a. Answer The expression for the perimeter is P = 6 w Explanation Given Information: The length l of the triangle is twice the breadth w . Formula: Perimeter P = 2 ( l + w ) Calculation: As the length is twice the breadth, So l = 2 w Thus P = 2 ( l + w ) = 2 ( 2 w + w ) = 2 ( 3 w ) Hence P = 6 w b. To determine To Make: An expression in terms of width w for the area. Answer The expression for the area is A=2 w 2 Explanation Given Information: The length l of the triangle is twice the width w . Formula: Area A = l w Calculation: As the length is twice the breadth, So l = 2 w Thus A = l w = ( 2 w ) w =2 w 2 c. To determine To Write: Perimeter and area for various values of width in the table. Answer Width (feet) 1 2 4 8 16 Perimeter (feet) 6 12 24 48 96 Area (square feet) 2 8 32 128 512 Explanation Given Information: The values of width 1, 2, 4, 8 and 16 Calculation: As we found in part (a), part (b) The perimeter and area are given respectively by P = 6 w and A = 2 w 2 So, for w = 1 , P = 6 ( 1 ) = 6 A = 2 ( 1 2 ) = 2 And for w = 2 , P = 6 ( 2 ) = 12 A = 2 ( 2 2 ) = 12 For w = 4 , P = 6 ( 4 ) = 24 A = 2 ( 4 2 ) = 32 And for w = 8 , P = 6 ( 8 ) = 48 A = 2 ( 8 2 ) = 128 w = 16 , P = 6 ( 16 d. To determine To Describe: How doubling the width affects perimeter and area. Answer On doubling the width, the perimeter becomes double and area becomes four times. Explanation Given Information: Perimeter P = 6 w and Area A = 2 w 2 , where w is the width. Calculation: When w changes to 2 w , perimeter becomes 6 ( 2 w ) = 2 ( 6 w ) = 2 P And area becomes ( ( 2 w ) 2 ) = 2 ( 4 w 2 ) = 4 ( 2 w 2 ) 4 A

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Question

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 2.3, Problem 36E

To determine

To Make: An expression in terms of width w for the perimeter.

Expert Solution

Answer to Problem 36E

The expression for the perimeter is P = 6w

Explanation of Solution

Given Information:

The length l of the triangle is twice the breadth w .

Formula:

Perimeter P = 2(l+w)

Calculation: As the length is twice the breadth,

So l = 2w

Thus P = 2(l+w)

= 2(2w+w)=2(3w)

Hence P = 6w

To determine

To Make: An expression in terms of width w for the area.

Expert Solution

Answer to Problem 36E

The expression for the area is A=2w2

Explanation of Solution

Given Information:

The length l of the triangle is twice the width w .

Formula:

Area A = lw

Calculation: As the length is twice the breadth,

So l = 2w

Thus A = lw

=(2w)w=2w2

To determine

To Write: Perimeter and area for various values of width in the table.

Expert Solution

Answer to Problem 36E

Width (feet)	1	2	4	8	16
Perimeter (feet)	6	12	24	48	96
Area (square feet)	2	8	32	128	512

Explanation of Solution

Given Information:

The values of width 1, 2, 4, 8 and 16

Calculation:

As we found in part (a), part (b)

The perimeter and area are given respectively by P = 6w and A=2w2

So, for

w = 1 , P = 6(1) = 6

A = 2(12) = 2

And for

w = 2 , P = 6(2) = 12A = 2(22) = 12

For

w = 4 , P = 6(4) = 24A = 2(42) = 32

And for

w = 8 , P = 6(8) = 48A = 2(82) = 128w = 16 , P = 6(16

To determine

To Describe: How doubling the width affects perimeter and area.

Expert Solution

Answer to Problem 36E

On doubling the width, the perimeter becomes double and area becomes four times.

Explanation of Solution

Given Information:

Perimeter P = 6w and Area A=2w2 , where w is the width.

Calculation:

When w changes to 2w , perimeter becomes 6(2w) = 2(6w) = 2P

And area becomes ((2w)2)=2(4w2)=4(2w2)

Chapter 2.3, Problem 35E

Chapter 2.3, Problem 37E

Chapter 2 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 2.1 - Prob. 1PSQ Ch. 2.1 - Prob. 2PSQ Ch. 2.1 - Prob. 3PSQ Ch. 2.1 - Prob. 4PSQ Ch. 2.1 - Prob. 5PSQ Ch. 2.1 - Prob. 6PSQ Ch. 2.1 - Prob. 7PSQ Ch. 2.1 - Prob. 8PSQ Ch. 2.1 - Prob. 9PSQ Ch. 2.1 - Prob. 10PSQ

Ch. 2.1 - Prob. 11PSQ Ch. 2.1 - Prob. 1C Ch. 2.1 - Prob. 2C Ch. 2.1 - Prob. 3C Ch. 2.1 - Prob. 4C Ch. 2.1 - Prob. 5C Ch. 2.1 - Prob. 6C Ch. 2.1 - Prob. 7C Ch. 2.1 - Prob. 8C Ch. 2.1 - Prob. 9C Ch. 2.1 - Prob. 10C Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - 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