Question

WILL GIVE BRAINLIEST IF YOU HELP! 20 POINTS!
David and his dad were going camping. When they went into the garage to get their tent they noticed it was ripped. David’s dad said he could fix the tent. He would buy some new material to cover the one rectangular side that ripped.

How much material does he need to cover the one rectangular side of the tent with the rip? Justify your answer using equations/formulas, models, and/or words to explain your mathematical reasoning.
Answer: To figure out how much material David and his dad need to cover we must figure out the area of where the damage is located. First off let’s figure out the dimensions of the 2 rectangles.
Their length is 9 ft and their width is 6 ft. The other rectangle, the base, has a width of 7 and a length of 9.
To begin solving we must know the formula of finding the area of a rectangle and that would be A = lw.
A = 9 * 6 = 54 square ft.


If David’s dad wanted to re-cover the whole tent including the bottom, how much material would he need? Justify your answer using equations/formulas, models, and/or words to explain your mathematical reasoning.
Answer: To figure out this question we must find the area of each side of the tent. Both triangles have the same dimensions. A height of 6 ft, a base of 7 ft and a side of 6 ft. Of the 2 rectangles on the side they are also the sam. Length of 9 ft and width of 6 ft. The other being the base which has a width of 7 and a length of 9.
base rectangle area; 9 ft * 7 ft = 63 square feet.
triangle area; (6 ft * 7 ft) / 2 = 21 square feet * 2 = 42 square feet.
We already solved the area of the side rectangles in part A so now all we have to do is multiple it by 2 to get the area of both sides.
54 square feet * 2 = 108 square feet
and then we get the total by adding them all together.
63 sqft + 42 sqft + 108 sqft = 213 sqft


( this one is the last part i need help with! ^^ )
What is the volume of the tent? Justify your answer using equations/formulas, models, and/or words to explain your mathematical reasoning. (You need to find the area of one of the triangular bases, and then you can take that measurement and multiply it with the height of the entire prism.) V=Bh, where B = area of one of the triangular bases
Answer:

Answer 1

Answer:

Part 1) $54\ ft^{2}$

Part 2) The surface area is $SA=(7\frac{\sqrt{95}}{2}+171)\ ft^{2}$  or  $SA=205.11\ ft^{2}$

Part 3) $V=63\frac{\sqrt{95}}{4}\ ft^{3}$ or  $V=153.51\ ft^{3}$

Step-by-step explanation:

Part 1) How much material does he need to cover the one rectangular side of the tent with the rip?

we know that

The area of a rectangle is equal to

$A=LW$

we have

$L=9\ ft$

$W=6\ ft$

substitute the values

$A=(9)(6)=54\ ft^{2}$            

Part 2) If David’s dad wanted to re-cover the whole tent including the bottom, how much material would he need?

we know that

The surface area of a triangular prism (the tent)  is equal to

$SA=2B+PL$

where

B is the area of the triangular face

P is the perimeter of the triangular face

L is the length of the triangular prism        

Find the area of the triangular face B

Note The height of the triangle cannot be equal to 6 ft. The height of a right triangle cannot be equal to the hypotenuse. The height must be calculated by applying Pythagoras theorem

$B=\frac{1}{2}bh$  

we have

$b=7\ ft$

Find the height applying the Pythagoras Theorem

substitute the values

$h^{2}=6^{2}-(7/2)^{2}$

$h^{2}=95/4$

$h=\frac{\sqrt{95}}{2}\ ft$

substitute

$B=\frac{1}{2}(7)(\frac{\sqrt{95}}{2})$  

$B=7\frac{\sqrt{95}}{4}\ ft^{2}$  

Fin the perimeter of the triangular base P

$P=6+6+7=19\ ft$

Find the surface area SA

$SA=2(7\frac{\sqrt{95}}{4})+(19)(9)$

$SA=(7\frac{\sqrt{95}}{2}+171)\ ft^{2}$ -----> exact value

or

$SA=205.11\ ft^{2}$ -----> approximate value

Part 3) What is the volume of the tent?

we know that

The volume of the tent is equal to

$V=BL$

where

B is the area of the triangular face

L is the length of the tent

we have

$B=7\frac{\sqrt{95}}{4}\ ft^{2}$  

$L=9\ ft$

substitute

$V=(7\frac{\sqrt{95}}{4})(9)$

$V=63\frac{\sqrt{95}}{4}\ ft^{3}$ -----> exact value

or

$V=153.51\ ft^{3}$ -----> approximate value