Using the Pythagorean Theorem: Calculation using the diagonal

Let's focus on triangle BCD in order to find side DC.

We'll use the Pythagorean theorem and input the known data:

$BC^2+DC^2=BD^2$

$6^2+DC^2=10^2$

$DC^2=100-36=64$

Let's now remove the square root:

$DC=8$

Since in a rectangle each pair of opposite sides are equal to each other, we know that:

$DC=AB=8$

$BC=AD=6$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$8+6+8+6=16+12=28$

Let's focus on triangle BCD in order to find side BC.

We'll use the Pythagorean theorem using our values:

$BC^2+DC^2=BD^2$

$BC^2+24^2=25^2$

$BC^2=625-576=49$

Let's now remove the square root:

$BC=7$

Since each pair of opposite sides are equal to each other in a rectangle, we can state that:

$DC=AB=24$

$BC=AD=7$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$24+7+24+7=14+48=62$

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:

$15\times8=120$

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 $\boxed{30}$.

To solve the problem of finding the perimeter of triangle ABD, we will apply the following steps:

• Step 1: Identify the given dimensions of the rectangle.
• Step 2: Calculate the length of the diagonal BD using the Pythagorean theorem.
• Step 3: Sum the sides of triangle ABD to find its perimeter.

Now, let's work through each step:

Step 1: We know from the problem that AB = 15 and AD = 8.

Step 2: The triangle ABD is a right triangle with AB and AD as the legs, and BD as the hypotenuse. Therefore, by the Pythagorean theorem:

Calculating these squares gives:

Taking the square root of both sides, we find:

Step 3: Now, calculate the perimeter of triangle ABD.

Therefore, the perimeter of triangle ABD is $40$.

To solve this problem, we'll utilize the properties of rectangles and the Pythagorean theorem:

• Step 1: Determine the full diagonal of the rectangle using Pythagorean theorem: $AC = \sqrt{AB^2 + AD^2}$.
• Step 2: Since $O$ is the midpoint of the diagonal $AC$, $BO = \frac{AC}{2}$.

Now, let's calculate step-by-step:

Step 1: We know $AB = 8$ and $AD = 6$, therefore, using the Pythagorean theorem:

$AC = \sqrt{AB^2 + AD^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

Step 2: Since the diagonals bisect each other, the length of $BO$ is half of $AC$:

$BO = \frac{AC}{2} = \frac{10}{2} = 5$

Therefore, the solution to the problem is $BO = 5$.

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:

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

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.