Using the Pythagorean Theorem: Calculate The Missing Side based on the formula

To answer this question, we must know the Pythagorean Theorem

The theorem allows us to calculate the sides of a right triangle.

We identify the sides:

ab = a = 5
bc = b = ?

ac = c = 13

We replace the data in the exercise:

5²+?² = 13²

We swap the sections

?²=13²-5²

?²=169-25

?²=144

?=12

To solve the exercise, it is necessary to know the Pythagorean Theorem:

A²+B²=C²

We replace the known data:

2²+B²=7²

4+B²=49

We input into the formula:

B²=49-4

B²=45

We find the root

B=√45

This is the solution. However, we can simplify the root a bit more.

First, let's break it down into prime numbers:

B=√(9*5)

We use the property of roots in multiplication:

B=√9*√5

B=3√5

This is the solution!

We use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

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:

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

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.

We calculate the perimeter of the triangle:

$12+4\sqrt{5}=4+AC+BC$

As we want to find the hypotenuse (BC), we isolate it:

$12+4\sqrt{5}-4-AC=BC$

$BC=8+4\sqrt{5}-AC$

Then calculate AC using the Pythagorean theorem:

$AB^2+AC^2=BC^2$

$4^2+AC^2=(8+4\sqrt{5}-AC)^2$

$16+AC^2=(8+4\sqrt{5})^2-2\times AC(8+4\sqrt{5})+AC^2$

We then simplify the two:$AC^2$

$16=8^2+2\times8\times4\sqrt{5}+(4\sqrt{5})^2-2\times8\times AC-2AC4\sqrt{5}$

$16=64+64\sqrt{5}+16\times5-16AC-8\sqrt{5}AC$

$16AC+8\sqrt{5}AC=64+64\sqrt{5}+16\times5-16$

$AC(16+8\sqrt{5})=128+64\sqrt{5}$

$AC=\frac{128+64\sqrt{5}}{16+8\sqrt{5}}=\frac{8(16+8\sqrt{5})}{16+8\sqrt{5}}$

We simplify to obtain:

$AC=8$

Now we can replace AC with the value we found for BC:

$BC=8+4\sqrt{5}-AC$

$BC=8+4\sqrt{5}-8=4\sqrt{5}$