Question

A rectangular pyramid. The rectangular base has a length of 10 inches and a width of 6 inches. 2 triangular sides have a base of 10 inches and a height of 5.6 inches. 2 triangular sides have a base of 5 inches and a height of 7.1 inches.
The base of the rectangular pyramid shown has an area of square inches
A triangular face with a base of 10 inches has an area of square inches.
A triangular face with a base of 5 inches has an area of square inches.
The total surface area of the pyramid is square inches.

Answer 1

Answer:

hi

Step-by-step explanation:

A rectangular pyramid. The rectangular base has a length of 10 inches and a width of 6 inches. 2 triangular sides have a base of 10 inches and height of 5.6 inches. 2 triangular sides have a base of 5 inches and height of 7.1 inches.

The base of the rectangular pyramid shown has an area of  

✔ 50

square inches

A triangular face with a base of 10 inches has an area of  

✔ 28

square inches.

A triangular face with a base of 5 inches has an area of  

✔ 17.75

square inches.

The total surface area of the pyramid is  

✔ 141.5

square inches.

Answer 2

Answer:

Total surface area is $151.5in^{2}$

Step-by-step explanation:

This problem is on the mensuration of solid shapes, rectangular pyramid.

We can solve for the total surface area by solving the area of individual parts of the pyramid

1. The area of the rectangular base

$Area = Length * breadth$

$Area = 10*6= 60in^{2}$

2. Area of 1st triangular side

$area= \frac{1}{2}bh$

Areas = $\frac{1}{2} 10*5.6$

Area = $\frac{1}{2}*56= 28in^{2}$

for the two sides the area is = $28*2= 56in^{2}$

3. Area of 2nd triangular side

$area= \frac{1}{2} bh\\area= \frac{1}{2} 5*7.1\\area= \frac{1}{2} *35.5\\area= 17.75in^{2}$

for the two sides the area is $17.75*2= 35.5in^{2}$

Therefore the total surface area is

Areas of rectangular base plus 1st and 2nd triangular sides

Total surface area = $60+56+35.5= 151.5in^{2}$