Exercise 36 Page 714

We are given a regular square pyramid with the surface area of 182 square centimeters, and the lateral area of 118 square centimeters. Note that the surface area of a pyramid and the lateral area differ only by the area of the base of the pyramid. Therefore, the area of the base is 64 square centimeters. The base is a square with the length of a side of the base s. Now, let's use the formula for the area of a square.

Therefore, s=8 centimeters.

The lateral area of the regular pyramid is given by the formula L.A.= 12Pl, where P is the perimeter of the base. Since the base is a square with the length of side s=8 centimeters, its perimeter is P=4* 8=32 inches. We can use this information to find the slant height, l.

L.A.=1/2Pl

SubstituteValues

Substitute values

118=1/2( 32)l

▼

Solve for l

Multiply

Multiply

118=1/2* 32l

MoveRightFacToNumOne

1/b* a = a/b

118=32l/2

CalcQuot

Calculate quotient

118=16l

RearrangeEqn

Rearrange equation

16l=118

DivEqn

.LHS /16.=.RHS /16.

l=118/16

CalcQuot

Calculate quotient

l=7.375

RoundDec

Round to 1 decimal place(s)

l≈ 7.4

Therefore, the slant height of the pyramid is about 7.4 centimeters. To find the height of the pyramid, h, let's analyze △ ABC.

Segment AC is the height of the pyramid, BC is the slant height, and AB is half of the base edge. This tells us that AB= 82=4 centimeters. Now, let's use the Pythagorean Theorem for right triangle △ ABC to find the height, h.

AB^2+AC^2=BC^2

SubstituteValues

Substitute values

4^2+ h^2= 7.4^2

▼

Solve for h

CalcPow

Calculate power

16+h^2=54.76

SubEqn

LHS-16=RHS-16

h^2=38.76

SqrtEqn

sqrt(LHS)=sqrt(RHS)

h=sqrt(38.76)

UseCalc

Use a calculator

h=6.225752...

RoundDec

Round to 1 decimal place(s)

h≈ 6.2

Finally, the height of the pyramid is about 6.2 centimeters.