Answer Length = 12 cm, Width = 6 cm and Height = 18 cm Explanation Given Information: Surface Area = 792 cm 2 , Length l = 2 × w i d t h , Height h = 3 × w i d t h Formula Used: Surface Area of Cuboid = 2 × ( L e n g t h × B a s e + B a s e × H e i g h t + H e i g h t × L e n g t h ) Calculation: Let length = l Width = w Height = h. So according to given information, Length l = 2 × w i d t h Height h = 3 × w i d t h Surface Area = 792 cm 2 Surface Area of Solid = 2 × ( L e n g t h × B a s e + B a s e × H e i g h t + H e i g h t × L e n g t h ) Put the value of Area, Length, Base and Height to find dimensions 792 cm 2 = 2 × ( 2 w × w + w × 3 w + 3 w × 2 w ) cm 2 So, 792 cm 2 = 2 × ( 2 w 2 + 3 w 2 + 6 w 2 ) cm 2 So, 792 cm 2 = 22 w 2 cm 2 Hence, w 2 = 792 22 So, w 2 = 36 Hence, Width w = 6 cm Length l = 2 × w i d t h Length l = 2 × 6 = 12 cm Height h = 3 × w i d t h Height h = 3 × 6 = 18 cm Length = 12 cm, Width = 6 cm and Height = 18 cm

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Question

Chapter 10.5, Problem 31E

To determine

To findthe dimensions.

Expert Solution & Answer

Answer to Problem 31E

Length = 12 cm, Width = 6 cm and Height = 18 cm

Explanation of Solution

Given Information: Surface Area = 792 cm², Length l = 2×width , Height h = 3×width

Formula Used:Surface Area of Cuboid = 2×(Length×Base + Base×Height + Height×Length)

Calculation:

Let length = l

Width = w

Height = h.

So according to given information,

Length l = 2×width

Height h = 3×width

Surface Area = 792 cm²

Surface Area of Solid = 2×(Length×Base + Base×Height + Height×Length)

Put the value of Area, Length, Base and Height to find dimensions

792 cm²= 2×(2w×w + w×3w+ 3w×2w) cm²

So,

792 cm²= 2×(2w2+ 3w2+ 6w2) cm²

So,

792 cm²= 22w2 cm²

Hence,

w2=79222

So,

w2=36

Hence,

Width w=6 cm

Length l = 2×width

Length l = 2×6 = 12 cm

Height h = 3×width

Height h = 3×6 = 18 cm

Length = 12 cm, Width = 6 cm and Height = 18 cm

Chapter 10.5, Problem 30E

Chapter 10.6, Problem 1C

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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