Question

question 12
calculate the area of triangle abc with altitude cd, given a(-3, -4), b(-6, 2), c(0, 0), and d(-4, -2).
14 square units
15 square units
18 square units
20 square units
question 13
towers a and b are located 1.7 miles apart. a cell phone user is 5.3 miles from tower a.
if ( x = 82.1 ), what is the distance between tower b, and the cell phone user? round your answer to the nearest tenth of a mile.
4.7 miles
5.6 miles
6.3 miles
6.4 miles

Explanation:

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Question 12

Step1: Find length of base AB

Use distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$AB=\sqrt{(-6-(-3))^2+(2-(-4))^2}=\sqrt{(-3)^2+(6)^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}$

Step2: Find length of altitude CD

Use distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$CD=\sqrt{(-4-0)^2+(-2-0)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$

Step3: Calculate triangle area

Area formula: $\frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area}=\frac{1}{2} \times 3\sqrt{5} \times 2\sqrt{5}=\frac{1}{2} \times 6 \times 5=15$

Step1: Apply Law of Cosines

Let $a=1.7$, $b=5.3$, $C=82.1^\circ$, find side $c$.
Law of Cosines: $c^2=a^2+b^2-2ab\cos(C)$

Step2: Substitute values into formula

$c^2=(1.7)^2+(5.3)^2-2(1.7)(5.3)\cos(82.1^\circ)$
$c^2=2.89+28.09-18.02\times0.137$
$c^2=30.98-2.47=28.51$

Step3: Solve for c

$c=\sqrt{28.51}\approx5.3$

Answer:

15 square units

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Question 13