Which of the following is equivalent to the expression below?

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

Is 104.12 greater than 104.002

Liono4ka [1.6K]

Yes, 104.12 is greater than 104.002

because 104.12 rounds up to 104.1
and 104.002 rounds up to 104.0

8 0

3 years ago

Round to the nearest thousandth 0.416

Liula [17]

Here is the value of each decimal digit value

.4⇒ tenths place

.01⇒hundredths place

.006,thousandths place

So .416 is .416 rounded since there is nothing to round to

In other words, its already rounded to the nearest thousandths
Its kinda a trick question to make sure you know your place values

3 0

3 years ago

The designer sketched a model of the pond. The model drawing had a circumference of 24 inches. What is the AREA of the model she

Delicious77 [7]

Answer:

Area of model pond = 45.6 inch² (Approx.)

Step-by-step explanation:

Given:

Circumference of circular pond = 24 inches

Find:

Area of model pond

Computation:

Circumference of circle = 2πr

Circumference of circular pond = 2πr

24 = 2[22/7][r]

Radius r = [24 x 7] / [2 x 22]

Radius r = 3.81 inch (Approx.)

Area of circle = πr²

Area of model pond = πr²

Area of model pond = (22/7)(3.81)²

Area of model pond = [3.1428][14.5161]

Area of model pond = 45.6 inch² (Approx.)

6 0

3 years ago

The mass of a sheet of metal varies jointly with its area and its thickness. A sheet of metal of area 250cm2 and thickness 1mm h

noname [10]

The Formula of the relation is M = 8AT.

The mass of the metal sheet is 960 g.

The expression below shows the variation between the mass of the sheet (M), Area of the Sheet(A), and thickness (T).

Proportionality:

M ∝ AT

Removing the proportionality sign,

M = KAT

Where:

K = constant of proportionality.

make K the subject of the equation:

K = M/KT.................... Equation 1

From the question,

Given:

M = 200 g
A = 250 cm²
T = 1 mm = 0.1 cm

Susbtitute into equation 1

K = 200/(250×0.1)
K = 8 g/cm³

Formula:

M = 8AT................. equation 2

Hence the formula of the relation is M = 8AT

If,

A = 400 cm²
T = 3 mm = 0.3

Substitute these values into equation 2

M = 8(400)(0.3)
M = 960 g

Hence, the mass of the metal sheet is 960 g

Learn more about joint variation here: brainly.com/question/26047253

6 0

2 years ago