Perimeter of a Rectangle: Using Pythagoras' theorem

A rectangle has two pairs of equal opposite sides.

That is:

BC=AD=4

AB=DC

In an equilateral triangle, all sides are equal, therefore:
EC=CD=DE

We know that EC=8, so:

EC=CD=DE=8

We know that:

AB=DC

Therefore:

AB=DC=8

Remember that the perimeter of a rectangle is equal to the sum of all its sides, therefore:

AB+BC+DC+AD

We substitute in all its known sides:

8+4+8+4=

24

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:

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

Let's solve for the perimeter of the rectangle by following these steps:

• First, identify the given information: The diagonal $AC$ is 10 units and the height $BC$ is 6 units.
• Next, use the Pythagorean theorem to calculate the missing side (the base), $AB$. Given that $AB^2 + BC^2 = AC^2$, we can write:
• Step 1: Substitute known values into the equation: $AB^2 + 6^2 = 10^2$.
• Step 2: Calculate $AB^2$:
$AB^2 + 36 = 100\\ AB^2 = 100 - 36\\ AB^2 = 64\\ AB = \sqrt{64} = 8$

We have found the base $AB = 8$ units.

Now, calculate the perimeter $P$ using the perimeter formula $P = 2(l + w)$, where $l$ is the length (AB = 8) and $w$ is the width (BC = 6).

Therefore, the perimeter of the rectangle is 28 units.

The correct choice is 3: 28.

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

$AD=BC=11$

We can calculate side FC:

$11-3=FC$

$8=FC$

Let's focus on triangle FCE and calculate side CE using the Pythagorean theorem:

$CF^2+CE^2=FE^2$

Let's substitute the known values into the formula:

$8^2+CE^2=17^2$

$64+CE^2=289$

$CE^2=289-64$

$CE^2=225$

Let's take the square root:

$CE=15$

Since CE equals AB and in a rectangle every pair of opposite sides are equal to each other, we can claim that:

$CE=AB=CD=15$

Now we can calculate the perimeter of the rectangle:

$11+15+11+15=22+30=52$

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

$BC=AD=6$

Therefore:

$$$AD-AE=ED$

Let's substitute the known data into the formula:

$6-3=ED$

$3=ED$

Let's focus on triangle EDF and find side DF using the Pythagorean theorem:

$ED^2+DF^2=EF^2$

Let's substitute the known data into the formula:

$3^2+DF^2=5^2$

$9+DF^2=25$

$DF^2=25-9$

$DF^2=16$

Let's find the square root:

$DF=4$

Side DC=8

Since in a rectangle every pair of opposite sides are equal to each other:

$AD=BC=6$

$AB=CD=8$

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

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

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

• Step 1: Identify the given information.
• Step 2: Apply the Pythagorean theorem to find the missing side.
• Step 3: Use the perimeter formula for rectangles.

Now, let's work through each step:

Step 1: The given information includes the base ($AB = 4$) and the diagonal ($AC = 5$).

Step 2: Apply the Pythagorean theorem. The diagonal acts as the hypotenuse of a right triangle with sides being the rectangle's base and height. Using the theorem:

$(AB)^2 + (AD)^2 = (AC)^2$

$4^2 + (AD)^2 = 5^2$

$16 + (AD)^2 = 25$

Solving for $(AD)^2$:

$(AD)^2 = 25 - 16 = 9$

Thus, $AD = \sqrt{9} = 3$.

Step 3: Calculate the perimeter using the formula for the perimeter of a rectangle:

$P = 2 \times (\text{length} + \text{width})$.

Inserting the values $AB = 4$ and $AD = 3$:

$P = 2 \times (4 + 3) = 2 \times 7 = 14$.

Therefore, the solution to the problem is $14$.

Based on the given data, we can claim that:

$OF=OE=3$

$EF=6$

$AB=EF=DC=6$

We'll find side BF using the Pythagorean theorem in triangle BFO:

$OF^2+BF^2=BO^2$

Let's substitute the known values into the formula:

$3^2+BF^2=5^2$

$9+BF^2=25$

$BF^2=25-9$

$BF^2=16$

Let's take the square root:

$BF=4$

Since the triangles overlap:

$BF=DE=4=FC$

From this, we can calculate side BC:

$BC=4+4=8$

Since in a rectangle, each pair of opposite sides are equal to each other, we can claim that AD also equals 8

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

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

In the first step, we must find the length of EC, which we will identify with an X.

We know that the perimeter of a rectangle is the sum of all its sides (AE+EC+CF+FA),

Since in a rectangle the opposite sides are equal, the formula can also be written like this: 2AE=2EC.

We replace the known data:

$2\times8+2X=24$

$16+2X=24$

We isolate X:

$2X=8$

and divide by 2:

$X=4$

Now we can use the Pythagorean theorem to find EB.

(Pythagoras: $A^2+B^2=C^2$)

$EB^2+4^2=5^2$

$EB^2+16=25$

We isolate the variable

$EB^2=9$

We take the square root of the equation.

$EB=3$

The area of a parallelogram is the height multiplied by the side to which the height descends, that is$AB\times EC$.

$AB=\text{ AE}+EB$

$AB=8+3=11$

And therefore we will apply the area formula:

$11\times4=44$

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:

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

To solve this problem, we will use the information in the diagram and relevant geometric principles.

Firstly, we consider the triangle $\triangle BCD$ within the rectangle:

• $\overline{BC} = 12$ is one leg of the triangle.
• $\overline{BD} = 13$ is the hypotenuse given in the problem.

Using Pythagoras' Theorem $(\overline{BD})^2 = (\overline{BC})^2 + (\overline{CD})^2$:

$13^2 = 12^2 + (\overline{CD})^2$

$169 = 144 + (\overline{CD})^2$

Subtract 144 from both sides to solve for $(\overline{CD})^2$:

$169 - 144 = (\overline{CD})^2$

$25 = (\overline{CD})^2$

Taking the square root gives:

$\overline{CD} = \sqrt{25} = 5$

The rectangle ABCD's other side, $\overline{AB} = \overline{CD} = 5$, since opposite sides of a rectangle are equal.

Using the perimeter formula for a rectangle:

$P = 2 \times (\overline{AB} + \overline{BC})$

Substitute the known lengths:

$P = 2 \times (5 + 12)$

$P = 2 \times 17$

$P = 34$

Therefore, the perimeter of rectangle ABCD is $\textbf{34}$.

Let's begin by observing triangle FCE and calculate side FC using the Pythagorean theorem:

$EC^2+FC^2=EF^2$

Let's begin by substituting all the known values into the formula:

$8^2+FC^2=10^2$

$64+FC^2=100$

$FC^2=100-64$

$FC^2=36$

Let's take the square root:

$FC=6$

Since we know that the triangles overlap:

$\frac{AD}{FC}=\frac{DE}{CE}=\frac{AE}{FE}$

Let's again substitute the known values into the formula:

$\frac{AD}{6}=\frac{16}{8}$

$AD=2\times6=12$

Finally let's calculate side CD:

$16+8=24$

Since in a rectangle each pair of opposite sides are equal, we can calculate the perimeter of rectangle ABCD as follows:

$12+24+12+24=24+48=72$

Observe triangle BCE and proceed to calculate side EC using the Pythagorean theorem:

$BC^2+EC^2=BE^2$

Insert the known values into the theorem:

$6^2+EC^2=10^2$

$36+EC^2=100$

$EC^2=100-36$

$EC^2=64$

Determine the square root:

$EC=8$

Calculate the area of triangle BCE:

$S=\frac{BC\times EC}{2}$

Insert the known values once again:

$S=\frac{6\times8}{2}=\frac{48}{2}=24$

According to the given data, the area of triangle BCE is one-third of rectangle ABCD's area, therefore:

$24=\frac{1}{3}$

Multiply by 3:

$S=3\times24=72$

The area of the rectangle equals 72

Now let's determine side CD

We know that the area of a rectangle equals the length multiplied by the width, meaning:

$S=BC\times DC$

Insert the known values in the formula:

$72=6\times CD$

Divide both sides by 6:

$CD=12$

Given that in a rectangle opposite sides are equal, AB also equals 12

Proceed to calculate the perimeter of the rectangle ABCD:

$12+6+12+6=24+12=36$

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

• Step 1: Use the Pythagorean theorem to find $AB$.
• Step 2: Calculate the perimeter of the rectangle $ABCD$.

Now, let's work through each step:
Step 1: For triangle $ACD$, where $AC = 8$ and $AD = 4$, apply the Pythagorean theorem:

Step 2: The perimeter of rectangle $ABCD$ is given by:

Therefore, the solution to the problem is $8 + 16\sqrt{3}$.

Since the triangles are equal to each other, we can claim that:

$AE=FC$

$AG=CH=8$

$EG=FH=10$

Now let's calculate side AB:

$8+5=13$

Since in a rectangle each pair of opposite sides are equal to each other:

$AB=CD=13$

We can also claim that:

$DH=13-8=5$

Side EF is also equal in length to sides AB and CD which are equal to 13

Now let's calculate side FC using the Pythagorean theorem in triangle FCH:

$HC^2+FC^2=HF^2$

Let's input the known data:

$8^2+FC^2=10^2$

$64+FC^2=100$

$FC^2=100-64$

$FC^2=36$

Let's take the square root:

$FC=6$

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

$13+6+13+6=26+12=38$