Question

find the area of the given triangle.$a = \frac{1}{2} \times base \times heightLXB0a = \frac{1}{2}(20h)LXB140 = \frac{1}{2} \times 16 \times hLXB25 = h$the height of the triangle is 5 m.for exercises 1-4, find the area of the given triangle. show your work.1.2.3.4.for exercises 5-6, find the missing dimension of the given triangle. show your work.5. $a = 54 \\, ft^2$6. $a = 40 \\, m^2$

Explanation:

Step1: Use triangle area formula

The area of a triangle is $A=\frac{1}{2} \times \text{base} \times \text{height}$
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Problem 1

Step1: Substitute base=12 ft, height=8 ft

$A=\frac{1}{2} \times 12 \times 8$

Step2: Calculate the product

$A=6 \times 8 = 48$
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Problem 2

Step1: Identify base and height

For right triangle, legs are base/height: base=8 in, height=15 in

Step2: Substitute into area formula

$A=\frac{1}{2} \times 8 \times 15$

Step3: Calculate the product

$A=4 \times 15 = 60$
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Problem 3

Step1: Substitute base=5 m, height=6 m

$A=\frac{1}{2} \times 5 \times 6$

Step2: Calculate the product

$A=\frac{1}{2} \times 30 = 15$
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Problem 4

Step1: Substitute base=12 cm, height=3 cm

$A=\frac{1}{2} \times 12 \times 3$

Step2: Calculate the product

$A=6 \times 3 = 18$
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Problem 5

Step1: Rearrange area formula for base

$\text{base}=\frac{2A}{\text{height}}$

Step2: Substitute $A=54\ \text{ft}^2$, height=6 ft

$\text{base}=\frac{2 \times 54}{6}$

Step3: Calculate the value

$\text{base}=\frac{108}{6}=18$
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Problem 6

Step1: Rearrange area formula for height

$\text{height}=\frac{2A}{\text{base}}$

Step2: Substitute $A=40\ \text{m}^2$, base=10 cm

$\text{height}=\frac{2 \times 40}{10}$

Step3: Calculate the value

$\text{height}=\frac{80}{10}=8$

Answer:

1. $48\ \text{ft}^2$
2. $60\ \text{in}^2$
3. $15\ \text{m}^2$
4. $18\ \text{cm}^2$
5. $18\ \text{ft}$
6. $8\ \text{cm}$