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2 years ago

Find the area of quadrilateral ABCD. [Hint: the diagonal divides the quadrilateral into two triangles.]

Mathematics

1 answer:

Vadim26 [7]2 years ago

7 0

Answer:

B) 28.53 unit²

Step-by-step explanation:

The diagonal AD divides the quadrilateral in two triangles:

Triangle ABD
Triangle ACD

Area of Quadrilateral will be equal to the sum of Areas of both triangles.

i.e.

Area of ABCD = Area of ABD + Area of ACD

Area of Triangle ABD:

Area of a triangle is given as:

$Area = \frac{1}{2} \times base \times height$

Base = AB = 2.89

Height = AD = 8.6

Using these values, we get:

$Area = \frac{1}{2} \times 2.89 \times 8.6 = 12.43$

Thus, Area of Triangle ABD is 12.43 square units

Area of Triangle ACD:

Base = AC = 4.3

Height = CD = 7.58

Using the values in formula of area, we get:

$Area = \frac{1}{2} \times 4.3 \times 7.58 = 16.30$

Thus, Area of Triangle ACD is 16.30 square units

Area of Quadrilateral ABCD:

The Area of the quadrilateral will be = 12.43 + 16.30 = 28.73 units²

None of the option gives the exact answer, however, option B gives the closest most answer. So I'll go with option B) 28.53 unit²

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There are five sales agents in a certain real estate office. One month Andy sold twice as many properties as Ellen, Bob sold 3 m

Novosadov [1.4K]

Answer:

Carry

Step-by-step explanation:

Here let’s take any arbitrary value for the number of properties sold by Ellen.

Let the number of properties sold by Ellen he 10. We were told Andy sold twice of this, this means he sold 20 properties

Bob sold 3 more than Ellen, meaning he sold 13.

Carry sold twice of Bob meaning he sold 26

Dora sold the addition of Bob and Ellen = 13 + 10 = 23

Carry sold 26 and this makes him the highest seller.

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2 years ago

A karate center has 120 students. The director wants to set a goal to motivate her instructors to increase student enrollment. U

iVinArrow [24]

A) Plan A requires for a percentage increase of a number of students. This means that year after year the number of new students will increase. Plan B requires for a constant number of new students each year. This means that year after year the percentage increase would get smaller.

B) To solve this problem we will use formula for a growth of population:
$final = initial * (1+percentage)^{t}$
Where:
final = final number of students
initial = initial number of students
percentage = requested percentage increase
t = number of years

We can insert numbers and solve for t:
$240= 120* (1+0.12)^{t} \\ 2=1.12^{t} \\ log2=log(1.12^{t} ) \\ log2=t*log1.12 \\ t= \frac{log2}{log1.12} \\ t=6.12years$

For Plan B we can use simple formula
increase = 120
increase per year = 20
number of years = increase / (increase per year) = 120 / 20 = 6 years

Plan B is better to double the <span>enrollment.

C)We use same steps as in B) to solve this.
</span>

$360= 120* (1+0.12)^{t} \\ 3=1.12^{t} \\ log3=log(1.12^{t} ) \\ log3=t*log1.12 \\ t= \frac{log3}{log1.12} \\ t=9.69years$

For Plan B we can use simple formula
increase = 240
increase per year = 20
number of years = increase / (increase per year) = 240 / 20 = 12 years

Plan A is better to triple the enrollment.

3 0

2 years ago

The grade point average (GPA) of the students at Lakeview High School is normally distributed with a mean of 3.1 and a standard

Morgarella [4.7K]

Approximately 1718 have a score within that range.

We calculate the z-score for each end of this spectrum:

z = (X-μ)/σ = (2.5-3.1)/0.3 = -0.6/0.3 = -2

Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this z-score is 0.0228.

For the upper end:
z = (3.7-3.1)/0.3 = 0.6/0.3 = 2

Using a z-table, we see that the area to the left of, less than, this z-score is 0.9772.

The probability between these is given by subtracting these:
0.9772 - 0.0228 = 0.9544.

This means the proportion of people that should fall between these is 0.9544:
0.9544*1800 = 1717.92 ≈ 1718

4 0

1 year ago

A bucket that has a mass of 30 kg when filled with sand needs to be lifted to the top of a 30 meter tall building. You have a ro

Rom4ik [11]

Answer:

765 J

Step-by-step explanation:

We are given;

Mass of bucket = 30 kg

Mass of rope = 0.3 kg/m

height of building= 30 meter

Now,

work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand

Or W = W1 + W2

Work done in lifting the rope is given as,

W1 = Force x displacement

W1 = (30,0)∫(0.2x .dx)

Integrating, we have;

W1 = [0.2x²/2] at boundary of 30 and 0

W1 = 0.1(30²)

W1 = 90 J

work done in lifting the sand is given as;

W2 = (30,0)∫(F .dx)

F = mx + c

Where, c = 30 - 15 = 15

m = (30 - 15)/(30 - 0)

m = 15/30 = 0.5

So,

F = 0.5x + 15

Thus,

W2 = (30,0)∫(0.5x + 15 .dx)

Integrating, we have;

W2 = (0.5x²/2) + 15x at boundary of 30 and 0

So,

W2 = (0.5 × 30²)/2) + 15(30)

W2 = 225 + 450

W2 = 675 J

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,

W = 90 + 675

W = 765 J

5 0

1 year ago

A math exercise states that the circumference of a certain circle is 23.55 inches with diameter d and pi rounded to 3.14. Larry

irinina [24]

Answer:

None of them are correct.

Diameter of the Circle is 7.5 inches

Step-by-step explanation:

Given;

Circumference of the Circle = 23.55

$\pi = 3.14$

Formula:

$\textrm {Circumference of the Circle} = 2\pi r$

Substituting the values

$23.55= 2\times 3.14\times r$

$23.55=7.28r$

$\therefore r= \frac{23.55}{7.28}$

$\therefore r= 3.23489$

Now diameter is twice the radius

$diameter = 2r$

$\therefore diameter = 2 \times 3.23489$

Diameter = 7.5 inches

4 0

1 year ago