Question

emma designed two triangular sails for a boat. answer parts a-c belowa. what is the area of sail a?the area of sail a is $\boldsymbol{36} \text{m}^2$.(type an integer or decimal rounded to two decimal places as needed.)b. what is the area of sail b?the area of sail b is $\boldsymbol{square}$ $\boldsymbol{
abla}$(type an integer or decimal rounded to two decimal places as needed.)(unit options: $\text{m}^3$, $\text{m}^2$, m)

Explanation:

Step1: Identify Sail A's legs

Sail A is a right isosceles triangle (45-45-90). Hypotenuse = 12 m. For such triangles, leg length $l = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$ m.

Step2: Calculate area of Sail A

Area of right triangle: $\frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$.
$\text{Area}_A = \frac{1}{2} \times 6\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 72 = 36$ $\text{m}^2$

Step3: Identify Sail B's legs

Sail B is a 30-60-90 right triangle, hypotenuse = 8.7 m. Shorter leg (opposite 30°): $s = \frac{8.7}{2} = 4.35$ m. Longer leg: $l = s\sqrt{3} = 4.35\sqrt{3}$ m.

Step4: Calculate area of Sail B

$\text{Area}_B = \frac{1}{2} \times 4.35 \times 4.35\sqrt{3}$
$\text{Area}_B = \frac{1}{2} \times 4.35^2 \times 1.732 \approx \frac{1}{2} \times 18.9225 \times 1.732 \approx 16.47$ $\text{m}^2$

Answer:

a. $36$ $\text{m}^2$
b. $16.47$ $\text{m}^2$