Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations

7 answers
Question:

Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.
Question 18 options:
552 units2
456 units2
360 units2
586 units2

Answers

340 units2 is the correct answer.

Step-by-step explanation:

340 units2

Step-by-step explanation:

draw perpendicular line from upper left corner of trapezoid to cut bottom line(=9) into two part having length 3 and 6. Now a triangle and rectangle are formed.

The triangle has base=3, angle 60 degrees and angle of 90 degrees.

Finding third angle:

180-60-90= 30

finding other two sides of triangle by using law of sines:

a/sinA=b/sinB

3/sin30=b/sin60

b=3sin60/sin30

b=3(0.866)/0.5

b=5.2

b is the perpendicular line we drew

a/sinA=b/sinB

3/sin30=c/sin90

c=3sin90/sin30

c=3(1)/0.5

c=6

c is the hypotenuse of the triangle

Now Finding the perimeter of base:

The perimeter of the base=sum of the lengths of sides.

p=6+9+6+5.2

 =26.2

area of trapezoid:

A= (a+b)h/2

  = (6+9)(5.2)/2

  =39

total surface area of prism:

TSA= ph + 2A

where h is height of prism = 10

TSA= 26.2(10) + 2(39)

       =340

Hence total surface area of the given prism is 340 units square!

correct answer is 456 sq units.

Step-by-step explanation:

Let us have a look at the formula for Surface Area of a prism:

[tex]A =p \times h+2 \times B[/tex]

Where p is the perimeter of base

h is the height of prism

and B is the base area of prism.

Given that:

h = 7.5 units

Hypotenuse of prism's base = 20 units

One of the Other sides = 12  units

Pythagorean theorem can be used to find the 3rd side of right angled base.

Square of hypotenuse = Sum of squares of other two sides

[tex]20^2=12^2+side^2\\\Rightarrow 400=144+side^2\\\Rightarrow side =\sqrt{256}\\\Rightarrow side =16\ units[/tex]

Area of base = area of right angled triangle:

[tex]B = \dfrac{1}{2} \times \text{Base Length} \times \text{Perpendicular Length}\\\Rightarrow B = \dfrac{1}{2} \times 16\times 12 = 96\ sq\ units[/tex]

Perimeter [tex]\times[/tex] height = (12+20+16) [tex]\times[/tex] 7.5 = (48) [tex]\times[/tex] 7.5 = 360 sq units

Now putting the values in formula:

Surface area, A = 360+96 = 456 sq units

So, correct answer is 456 sq units.

Option (4)

Step-by-step explanation:

Surface area of a prism = 2B + P×h

where B = Area of the triangular base

P = perimeter of the triangular base

h = height of the prism

B = [tex]\frac{1}{2}(\text{leg 1})(\text{leg 2})[/tex]

Since, (Hypotenuse)² + (Leg 1)² + (Leg 2)² [Pythagoras theorem]

(20)² = (12)² + (Leg 2)²

Leg 2 = [tex]\sqrt{400-144}[/tex]

          = 16 units

Therefore, B = [tex]\frac{1}{2}\times 12\times 16[/tex]

                     = 96 units²

P = 12 + 16 + 20

P = 48 units

h = 7.5 units

Surface area of the prism = 2(96) + (48×7.5)

                                           = 192 + 360

                                           = 552 units²

Therefore, surface area of the given triangular prism = 552 units²

Option (4) will be the answer.

Phillies game is on Friday

Step-by-step explanation:

The answer is "340 units²"

Step-by-step explanation:

In the given question, an attachment file is missing, which can be defined as follows:

[tex]\therefore \tan \theta= \frac{height}{base}\\\\\to \tan 60^{\circ}=\frac{h}{3}\\\\\to \sqrt{3}=\frac{h}{3}\\\\\to 3\sqrt{3}=h\\\\\to h = 3\sqrt{3} units\\\\[/tex]

The area of trapezoid prism:

[tex](a)= (b_1+b_2)h+ph[/tex]

value:

[tex]b_1=6 \ units\\b_2=9 \units\\h=10 \ units\\\\\to P=6+9+3\sqrt{3}+\sqrt{3^2+(3\sqrt{3})^2}\\\\\to P=15+3\sqrt{3}+\sqrt{9+27}\\\\\to P=15+3\sqrt{3}+\sqrt{36}\\\\\to P=21+3\sqrt{3} \ unit\\\\\bold{ Area= (6+9) \times 3\sqrt{3}+(21+3\sqrt{3})\times 10}\\\\\to Area= (15) \times 3\sqrt{3}+(21+3\sqrt{3})\times 10\\\\\to Area= 45\sqrt{3}+210+30\sqrt{3}\\\\\to Area=339.9038\\\\\to Area=340 \ units^2[/tex]


[tex]Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calcul[/tex]

The answer would be the 3rd one

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