Question

ASAP !!!!! (05.06; 05.07 MC)

Answer 1

Answer:

Part A: Correct net is A

Part B: AB = 3 in; BC = 5 in; CD = 8.6 in

Part C: Total surface of prism is 115.2 $in^2$

Step-by-step explanation:

Part A :

The only net that works is Net A, because it would give proper coverage and closure for the triangular bases of the prism. The other option has one of the coverage for the triangular base in the wring location.

Part B:

Side AB should be in accord with the 3 in side of the right triangular base.

Side BC should agree with the 5 in hypotenuse of the triangular base.

Side CD should be in agreement with the 8.6 height of the prism.

Part C:

The surface area of the prism is the addition of the surface area of three rectangles (the lateral faces of the prism) plus the surface are of two right angle triangles (the bases of the prism). It can be calculated as shown below [notice area of all rectangles are calculated using Area = base x height]

Smallest rectangle: 8.6 in x 3 in = 25.8 $in^2$

Medium size rectangle: 8.6 in x 4 in = 34.4 $in^2$

Larger rectangle: 8.6 in x 5 in = 43 $in^2$

Area of each base (formula for triangle area = base x height /2) :

3 in x 4 in /2 = 12/2 $in^2$ = 6 $in^2$

Therefore the sum of two of this gives: 12 $in^2$

Now, the sum of all of the above gives:

Total surface area of the prism: (25.8 + 34.4 + 43 + 12) $in^2$ = 115.2 $in^2$

Answer 2

Part A: Correct net is A

Part B: AB = 3 in; BC = 5 in; CD = 8.6 in

Part C: Total surface of prism is 115.2 in^2