(a) Answer 1:24 Explanation Given: A set of nested tables. The surface of the tables will be 45 ∘ − 45 ∘ − 90 ∘ triangle . Each of the two congruent edges of the surface of the smallest table has a length of 24 inches. First draw smallest triangle. Let the scale of congruent side be 1:24. 1 unit = 24 in So, scale drawing would be (b) To determine To find: The length of two congruent edges of the surface of the largest triangle. Answer 48 in Explanation Given: A set of nested tables. The surface of the tables will be 45 ∘ − 45 ∘ − 90 ∘ triangle. The ratio of an edge length of the surface of the smallest table to a corresponding edge of the largest table is 1:2. The length of congruent edges of the surface of the smallest table is 24 in. The ratio of an edge length of the surface of the smallest table to a corresponding edge of the largest table is 1:2. Let the length of the surface of the largest table be x in. So, the ratio is 24: x which is equal to 1:2 Therefore, 24 x = 1 2 ⇒ x = 48 in (C) To determine To find: The length of two congruent edges of the surface of the middle sized table. Answer 36 in Explanation Given: A set of nested tables. The surface of the tables will be 45 ∘ − 45 ∘ − 90 ∘ triangle. The ratio of an edge length of the surface of the middle sized table to smallest table is 3:2. The length of congruent edges of the surface of the smallest table is 24 in. The ratio of an edge length of the surface of the middle sized table to smallest table is 3:2. Let the length of edge of the surface of themiddle sized table be x in. So, the ratio is x :24 which is equal to 3:2 Therefore, x 24 = 3 2 ⇒ x = 36 in (d) To determine To find: The length of thirdedge of the surface of the each table. Answer Smallest table = 34 in Middle size table = 51 in Largest table = 68 in Explanation Given: A set of nested tables. The surface of the tables will be 45 ∘ − 45 ∘ − 90 ∘ triangle. The scale of the each table. So, the ratio of the congruent sides and third side is 1 : 1 : 2 For smallest table, length of third side x, 1 2 = 24 x ⇒ x ≈ 34 in For middle sized table, length of third side y , 1 2 = 36 y ⇒ y ≈ 51 in For largest table, length of third side z , 1 2 = 48 z ⇒ y ≈ 68 in

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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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 10.1, Problem 23E

(a)

To determine

To find: The scale of drawing of the surface of the smallest table.

(a)

Expert Solution

Answer to Problem 23E

1:24

Explanation of Solution

Given: A set of nested tables. The surface of the tables will be 45∘−45∘−90∘ triangle.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 1

Each of the two congruent edges of the surface of the smallest table has a length of 24 inches.

First draw smallest triangle.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 2

Let the scale of congruent side be 1:24.

1 unit = 24 in

So, scale drawing would be

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 3

(b)

To determine

To find:The length of two congruent edges of the surface of the largest triangle.

(b)

Expert Solution

Answer to Problem 23E

48 in

Explanation of Solution

Given: A set of nested tables. The surface of the tables will be 45∘−45∘−90∘ triangle.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 4

The ratio of an edge length of the surface of the smallest table to a corresponding edge of the largest table is 1:2.

The length of congruent edges of the surface of the smallest table is 24 in.

The ratio of an edge length of the surface of the smallest table to a corresponding edge of the largest table is 1:2.

Let the length of the surface of the largest table be x in.

So, the ratio is 24: x which is equal to 1:2

Therefore, 24x=12⇒x=48 in

(C)

To determine

To find: The length of two congruent edges of the surface of the middle sized table.

(C)

Expert Solution

Answer to Problem 23E

36 in

Explanation of Solution

Given: A set of nested tables. The surface of the tables will be 45∘−45∘−90∘ triangle.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 5

The ratio of an edge length of the surface of the middle sized table to smallest table is 3:2.

The length of congruent edges of the surface of the smallest table is 24 in.

The ratio of an edge length of the surface of the middle sized table to smallest table is 3:2.

Let the length of edge of the surface of themiddle sized table be x in.

So, the ratio is x :24 which is equal to 3:2

Therefore, x24=32⇒x=36 in

(d)

To determine

To find: The length of thirdedge of the surface of the each table.

(d)

Expert Solution

Answer to Problem 23E

Smallest table = 34 in

Middle size table = 51 in

Largest table = 68 in

Explanation of Solution

Given: A set of nested tables. The surface of the tables will be 45∘−45∘−90∘ triangle.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 6

The scale of the each table.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10.1, Problem 23E , additional homework tip 7

So, the ratio of the congruent sides and third side is 1:1:2

For smallest table, length of third side x, 12=24x⇒x≈34 in

For middle sized table, length of third side y , 12=36y⇒y≈51 in

For largest table, length of third side z , 12=48z⇒y≈68 in

Chapter 10.1, Problem 22E

Chapter 10.1, Problem 24E

Chapter 10 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 10.1 - Prob. 1C Ch. 10.1 - Prob. 2C Ch. 10.1 - Prob. 3C Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E

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