Question

Find the area of the triangle Can you please show me step by step so I understand?

Answer 1

To solve this question we will use the following diagram:

Notice that triangles ABD and BCD are right triangles.

Using the Pythagorean theorem we get:

`\begin{gathered} x^2+7^2=11^2, \\ x^2+y^2=9^2\text{.} \end{gathered}`

Simplifying the above equations we get:

`\begin{gathered} x^2+49=121, \\ x^2+y^2=81 \end{gathered}`

Subtracting 49 from x²+49=121 we get:

`\begin{gathered} x^2+49-49=121-49, \\ x^2=72. \end{gathered}`

Therefore:

`x=\sqrt[]{72}=\sqrt[]{36\cdot2}=6\sqrt[]{2}\text{.}`

Substituting x²=72 in x²+ y²=81 we get:

`72+y^2=81.`

Subtracting 72 from the above equation we get:

`\begin{gathered} 72+y^2-72=81-72, \\ y^2=9. \end{gathered}`

Therefore:

`y=3.`

Therefore the measure of the base of the original triangle is:

`y+7=3+7=10,`

and its height is:

`x=6\sqrt[]{2}\text{.}`

Using the formula for the area of a triangle:

`A=(bh)/(2),`

we get:

`A=\frac{10\cdot6\sqrt[]{2}}{2}=\frac{60\sqrt[]{2}}{2}=30\sqrt[]{2}\text{.}`

Answer: The area of the original triangle is:

`30\sqrt[]{2}\text{ square units.}`

ALTERNATIVE SOLUTION.

To be able to use trigonometric functions first we have to determine at least one angle of the triangle. To do that, we will use the trigonometric function cosine, recall that:

`\cos \alpha=\frac{\text{adjacent side}}{\text{hypothenuse}}\text{.}`

Now, we will use triangle ADB and angle A, then:

`\cos A=(7)/(11)\text{.}`

Therefore:

`A=\cos ^(-1)((7)/(11)).`

Once we have the angle, now we use another trigonometric function, the sine:

`\sin \alpha=\frac{opposite\text{ side}}{\text{hypothenuse}}\text{.}`

Then:

`\begin{gathered} \sin A=(x)/(11), \\ x=\sin A(11)\text{.} \end{gathered}`

Once we have x we do the same to find y but with angle C:

`\begin{gathered} \cos C=(y)/(9), \\ y=\cos C*9. \end{gathered}`

Finally, we use the formula for the area of a triangle, and get:

`A=((7+y)x)/(2)=((7+3)*11\sin A)/(2)\approx42.5\text{ square units.}`