array}{l}\underline{\textsf{Area of a rectangle}}\\\\A=wl\\\\\textsf{where:}\\\phantom{ww}\bullet\; \textsf{ A is the area.}\\ \phantom{ww}\bullet\;\textsf{ w is the width.}\\ \phantom{ww}\bullet\;\textsf{ l is the length.}\end{array}}$

In this case:

• w = 2
• l = 3

Substitute the values of w and l into the area equation and solve for A:

$A=2 \cdot 3\\\\A=6\; \sf square\;units$

Therefore, the area of the rectangle is:

$\Large\boxed{\boxed{\textsf{Area of rectangle:}\quad 6\;\sf units^2}}$

$\dotfill$

Area of the regular polygon

The formula for the area of a regular polygon is:

$\boxed{\begin{array}{l}\underline{\textsf{Area of a regular polygon}}\\\\A=\dfrac{n\cdot s\cdot a}{2}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ n is the number of sides.}\\ \phantom{ww}\bullet\;\textsf{ s is the length of one side.}\\ \phantom{ww}\bullet\;\textsf{ a is the apothem.}\end{array}}$

In this case:

• n = 5
• s = 3
• a = 2

Substitute the values of n, s and a into the area equation and solve for A:

$A=\dfrac{5 \cdot 3 \cdot 2}{2}\\\\\\A=\dfrac{30}{2}\\\\\\A=15\; \sf square\;units$

Therefore, the area of the regular polygon is:

$\Large\boxed{\boxed{\textsf{Area of regular polygon:}\quad 15\;\sf units^2}}$

$\dotfill$

Area of the sector

The formula for the area of a sector is:

$\boxed{\begin{array}{l}\underline{\textsf{Area of a sector}}\\\\A=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{ r is the radius.}\\\phantom{ww}\bullet\;\;\textsf{ \theta is the angle measured in degrees.}\end{array}}$

In this case:

• θ = 66°
• r = 3
• π = 3.14

Substitute the values of θ r and π into the area equation and solve for A:

$A=\left(\dfrac{66^{\circ}}{360^{\circ}}\right) \cdot 3.14 \cdot 3^2\\\\\\A=\dfrac{11}{60}\cdot 3.14 \cdot 9\\\\\\A=\dfrac{11}{60}\cdot28.26\\\\\\A=5.181\\\\\\A=5.18\; \sf square\;units$

Therefore, the area of the sector is:

$\Large\boxed{\boxed{\textsf{Area of sector:}\quad 5.18\;\sf units^2}}$

$\dotfill$

Total area

To calculate the total area, sum the individual areas:

$\textsf{Total area}=3+6+15+5.18\\\\\textsf{Total area}=9+15+5.18\\\\\textsf{Total area}=24+5.18\\\\\textsf{Total area}=29.18\; \sf square\;units$

Therefore, the total area of the composite shape is:

$\Large\boxed{\boxed{\textsf{Total:}\quad 29.18\;\sf units^2}}$