Question

Jennifer got a box of chocolates. the box is a right triangular prism shaped box. it is 7 in long. and the triangular base measure 2in times 3in times 4in. what is the surface area of the box of chocolates?

Answer 1

Answer:

$A = 68.8 inches^{2}$

Step-by-step explanation:

Given : Jennifer got a box of chocolates. the box is a right triangular prism shaped box. it is 7 in long. and the triangular base measure 2in times 3in times 4in.

a = 2

b=3

c=4

To Find : what is the surface area of the box of chocolates?

Solution: The first thing we should know is the area of the triangular base.

 Triangular base area: 

$A1 =\sqrt{(s * (s-a) * (s-b) * (s-c)) }$

 Where, s is the semi-meter of the triangle:

$s =\frac{ (a + b + c) }{ 2}$

 a, b, c: sides of the triangle.

 Substituting:

 s = (2 + 3 + 4) /2=4.5

$A1 = \sqrt{ (4.5 * (4.5-2) * (4.5-3) * (4.5-4))}$

 A1 = 2.90

 Then, you must know the area of each rectangle associated with each side of the triangular base.

 Rectangle area 1: Ar1 = (a) * (l)

                              Ar1 = (2) * (7)

                              Ar1 = 14

 Rectangular area 2:   Ar2 = (b) * (l)

                                    Ar2 = (3) * (7)

                                    Ar2 = 21

 Rectangular area 3:   Ar3 = (a) * (l)

                                    Ar3 = (4) * (7)

                                    Ar3 = 28

 Finally the surface area is: A = Ar1 + Ar2 + Ar3 + 2 * A1

                                             A = (14) + (21) + (28) + 2 * (2.90)

                                            $A = 68.8 inches^{2}$

 Hence the surface area of the box of chocolates is 

$A = 68.8 inches^{2}$