Using the Pythagorean Theorem: Calculating area

To solve this problem, we'll follow these steps:

• Step 1: Use the Pythagorean Theorem to calculate the length of $AD$.
• Step 2: Calculate the area of triangle $\triangle BCD$.

Step 1: Given $AB = 12$ and $AC = 13$, we use the Pythagorean Theorem to find $AD$.

Step 2: Knowing the sides $AD = 5$ (height of the rectangle) and $AB = 12$ (base of the rectangle), triangle $\triangle BCD$ will have the base $BC = 12$ and the height $BD = 5$.

The area of triangle $\triangle BCD$ is:

Therefore, the area of triangle $\triangle BCD$ is 30.

Let's solve this step-by-step:

• Step 1: Identify the given information.
  We know the rectangle $ABCD$, is divided by its diagonal $AC$. The length $AB$ is $12$, and the diagonal $AC$ is $13$.
• Step 2: Apply Pythagorean theorem to find $BC$, which acts as the height.
  Using the Pythagorean theorem in $\triangle ABC$ gives us: $AC = \sqrt{AB^2 + BC^2}$ Given $AC = 13$ and $AB = 12$, we set up the equation: $13 = \sqrt{12^2 + BC^2}$ Squaring both sides leads to: $169 = 144 + BC^2$ $BC^2 = 169 - 144 = 25$ Thus, $BC = \sqrt{25} = 5$.
• Step 3: Calculate the area of $\triangle ABC$.
  The area can be found using the formula: $\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC$ $= \frac{1}{2} \times 12 \times 5$ $= \frac{1}{2} \times 60 = 30$

Therefore, the area of triangle $ABC$ is $\boxed{30}$.

We will find side DC by using the Pythagorean theorem in triangle DBC:

$BC^2+CD^2=BD^2$

Let's substitute the known data:

$8^2+CD^2=17^2$

$CD^2=289-64=225$

Let's take the square root:

$CD=15$

Now we have the length and width of rectangle ABCD and we'll calculate the area:

$15\times8=120$

According to the given information, we can claim that:

$BD=2BO=8.5\times2=17$

Now let's look at triangle ABD to calculate side AB

$AB^2+AD^2=BD^2$

Let's input the known data:

$AB^2+8^2=17^2$

$AB^2=289-64=225$

We'll take the square root

$AB=15$

Now let's calculate the area of triangle ABD:

$\frac{15\times8}{2}=\frac{120}{2}=60$

We will use the Pythagorean theorem in order to find the side DC:

$(BC)^2+(DC)^2=(DB)^2$

We begin by inserting the existing data into the theorem:

$7^2+(DC)^2=25^2$

$49+DC^2=625$

$DC^2=625-49=576$

Finally we extract the root:

$DC=\sqrt{576}=24$

When writing the name of a polygon, the letters will always be in the order of the sides:

This is a rectangle ABCD:

This is a rectangle ABDC:

Always go in order, and always with the right corner to the one we just mentioned.

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.

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

To find the area,

first, the height of the parallelogram must be found.

To conclude, let's take a look at triangle EBC.

Since we know it is a right triangle (since it is the height of the parallelogram)

the Pythagorean theorem can be used:

$a^2+b^2=c^2$

In this case: $EB^2+EC^2=BC^2$

We place the given information: $2^2+EC^2=5^2$

We isolate the variable:$EC^2=5^2+2^2$

We solve:$EC^2=25-4=21$

$EC=\sqrt{21}$

Now all that remains is to calculate the area.

It is important to remember that for this, the length of each side must be used.
That is, AE+EB=2+7=9

$\sqrt{21}\times9=41.24$

To find the area of the trapezoid, you must remember its formula:$\frac{(base+base)}{2}+\text{altura}$We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$ In triangle AFG

We replace:

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We will do the same process with side DB in triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

From here there are two ways to finish the exercise:

1. Calculate the area of the trapezoid GFBD, prove that it is equal to the trapezoid EGDC and add them up.

2. Use the data we have revealed so far to find the parts of the trapezoid EFBC and solve.

Let's start by finding the height of GD:

$GD=AD-AG=8-3=5$

Now we reveal that EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts then:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

We replace the data in the trapezoid formula:

$\frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95$

To find the perimeter of the trapezoid, all its sides must be added:

We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$in the triangle AFG

We replace

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We perform the same process with the side DB of the triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding FB:

$FB=AB-AF=17-5=12$

Now we reveal EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts so:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

All that's left is to calculate:

$30+8+12\times2=30+8+24=62$

To find the missing side, we use the Pythagorean theorem in the upper triangle.

Since the triangle is isosceles, we know that the length of both sides is 7.

Therefore, we apply Pythagoras$A^2+B^2=C^2$$7^2+7^2=49+49=98$

Therefore, the area of the missing side is:$\sqrt{98}$

The area of a rectangle is the multiplication of the sides, therefore:

$\sqrt{98}\times10=98.99\approx99$