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sdas [7]
4 years ago
11

TIMED HELP PLEASE!!!! What is the area of triangle RST? ____ square units

Mathematics
2 answers:
Vikentia [17]4 years ago
8 0
The base must be perpendicular to the height for area= bh/2.

We can use RS as the base and TU as the height.
6*3/2
18/2
9

Final answer: 9 units^2
Alexxandr [17]4 years ago
6 0

Answer:

The coordinate of triangle RST from the figure are;

R = (-3,2), S=(3,2) and T=(-1,-1). also the coordinate of U = (-1, 2).

Distance Formula: It is used to determine the distance between two points with the coordinates (x_{1},y_{1}) and (x_{2},y_{2}) i.e,

Distance = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Now, using above formula to find the sides of a given triangle:

Calculate the length of RS , where R=(-3,2) and S=(3,2);

RS=\sqrt{(3-(-3))^2+(2-2)^2} or

RS=\sqrt{(3+3)^2+(0)^2}

Simplify: we get

RS=\sqrt{36} =6 unit.

Similarly, for TU, where T=(-1.-1) and U=(-1,2).

then:

TU=\sqrt{(-1-(-1))^2+(2-(-1))^2} or

TU=\sqrt{(-1+1)^2+(2+1)^2} or

Simplify:

TU=\sqrt{0+9} =\sqrt{9} =3 unit.

Since, we have to calculate the Area of triangle RST.

To, find the area of a triangle, multiply the base by the height and then divide it by 2.

i.e,

Area of triangle = \frac{b\cdot h}{2} where b is the base and h is the height of the triangle.

Here, in the given triangle RST, the base of the triangle = RS and the height of the triangle= TU.

Area of \triangle RST = \frac{RS \cdot TU}{2}

Substitute the value of RS = 6 unit and TU= 3 unit in the above formula;

Area of \triangle RST = \frac{6\cdot3}{2}

Simplify:

Area of \triangle RST=3\cdot 3=9 square unit.

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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman
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Answer:

W\approx 8.72 and L\approx 15.57.

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})

\text{Area}=W\cdot L+\frac{\pi}{8}W^2

Now, we will solve for L is terms W from perimeter equation as:

L=38-(W+\frac{\pi }{2}W)

Substitute this value in area equation:

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2

Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

A'=38-2W-\pi W+\frac{\pi}{4}W    

A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

W=8.723210

W\approx 8.72

Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

L\approx 15.57

Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

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