Using the Pythagorean Theorem: Using additional geometric shapes

The rectangle ABCD is shown below.

$BD=25,BC=7$

Calculate the area of the rectangle.

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:

$DC=\sqrt{576}=24$

168

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

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

No

The trapezoid ABCD and the rectangle ABGE are shown in the figure below.

Given in cm:

AB = 5

BC = 5

GC = 3

Calculate the area of the rectangle ABGE.

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:

$5\times4=20$

20

The trapezoid DECB forms part of triangle ABC.

AB = 6 cm
AC = 10 cm

Calculate the area of the trapezoid DECB, given that DE divides both AB and AC in half.

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.

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

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

18

Using the rhombus in the drawing:

Calculate the area?

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:

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

24

Given the rectangle ABCD

It is known that:

AB=4

AD=3

What is the length of the diagonal BD?

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:

$BD=\sqrt{25}=5$

$5$

ABC is an isosceles triangle.

AD is the height of triangle ABC.

AF = 5

AB = 17
AG = 3

AD = 8

What is the perimeter of the trapezoid EFBC?

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:

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

62

Given that the triangle ABC is isosceles,
and inside it we draw EF parallel to CB:

AF=5 AB=17
AG=3 AD=8
AD the height in the triangle

What is the area of the trapezoid EFBC?

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:

$\frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95$

95

Shown below is the rectangle ABCD.

Given in cm:

AK = 5

DK = 4

The area of the rectangle is 24 cm².

Calculate the side AB.

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

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

Let's input the given data:

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

$AD^2+16=25$

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

$AD^2=25-16$

$AD^2=9$

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 input the given data:

$24=3\times AB$

We'll divide both sides by 3:

$AB=8$

8

ABCD is a parallelogram
BFCE is a deltoid

What is the area of the parallelogram ABCD?

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.

$10\times11\approx110$

$\approx110$

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

False

Look at the following square:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

No

Is the sum of the two diagonals in the above square greater than the sum of the 3 sides of the square?

No

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

No

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

No

ABCD is a square with a side length of 8 cm.

EB = 10

What is the area of the parallelogram EBFC?

112 cm²

Look at the isosceles trapezoid ABCD below.

DF = 2 cm
AD =$\sqrt{20}$ cm

Calculate the area of the trapezoid given that the quadrilateral ABEF is a square.

24

Given ABCD deltoid AB=AC DC=BD

The diagonals of the deltoid intersect at the point O

Given in cm AO=12 OD=4

The area of the deltoid is equal to 48 cm².

Calculate the side CD

5 cm

The trapezoid ABCD is drawn inside a rectangle.

DC = 12 cm
BK = 3 cm
Height of the trapezoid (H) = 4

Calculate the area of the trapezoid.

36

The rectangle ABCD is shown below.

BC = 5

AB = 12

Calculate the diagonal of the rectangle.

$13$