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

We use the Pythagorean theorem:

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

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!

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 this problem, we'll perform the following steps:

• Step 1: Identify the sides of the triangle. The problem states $AB = 5$ and $AC = 3$. Since triangle ABC is a right triangle, $AB = 5$ is the hypotenuse.

• Step 2: Apply the Pythagorean Theorem, which states $a^2 + b^2 = c^2$, where $c$ is the hypotenuse.

• Step 3: In this case, set $CA = 3$, $BC = x$, and $AB = 5$. Substitute these into the theorem:
  $3^2 + x^2 = 5^2$

• Step 4: Calculate $3^2 = 9$ and $5^2 = 25$. Thus, the equation becomes:
  $9 + x^2 = 25$

• Step 5: Rearrange to solve for $x^2$:
  $x^2 = 25 - 9$
  $x^2 = 16$

• Step 6: Solve for $x$ by taking the square root:
  $x = \sqrt{16} = 4$

Therefore, the length of side BC is $4$.

To solve the given problem, we will use the Pythagorean Theorem, which is stated as:

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

Here, $c$ represents the hypotenuse, and $a$ and $b$ are the legs of the right triangle. Based on our problem:

• $c = 7$ (the hypotenuse)
• $a = 5$ (one of the legs)
• $b = X$ (the missing side we need to find)

Applying the Pythagorean theorem, we get:

$5^2 + X^2 = 7^2$

Calculating the squares, we have:

$25 + X^2 = 49$

Subtracting 25 from both sides to isolate $X^2$:

$X^2 = 49 - 25$

$X^2 = 24$

Taking the square root of both sides to solve for $X$:

$X = \sqrt{24}$

Therefore, the solution to the problem is $X = \sqrt{24}$, which corresponds to choice 1 in the multiple-choice answers.

To find the length of side $AB$ in the right triangle $\triangle ABC$, we will use the Pythagorean Theorem. The theorem is given by:

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

In this equation, $c$ is the hypotenuse, and $a$ and $b$ are the other two sides. Based on the given information, we have:

• $AC = 6$ is one leg.
• $BC = 9$ is the hypotenuse.
• $AB$ is the unknown leg we want to find.

Substituting the known values into the Pythagorean Theorem:

$AB^2 + 6^2 = 9^2$

Simplifying the equation:

$AB^2 + 36 = 81$

Subtract 36 from both sides to solve for $AB^2$:

$AB^2 = 81 - 36$

$AB^2 = 45$

To find $AB$, take the square root of both sides:

$AB = \sqrt{45}$

Therefore, the length of side $AB$ is $\sqrt{45}$.

To find the length of side AB in the right triangle ABC:

• Step 1: Identify the known lengths. We know $BC = 6$ and $AC = \sqrt{72}$.
• Step 2: Apply the Pythagorean Theorem: $AB^2 + BC^2 = AC^2$.
• Step 3: Plug in the known values: $AB^2 + 6^2 = (\sqrt{72})^2$.
• Step 4: Since $(\sqrt{72})^2 = 72$, the equation becomes $AB^2 + 36 = 72$.
• Step 5: Subtract 36 from both sides, yielding: $AB^2 = 36$.
• Step 6: Take the square root of each side: $AB = \sqrt{36} = 6$.

Thus, the length of side AB is $6$.

To find the length of side $AB$ in the right triangle $ABC$, we will use the Pythagorean theorem:

Since $BC$ is the hypotenuse, $BC = 10$, and one leg $AC = 5$, we can express the Pythagorean theorem as:

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

Substituting the known values gives us:

$AB^2 + 5^2 = 10^2$

Simplifying further:

$AB^2 + 25 = 100$

Subtracting 25 from both sides results in:

$AB^2 = 75$

Taking the square root of both sides, we obtain:

$AB = \sqrt{75}$

Therefore, the length of side $AB$ is $\sqrt{75}$.

To solve the problem, we will use the Pythagorean Theorem to calculate the length of side $AB$. According to the theorem, $AB^2 + AC^2 = BC^2$.

Given:

• Hypotenuse $BC = 2$
• One leg $AC = 1$

We need to find $AB$: $AB^2 + 1^2 = 2^2$

Solve for $AB^2$: $AB^2 + 1 = 4$

By subtracting 1 from both sides: $AB^2 = 4 - 1 = 3$

Taking the square root of both sides gives us: $AB = \sqrt{3}$

Therefore, the length of side $AB$ is $\sqrt{3}$.

To solve for $BC$ in the right triangle $\triangle ABC$, we follow these steps:

• Identify the given lengths: $AB = 6$ and $AC = 20$.
• Apply the Pythagorean Theorem: $AB^2 + BC^2 = AC^2$.
• Substitute known values into the formula:

$6^2 + BC^2 = 20^2$

Simplify the left side:

$36 + BC^2 = 400$

Subtract 36 from both sides:

$BC^2 = 400 - 36$

$BC^2 = 364$

Solve for $BC$ by taking the square root of both sides:

$BC = \sqrt{364}$

Thus, the length of $BC$ is $\sqrt{364}$.

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:

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

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