Examples with solutions for Area of a Deltoid: Using additional geometric shapes

Exercise #1

In an amusement park with a rectangle shape, they decided to place part of the floor of its surface (referring to the shape of the deltoid).

The length of the tile is 3 meter and its width 2 meter.

The length of the garden is 10 meters and its width 6 meters.

Calculate how many tiles you will need to use to complete the deltoid shape.

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Video Solution

Step-by-Step Solution

Let's solve the problem step by step:

  • Calculate the area of the rectangular garden:
    The garden has a length of 10 meters and a width of 6 meters. Thus, the area is given by: Area of rectangle=10m×6m=60m2. \text{Area of rectangle} = 10 \, \text{m} \times 6 \, \text{m} = 60 \, \text{m}^2.

  • Consider the deltoid shape:
    The provided image suggests the deltoid is inscribed within the rectangle. If we assume the deltoid is two congruent triangles making up part of the rectangle, let's find the area of each triangle.

  • Area of each triangle of the deltoid:
    Assume two symmetrical triangles split the rectangle, each covering 30 m² (half of the rectangle). Hence, the deltoid area is the total area: Area of deltoid=602m2=30m2. \text{Area of deltoid} = \frac{60}{2} \, \text{m}^2 = 30 \, \text{m}^2.

  • Calculate the area of one tile:
    The tile dimensions are 3 meters by 2 meters, so the area is: Area of one tile=3m×2m=6m2. \text{Area of one tile} = 3 \, \text{m} \times 2 \, \text{m} = 6 \, \text{m}^2.

  • Determine the number of tiles needed:
    Divide the deltoid's area by the area of one tile: Number of tiles=30m26m2=5. \text{Number of tiles} = \frac{30 \, \text{m}^2}{6 \, \text{m}^2} = 5.

Therefore, the number of tiles needed to complete the deltoid shape is 5.

Answer

5

Exercise #2

A deltoid-shaped stage is to be built in a rectangular field.

The length of the field is 30 m and the width is 20 m.

Determine the area of the stage shaded in orange?

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Video Solution

Step-by-Step Solution

We can calculate the area of rectangle ABCD as follows:

20×30=600 20\times30=600

Now let's divide the deltoid along its length and width and add the following points:

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PMNK=PN×MK2=20×302=6002=300 PMNK=\frac{PN\times MK}{2}=\frac{20\times30}{2}=\frac{600}{2}=300

Answer

300 m

Exercise #3

Below is a deltoid with a length 2 times its width and an area equal to 16 cm².


Calculate x.

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Video Solution

Step-by-Step Solution

Given the problem, we are tasked to find the value of x x for a deltoid where the length is twice the width and the area is given. Let's proceed as follows:

  • Step 1: In this deltoid problem, the diagonals correspond to length 2x 2x and width x x . The formula for the area of a deltoid in terms of its diagonals is A=12×d1×d2 A = \frac{1}{2} \times d_1 \times d_2 .
  • Step 2: Substitute the values. Thus, the area 16=12×(2x)×x 16 = \frac{1}{2} \times (2x) \times x .
  • Step 3: Simplify the equation: 16=12×2x2=x2 16 = \frac{1}{2} \times 2x^2 = x^2 .
  • Step 4: Solve for x x : We find x2=16 x^2 = 16 , so x=16 x = \sqrt{16} .
  • Step 5: Conclude x=4 x = 4 .

Therefore, the solution to the problem is x=4 x = 4 .

Answer

x=4 x=4

Exercise #4

ABCD is a kite.

AB = AD

ABD has an area of 30 cm².
EC is equal to 6 cm.
AE is equal to 5 cm.

Calculate the area of the kite.

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Video Solution

Step-by-Step Solution

To solve this problem, we'll use the properties of triangle and kite areas:

Firstly, we note that the area of triangle ABDABD is given as 30cm230 \, \text{cm}^2.

We recognize that triangles ABDABD and ADCADC together form the kite with diagonal ACAC, and triangles ADBADB and BCDBCD form diagonals where two triangles area will be half the kite’s complete area.

Now, find triangle ABEABE: 12×AE×BD=30    12×5×BD=30    BD=12cm \frac{1}{2} \times AE \times BD = 30 \implies \frac{1}{2} \times 5 \times BD = 30 \implies BD = 12 \, \text{cm}

Next, calculate diagonal ACAC (sum of segments): AE+EC=AC    5+6=AC    AC=11cm AE + EC = AC \implies 5 + 6 = AC \implies AC = 11 \, \text{cm}

The area of kite ABCDABCD from diagonals ACAC and BDBD: Areakite=12×AC×BD=12×11×12=66cm2 \text{Area}_{kite} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 11 \times 12 = 66 \, \text{cm}^2

Therefore, the solution to the problem is 66 cm².

Answer

66 cm²

Exercise #5

ABCD is a kite.

BD is the diagonal of a square that has an area equal to 36 cm².

AC=2x AC=2x

Express the area of the kite in terms of X.

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Video Solution

Step-by-Step Solution

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 BDBD using the side length.
  • Step 3: Use the kite area formula with diagonals BDBD and AC=2xAC=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 ss, can be calculated by taking the square root of the area:

s=36=6 cms = \sqrt{36} = 6 \text{ cm}

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

BD=s2BD = s \sqrt{2}. Given s=6s = 6, we compute:

BD=62 cmBD = 6\sqrt{2} \text{ cm}

Step 3: Now, we apply the formula for the area of a kite, which is 12×d1×d2\frac{1}{2} \times d1 \times d2, where d1=BD=62d1 = BD = 6\sqrt{2} and d2=AC=2xd2 = AC = 2x:

The area AA of the kite is:

A=12×62×2x=62x cm2A = \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 xx is 62x6\sqrt{2}x cm².

Answer

62x 6\sqrt{2}x cm²

Exercise #6

Given the deltoid ABCD

and the deltoid AFCE whose area is 20 cm².

The ratio between AO and OC is 1:3

the angle ADC⦠. is equal to the angle ACD⦠.

AD is equal to 8

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Calculate the area of the triangle CEF

Video Solution

Answer

15 cm²

Exercise #7

Given the deltoid ABCD and the circle that its center O on the diagonal BC

Area of the deltoid 28 cm² AD=4

What is the area of the circle?

S=28S=28S=28444AAABBBDDDCCCOOO

Video Solution

Answer

49π 49\pi cm².

Exercise #8

ABCD is a deltoid, EFBD is a square whose area is 25 cm²

Given that GC is equal to 7 cm

Calculate the area of the deltoid.

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Video Solution

Answer

30 cm²

Exercise #9

ABCD is a kite.

O is the center of the circle whose diameter is DE and which has an area of 36π 36\pi cm².

The area of a circle whose radius is AE is 5 times greater than the area of the circle O.

EC=32AE EC=\frac{3}{2}AE

Calculate the area of the kite.

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Video Solution

Answer

1805 180\sqrt{5} cm²

Exercise #10

ABCD is a kite

ABED is a trapezoid with an area of 22 cm².

AC is 6 cm long.

Calculate the area of the kite.

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Video Solution

Answer

613 6\sqrt{13} cm²

Exercise #11

ABCD is a kite.

BC is the radius of a circle with an area of 4π 4\pi cm.

Calculate the area of the kite.

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Video Solution

Answer

33 3\sqrt{3} cm²

Exercise #12

ABCD is a parallelogram and BCEF is a kite.

EG=2 EG=2

GC=3 GC=3

Calculate the area of the kite

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Video Solution

Answer

1826 18\sqrt{2}-6 cm²

Exercise #13

ABCD is a kite.

The area of triangle BCD is equal to 20 cm².

Calculate the area of the kite.

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Video Solution

Answer

32 cm²

Exercise #14

ABCD is a deltoid with an area of 58 cm².

DB = 4

AE = 3

What is the ratio between the circles that have diameters formed by AE and and EC?

S=58S=58S=58333AAABBBCCCDDDEEE4

Video Solution

Answer

3:26

Exercise #15

Given the triangle ABC and the deltoid ADEF

The height of the triangle is 4 cm

The base of the triangle is greater by 2 than the height of the triangle.

Segment FD cut to the middle

Calculate the area of the deltoid ADEF

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Video Solution

Answer

8 cm²