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3 years ago

14

Calculate the area of triangle ABC with altitude CD, given A (6,0) B(1,5) C (2,0) and D (4,2)

2 answers:

Elina [12.6K]3 years ago

7 0

So hmm check the picture below

the height or altitude is CD
and the base is AB

how long are those? well $\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 6}}\quad ,&{{ 0}})\quad 
% (c,d)
B&({{ 1}}\quad ,&{{ 5}})\\\\
C&({{ 2}}\quad ,&{{ 0}})\quad 
% (c,d)
D&({{ 4}}\quad ,&{{ 2}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
AB=\sqrt{(1-6)^2+(5-0)^2}\qquad \qquad CD=\sqrt{(4-2)^2+(2-0)^2}$

kirill115 [55]3 years ago

3 0

Answer:

10 square units

Step-by-step explanation:

add the distances and round

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3 years ago

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3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

wel

Answer:

$z=\frac{15\sqrt{3}}{2}$

Step-by-step explanation:

In leftmost triangle:

opposite side =y

adjacent=a

now, we can use trig formula

$tan(60)=\frac{y}{a}$

now, we can solve for y

$y=atan(60)$

In rightmost triangle:

adjacent=b

opposite=y

a+b=15

b=15-a

now, we can use trig formula

$tan(30)=\frac{y}{15-a}$

now, we can solve for y

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now, we can set them equal

and then we can solve for a

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$\sqrt{3}a\cdot \:3=\frac{\sqrt{3}\left(15-a\right)}{3}\cdot \:3$

$4\sqrt{3}a=15\sqrt{3}$

$a=\frac{15}{4}$

now, we can find b

$b=15-\frac{15}{4}$

$b=\frac{45}{4}$

now, we can use trig formula

$cos(30)=\frac{b}{z}$

now, we can find z

$z=\frac{\frac{45}{4}}{cos(30)}$

we can simplify it

and we get

$z=\frac{15\sqrt{3}}{2}$

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3 years ago