What is the area, in square units, of triangle ABC in the figure shown if points A, B, C and D are coplanar, angle D is a right angle, AC = 13, AB = 15 and DC = 5?

Guest Jan 20, 2019

#1**+6 **

Use Pythagorean Theorem to find DA. So 13^2-5^2=144, and the square root of 144 is 12. So DA must be 12.

Notice that AB has the length of 15 and DA has the length of twelve. Since some right triangles have a ratio of 3:4:5, side DB must have the length of 9

Since DC+BC is 9, and DC is 5, BC is 4.

So now we know the three sides of ABC, which is 13, 15, and 4. We can use Heron's formula to figure out the area.

Heron's formula is:

A=sqrt((S)(S-A)(S-B)(S-C))

S=(a+b+c)/2

We plug it sides into the 2nd equation, S=(13+15+4)/2, and we get S=16.

Then plug your answers into the first equation. A=sqrt((16)(16-15)(16-13)(16-4)), and we get A=24.

So the area of ABC is 24

CalculatorUser Jan 21, 2019

#2**+8 **

**What is the area, in square units, of triangle ABC in the figure shown if points A, B, C and D are coplanar, angle D is a right angle, AC = 13, AB = 15 and DC = 5?**

**1.**

\(\begin{array}{|rcll|} \hline DA^2+DC^2 &=& AC^2 \quad & | \quad DC=5,~ AC = 13 \\ DA^2+5^2&=& 13^2 \\ DA^2 &=& 13^2-5^2 \\ DA^2 &=& 144 \\ DA^2 &=& 12^2 \\ \mathbf{DA} &\mathbf{=}& \mathbf{12} \\ \hline \end{array} \)

**2.**

\(\begin{array}{|rcll|} \hline BD^2+DA^2 &=& AB^2 \quad & | \quad DA=12,~ AB = 15 \\ BD^2+12^2&=& 15^2 \\ BD^2 &=& 15^2-12^2 \\ BD^2 &=& 81 \\ BD^2 &=& 9^2 \\ \mathbf{BD} &\mathbf{=}& \mathbf{9} \\ \hline \end{array}\)

**3.**

\(\begin{array}{|rcll|} \hline BC &=& BD-DC \quad & | \quad BD=9,~ DC = 5 \\ BC &=& 9-5 \\ \mathbf{BC} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

**4.**

**\(\begin{array}{|rcll|} \hline A &=& \dfrac{BC\cdot DA}{2} \quad & | \quad BC=4,~ DA = 12 \\ A &=& \dfrac{4\cdot 12}{2} \\ A &=& 2\cdot 12 \\ \mathbf{A} &\mathbf{=}& \mathbf{24} \\ \hline \end{array}\)**

The area of triangle ABC is **24**

heureka Jan 21, 2019