Using the Pythagorean Theorem: Using additional geometric shapes

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:

$DC=\sqrt{576}=24$

Let's look at triangle BCD, let's calculate the diagonal by the Pythagorean theorem:

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

As we are given one side, we know that the other sides are equal to 4, so we will replace accordingly in the formula:

$4^2+4^2=BD^2$

$16+16=BD^2$

$32=BD^2$

We extract the root:$BD=AC=\sqrt{32}$

Now we calculate the sum of the diagonals:

$2\times\sqrt{32}=11.31$

Now we calculate the sum of the 3 sides of the square:

$4\times3=12$

And we reveal that the sum of the two diagonals is less than the sum of the 3 sides of the square.

11.31 < 12

To find the length of the diagonal $AC$ on a cuboid, we will use the Pythagorean theorem twice:

• Step 1: Calculate $AB$, the diagonal of the face of the cuboid:

We assume the cuboid's known dimensions, with width $8 \, \text{cm}$ and height $4 \, \text{cm}$. Assume $l$ for one dimension along the base.

Since $AB = \sqrt{w^2 + l^2}$, solving for AB means understanding both directions. As the length isn't given, we solve specifically for vertically so $AC$ depends on full volume space:

• Step 2: Calculate diagonal $AC$:

$AC = \sqrt{AB^2 + \text{height}^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \, \text{cm}$

In this way, we determine the length of diagonal $AC$ is $\sqrt{80}$ cm.

The correct choice corresponding to this calculation is Choice 3: $\sqrt{80}$ cm.

Let's calculate side BG using the Pythagorean theorem:

$BG^2+GC^2=BC^2$

We'll substitute the known data:

$BG^2+3^2=5^2$

$BG^2+9=25$

$BG^2=16$

$BG=\sqrt{16}=4$

Now we can calculate the area of rectangle ABGE since we have the length and width:

$5\times4=20$

Remember there are two options to calculate the area of a rhombus:

1: The diagonal multiplied by the diagonal divided by 2.

2: The base multiplied by the height.

In the question, we are only given the data for one of the diagonals and one of the sides, which means we cannot use either of the above formulas.

We need to find more data. Let's begin by finding the second diagonal:

Remember that the diagonals of a rhombus are perpendicular to one another, which means that they form a 90-degree angle.

Therefore, all the triangles in a rhombus are right-angled.

Now we can focus on the triangle where the side and the height are given, and we will calculate the third side using the Pythagorean theorem:

$a²+b²=c²$Insert the given data:

$3^2+x^2=5^2$$9+x^2=25$$x^2=25-9=16$$x=\sqrt{16}=4$

Now that we have found the second half of the diagonal, we can calculate the area of the rhombus by multiplying the two diagonals together.

Since the diagonals in a rhombus are perpendicular and cross each other, they are equal. Hence, our diagonals are equal:

$3+3=6$$4+4=8$Therefore, the area of the rhombus is:

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

DE crosses AB and AC, that is to say:

$AD=DB=\frac{1}{2}AB=\frac{1}{2}\times6=3$

$AE=EC=\frac{1}{2}AC=\frac{1}{2}\times10=5$

Now let's look at triangle ADE, two sides of which we have already calculated.

Now we can find the third side DE using the Pythagorean theorem:

$AD^2+DE^2=AE^2$

We substitute our values into the formula:

$3^2+DE^2=5^2$

$9+DE^2=25$

$DE^2=25-9$

$DE^2=16$

We extract the root:

$DE=\sqrt{16}=4$

Now let's look at triangle ABC, two sides of which we have already calculated.

Now we can find the third side (BC) using the Pythagorean theorem:

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

We substitute our values into the formula:

$6^2+BC^2=10^2$

$36+BC^2=100$

$BC^2=100-36$

$BC^2=64$

We extract the root:

$BC=\sqrt{64}=8$

Now we have all the data needed to calculate the area of the trapezoid DECB using the formula:

(base + base) multiplied by the height divided by 2:

Keep in mind that the height in the trapezoid is DB.

$S=\frac{(4+8)}{2}\times3$

$S=\frac{12\times3}{2}=\frac{36}{2}=18$

We will use the Pythagorean theorem in order to find BD:

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

Let's input the known data:

$BD^2=3^2+4^2$

$BD^2=9+16$

$BD^2=25$

We'll take the square root:

$BD=\sqrt{25}=5$

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:

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

Let's look at triangle ADK in order to calculate side AD:

$AD^2+DK^2=AK^2$

Now let's substitute in our values:

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

$AD^2+16=25$

We'll then move 16 to the other side and change the sign to the appropriate one:

$AD^2=25-16$

$AD^2=9$

Next, we'll take the square root and get:

$AD=3$

Since AD is a side of rectangle ABCD, we can calculate side AB as follows:

$S=AB\times AD$

Let's substitute in our values:

$24=3\times AB$

Finally, we'll divide both sides by 3:

$AB=8$

To determine which has the larger measurement, the triangle's perimeter or the circle's circumference, we need to compute both values.

Step 1: Calculate the perimeter of the Triangle
We are given two sides of the triangle: 6 and 5. Since it's implied to be a right triangle, we apply the Pythagorean theorem to find the third side, the hypotenuse $c$:

The perimeter $P$ of the triangle is:

Step 2: Calculate the circumference of the Circle
The circumference $C$ of a circle with radius $r$ is given by the formula:

Assuming the radius of the circle is equivalent to the '6' mentioned for the green line in the SVG:

Step 3: Compare the Triangle's Perimeter and the Circle's Circumference
We compare the values:

• Perimeter of the Triangle: $P \approx 18.81$
• Circumference of the Circle: $C \approx 37.7$

The circumference of the circle (37.7) is greater than the perimeter of the triangle (18.81).

Therefore, the circle has the larger measurement.

Conclusion: The circle has the larger perimeter.

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

• Step 1: Calculate the side length of the square using the given area.
• Step 2: Determine the length of $BD$ using the side length.
• Step 3: Use the kite area formula with diagonals $BD$ and $AC=2x$.

Now, let's work through each step:

Step 1: The area of the square is 36 cm². The side length of the square, denoted as $s$, can be calculated by taking the square root of the area:

$s = \sqrt{36} = 6 \text{ cm}$

Step 2: To find diagonal $BD$, we use the relationship for the diagonal of a square in terms of its side:

$BD = s \sqrt{2}$. Given $s = 6$, we compute:

$BD = 6\sqrt{2} \text{ cm}$

Step 3: Now, we apply the formula for the area of a kite, which is $\frac{1}{2} \times d1 \times d2$, where $d1 = BD = 6\sqrt{2}$ and $d2 = AC = 2x$:

The area $A$ of the kite is:

$A = \frac{1}{2} \times 6\sqrt{2} \times 2x = 6\sqrt{2}x \text{ cm}^2$

Therefore, the area of the kite in terms of $x$ is $6\sqrt{2}x$ cm².

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:

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

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

• Identify the given information.
• Use the geometric properties of a circle and a right triangle to find the radius.
• Express the area of the circle in terms of the given expression.

Now, let's work through each step:

Step 1: Given a circle with center $O$ and radii $AO$ and $OB$ such that $AO\perp OB$, each is a radius $r$, and $AB = \text{and}+2$.

Step 2: By the Pythagorean theorem, we know:

Step 3: Solving for the area of the circle:

The radius $r$ can be expressed by rearranging:

The area of the circle using this radius is:

Therefore, the expression for the area of the circle is $\frac{\pi}{2}[y^2+4y+4]$.

First, we must remember the formula for the area of a parallelogram:$\text{Lado }x\text{ Altura}$.

In this case, we will try to find the height CH and the side BC.

We start from the side

First, let's observe the small triangle EBG,

As it is a right triangle, we can use the Pythagorean theorem (

$A^2+B^2=C^2$)

$BG^2+4^2=5^2$

$BG^2+16=25$

$BG^2=9$

$BG=3$

Now, let's start looking for GC.

First, remember that the deltoid has two pairs of equal adjacent sides, therefore:$FC=EC=9$

Now we can also do Pythagoras in the triangle GCE.

$GC^2+4^2=9^2$

$GC^2+16=81$

$GC^2=65$

$GC=\sqrt{65}$

Now we can calculate the side BC:

$BC=BG+GT=3+\sqrt{65}\approx11$

Now, let's observe the triangle BGE and DHC

Angle BGE = 90°
Angle CHD = 90°
Angle CDH=EBG because these are opposite parallel angles.

Therefore, there is a ratio of similarity between the two triangles, so:

$\frac{HD}{BG}=\frac{HC}{GE}$

$\frac{HD}{BG}=\frac{7.5}{3}=2.5$

$\frac{HC}{EG}=\frac{HC}{4}=2.5$

$HC=10$

Now that there is a height and a side, all that remains is to calculate.

$10\times11\approx110$