Examples with solutions for Diagonals: Calculation using the diagonal

Exercise #1

AAABBBCCCDDD2524Calculate the perimeter of the rectangle ABCD.

Video Solution

Step-by-Step Solution

Let's focus on triangle BCD in order to find side BC.

We'll use the Pythagorean theorem using our values:

BC2+DC2=BD2 BC^2+DC^2=BD^2

BC2+242=252 BC^2+24^2=25^2

BC2=625576=49 BC^2=625-576=49

Let's now remove the square root:

BC=7 BC=7

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

DC=AB=24 DC=AB=24

BC=AD=7 BC=AD=7

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

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

Answer

62

Exercise #2

Look at the following rectangle:

AAABBBCCCDDD1213

Calculate the area of the triangle ABC.

Video Solution

Step-by-Step Solution

Let's solve this step-by-step:

  • Step 1: Identify the given information.
    We know the rectangle ABCD ABCD , is divided by its diagonal AC AC . The length AB AB is 12 12 , and the diagonal AC AC is 13 13 .
  • Step 2: Apply Pythagorean theorem to find BC BC , which acts as the height.
    Using the Pythagorean theorem in ABC\triangle ABC gives us: AC=AB2+BC2 AC = \sqrt{AB^2 + BC^2} Given AC=13 AC = 13 and AB=12 AB = 12 , we set up the equation: 13=122+BC2 13 = \sqrt{12^2 + BC^2} Squaring both sides leads to: 169=144+BC2 169 = 144 + BC^2 BC2=169144=25 BC^2 = 169 - 144 = 25 Thus, BC=25=5 BC = \sqrt{25} = 5 .
  • Step 3: Calculate the area of ABC\triangle ABC.
    The area can be found using the formula: Area of ABC=12×AB×BC \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC =12×12×5 = \frac{1}{2} \times 12 \times 5 =12×60=30 = \frac{1}{2} \times 60 = 30

Therefore, the area of triangle ABC ABC is 30\boxed{30}.

Answer

30

Exercise #3

Look at the following rectangle:

AAABBBCCCDDD158

Calculate the perimeter of the triangle ABD.

Video Solution

Step-by-Step Solution

To solve the problem of finding the perimeter of triangle ABD, we will apply the following steps:

  • Step 1: Identify the given dimensions of the rectangle.
  • Step 2: Calculate the length of the diagonal BD using the Pythagorean theorem.
  • Step 3: Sum the sides of triangle ABD to find its perimeter.

Now, let's work through each step:

Step 1: We know from the problem that AB = 15 and AD = 8.

Step 2: The triangle ABD is a right triangle with AB and AD as the legs, and BD as the hypotenuse. Therefore, by the Pythagorean theorem:

BD2=AB2+AD2=152+82 BD^2 = AB^2 + AD^2 = 15^2 + 8^2

Calculating these squares gives:

BD2=225+64=289 BD^2 = 225 + 64 = 289

Taking the square root of both sides, we find:

BD=289=17 BD = \sqrt{289} = 17

Step 3: Now, calculate the perimeter of triangle ABD.

Perimeter of ABD=AB+AD+BD=15+8+17=40 \text{Perimeter of ABD} = AB + AD + BD = 15 + 8 + 17 = 40

Therefore, the perimeter of triangle ABD is 4040.

Answer

40

Exercise #4

ABCD is a rectangle.

AC = 13

AB = 12

Calculate the length of the side BC.

Video Solution

Step-by-Step Solution

When writing the name of a polygon, the letters will always be in the order of the sides:

This is a rectangle ABCD:

This is a rectangle ABDC:

Always go in order, and always with the right corner to the one we just mentioned.

Answer

5

Exercise #5

Given the rectangle such that:

O is the intersection point of the diagonals of the rectangle.

Given: AD=6 , AB=8

Calculate the length of the section BO.

AAABBBCCCDDDOOO86

Video Solution

Step-by-Step Solution

To solve this problem, we'll utilize the properties of rectangles and the Pythagorean theorem:

  • Step 1: Determine the full diagonal of the rectangle using Pythagorean theorem: AC=AB2+AD2AC = \sqrt{AB^2 + AD^2}.
  • Step 2: Since OO is the midpoint of the diagonal ACAC, BO=AC2BO = \frac{AC}{2}.

Now, let's calculate step-by-step:

Step 1: We know AB=8AB = 8 and AD=6AD = 6, therefore, using the Pythagorean theorem:

AC=AB2+AD2=82+62=64+36=100=10 AC = \sqrt{AB^2 + AD^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10

Step 2: Since the diagonals bisect each other, the length of BOBO is half of ACAC:

BO=AC2=102=5 BO = \frac{AC}{2} = \frac{10}{2} = 5

Therefore, the solution to the problem is BO=5 BO = 5 .

Answer

5

Exercise #6

Below is the rectangle ABCD.

O is the intersection point of the diagonals of the rectangle.

AD = 8

BO = 8.5

Calculate the area of the triangle ABD.

AAABBBCCCDDDOOO8.58

Video Solution

Step-by-Step Solution

According to the given information, we can claim that:

BD=2BO=8.5×2=17 BD=2BO=8.5\times2=17

Now let's look at triangle ABD to calculate side AB

AB2+AD2=BD2 AB^2+AD^2=BD^2

Let's input the known data:

AB2+82=172 AB^2+8^2=17^2

AB2=28964=225 AB^2=289-64=225

We'll take the square root

AB=15 AB=15

Now let's calculate the area of triangle ABD:

15×82=1202=60 \frac{15\times8}{2}=\frac{120}{2}=60

Answer

60

Exercise #7

Look at the following rectangle:

AAABBBCCCDDD54

DC = 4

AC = 5

Calculate the area of the rectangle ABCD.

Video Solution

Answer

12

Exercise #8

Below is the rectangle ABCD.

O is the intersection point of the diagonals of the rectangle.

DC = 15

OC = 8.5

Calculate the area of the rectangle ABCD.

AAABBBCCCDDDOOO8.515

Video Solution

Answer

120

Exercise #9

Given the rectangle ABCD when

O is the intersection point of the diagonals of the rectangle.

Given: BC=5 , BO=6.5

Calculate the length of the section AB.

AAABBBCCCDDDOOO6.55

Video Solution

Answer

12