Exercise 35 Page 861

Practice makes perfect

a We are asked to sketch a square pyramid with a base edge of

3

units using isometric dot paper.

To draw the prism we will follow the steps below.

Mark a corner of the solid.
Draw $3$ units left and right from the corner.
Draw another two sides of the base using a dashed line.
Find the center of the base. This point is the point where the altitude of the pyramid meets the base.
Draw few units up from the center of the base.
Connect the top vertex of the pyramid with the base vertices.
Add some colors and shadings, if desired. We will!

b We are asked to make a table with the lateral areas of the pyramid for the slant heights of

1,

3,

and

9

units. The lateral area of a regular pyramid is

L = \frac{1}{2} P ℓ .

The variable

P

is the perimeter of the base, and

ℓ

is the slant height of the pyramid. In our case the perimeter is

P = 4 \cdot 3 = 12

units.

Option	I	II	III
Base Perimeter	$P = 12$
Slant Height	$ℓ = 1$	$ℓ = 3$	$ℓ = 9$
Lateral Area	$L = \frac{1}{2} P ℓ$
Lateral Area	$L = \frac{1}{2} (12) (1) = 6$	$L = \frac{1}{2} (12) (3) = 18$	$L = \frac{1}{2} (12) (9) = 54$

c We are asked to describe what happens to the lateral area of the pyramid if the slant height is tripled. Let's analyze the table from Part A.

Slant Height	$1$	$3$	$9$
Lateral Area	$6$	$18$	$54$

Now, let's rewrite the lateral areas of the bigger pyramids in terms of the smaller ones.

Slant Height	$1$	$1 \cdot 3 = 3$	$3 \cdot 3 = 9$
Lateral Area	$6$	$6 \cdot 3 = 18$	$18 \cdot 3 = 54$

This can tells us that if the slant height is tripled, the lateral area of the pyramid is also tripled.

d We are asked to make a conjecture about how the lateral area of a square pyramid is affected if both the slant height and the base edge are tripled. From Part C we deduced the following.

Conclusion from Part C
If the slant height is tripled, then the lateral area is tripled.

From the formula for the lateral area, $L = \frac{1}{2} P ℓ,$ we can also conclude that if we triple the base edge the perimeter of the base is tripled and this tells us that the lateral area is tripled. Therefore, if both the slant height and the base edge are tripled, then the lateral area is multiplied by $3 \cdot 3 = 9 .$

Conjecture
If both the slant height and the base edge are tripled, then the lateral area is multiplied by $3 \cdot 3 = 9 .$

Now, let's check our conjecture for a few cases.

Option	I	II	III
Base Edge	$s = 1$	$s = 3$	$s = 9$
Base Perimeter	$P = 4 s$
Base Perimeter	$P = 4 \cdot 1 = 4$	$P = 4 \cdot 3 = 12$	$P = 4 \cdot 9 = 36$
Slant Height	$ℓ = 1$	$ℓ = 3$	$ℓ = 9$
Lateral Area	$L = \frac{1}{2} P ℓ$
Lateral Area	$L = \frac{1}{2} (4) (1) = 2$	$L = \frac{1}{2} (12) (3) = 18$	$L = \frac{1}{2} (36) (9) = 162$

Note that $\frac{1 8}{2} = \frac{1 6 2}{1 8} = 9 .$ Therefore, we were able to confirm the conjecture for the above examples.

Subchapter links

Check Your Understanding p.859

Practice and Problem Solving p.859-861

H.O.T. Problems p.861

Standardized Test Practice p.862

Spiral Review p.862

Skills Review p.862

Exercise 35 Page 861

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