Question

Find the Area.
AND
6
6
9
First, find the area of the triangle.
D
6
6
9
Triangle Rectangle
[?] +
+
Half-circle
Remember:
A₁ = bh
=
Total Area
units²

Answer 1

Answer :

• Total area = 86.137 units^2 ( 3 d.p. )
• A(Δ) = 18 units^2
• A(rec) = 54 units^2
• A(half-circle) = 14.137 units^2 ( 3 d. p. )

Explanation :

area of a triangle is given by,

• A(t) = 1/2bh

here,

• b = 6
• h = 6
• A(t) = 1/2*6*6
• A(t) = 1/2*36
• A(t) = 18 units^2

area of a rectangle is given by,

• A(r) = bh

here,

• b = 9
• h = 6
• A(r) = 9*6
• A(r) = 54 units^2

area of a semi-circle is given by,

• A(h) = 1/2πr^2

here,

• r = 6/2
• r = 3
• A(h) = 1/2*π*(3)^2
• A(h) = 1/2*π*9
• A(h) = 14.137 units ^2 ( 3 d.p. )

therefore,the total area of the figure would be

• Total area = A(t) + A(r) + A(h) = (18 + 54 + 14.137) units^2
• Total area = 86.137 units^2 ( 3 d.p. )

Answer 2

Answer:

Triangle = 18 units²

Rectangle = 54 units²

Half-circle = 14.1 units² (nearest tenth)

Total Area = 86.1 units² (nearest tenth)

Step-by-step explanation:

Area of the triangle

The formula for the area of a triangle is:

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

Given:

• b = 6
• h = 6

Substitute the given values into the formula and solve for area:

$\textsf{Area of the triangle}=\dfrac{1}{2} \cdot 6 \cdot 6\\\\\\\textsf{Area of the triangle}=3 \cdot 6\\\\\\\textsf{Area of the triangle}=18\; \sf units^2$

Therefore, the area of the triangle is 18 units².

$\hrulefill$

Area of the rectangle

The formula for the area of a rectangle is:

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

Given:

• w = 6
• l = 9

Substitute the given values into the formula and solve for area:

$\textsf{Area of the rectangle}=6 \cdot 9\\\\\\\textsf{Area of the rectangle}=54\; \sf units^2$

Therefore, the area of the rectangle is 54 units².

$\hrulefill$

Area of the half-circle

The formula for the area of a half-circle is:

$\boxed{\begin{array}{l}\underline{\textsf{Area of a half-circle}}\\\\A=\dfrac{1}{2}\pi r^2\\\\\textsf{where:}\\\phantom{ww}\bullet\; \textsf{ A is the area.}\\ \phantom{ww}\bullet\;\textsf{ r is the radius.}\end{array}}$

The diameter of the half-circle is 6 units. As the radius of a circle is half its diameter, then the radius is r = 3.

Substitute r = 3 into the formula and solve for area:

$\textsf{Area of the half-circle}=\dfrac{1}{2}\cdot \pi \cdot 3^2\\\\\\\textsf{Area of the half-circle}=\dfrac{1}{2}\cdot \pi \cdot 9\\\\\\\textsf{Area of the half-circle}=\dfrac{9}{2}\pi\\\\\\\textsf{Area of the half-circle}=14.137166941...\\\\\\\textsf{Area of the half-circle}=14.1\; \sf units^2$

Therefore, the area of the semicircle is 14.1 units².

$\hrulefill$

Total Area

To find the total area of the composite shape, sum the individual areas:

$\textsf{Total Area} = 18+54+14.1\\\\\\\textsf{Total Area} = 86.1\;\sf units^2$

Therefore, the total area is 86.1 units² (rounded to the nearest tenth).