Area of a Triangle: Using Pythagoras' theorem

Let's calculate side BG using the Pythagorean theorem:

$BG^2+GC^2=BC^2$

We'll substitute the known data:

$BG^2+3^2=5^2$

$BG^2+9=25$

$BG^2=16$

$BG=\sqrt{16}=4$

Now we can calculate the area of rectangle ABGE since we have the length and width:

$5\times4=20$

The area of triangle ABC is equal to:

$\frac{AD\times BC}{2}=2x+16$

As we are given the area of the triangle, we can insert this data into BC in the formula:

$\frac{AD\times(BD+DC)}{2}=2x+16$

$\frac{AD\times(x+5+3)}{2}=2x+16$

$\frac{AD\times(x+8)}{2}=2x+16$

We then multiply by 2 to eliminate the denominator:

$AD\times(x+8)=4x+32$

Divide by: $(x+8)$

$AD=\frac{4x+32}{(x+8)}$

We rewrite the numerator of the fraction:

$AD=\frac{4(x+8)}{(x+8)}$

We simplify to X + 8 and obtain the following:

$AD=4$

We now focus on triangle ADC and by use of the Pythagorean theorem we should find X:

$AD^2+DC^2=AC^2$

Inserting the existing data:

$4^2+(x+5)^2=(\sqrt{65})^2$

$16+(x+5)^2\text{ }=65/-16$

$(x+5)^2=49/\sqrt{}$

$x+5=\sqrt{49}$

$x+5=7$

$x=7-5=2$

There are two ways to solve the exercise:

It is possible to drop a height from one of the vertices, as we know

In an equilateral triangle, the height intersects the base,

This creates a right triangle whose two sides are 6 and 3,

Using the Pythagorean theorem$A^2+B^2=C^2$

We can find the length of the missing side.

$3^2+X^2=6^2$

We convert the formula

$6^2-3^2=X^2$

$36-9=27$

Therefore, the height of the triangle is equal to:$\sqrt{27}$

From here we calculate with the usual formula for the area of a triangle.

$\frac{6\times\sqrt{27}}{2}=15.588$

Option B for the solution is through the formula for the area of an equilateral triangle:

$S=\frac{\sqrt{3}\times X^2}{4}$

Where X is one of the sides.

$\frac{\sqrt{3}\times6^2}{4}=\frac{62.353}{4}=15.588$