Question

WILL GIVE BRAINLIEST IF YOU HELP! 20 POINTS!

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