BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 1, Problem 2P
To determine

To express: The length of the hypotenuse as a function of the perimeter.

Expert Solution

Answer to Problem 2P

Solution:

The length of the hypotenuse is h=P22P+24.

Explanation of Solution

Given:

The altitude perpendicular to the hypotenuse a=12 cm

Formula used:

Area of the triangleA=12(xy) (1)

Pythagoras theorem x2+y2=h2 (2)

Where, h is a hypotenuse, and x, y is the sides of triangle

Perimeter of the triangle P=x+y+h (3)

(x+y)2=x2+y2+2xy

Calculation:

Consider the triangle as shown below in the Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 1, Problem 2P

From the Figure 1, A=12(ah)

Substitute a=12 in above equation,

A12(12h)=6h

From the equation (1),

A= 12(xy)

This implies xy=2A (4)

Consider the equation (x+y)2=x2+y2+2xy

Substitute the values of equation (1) and equation (2) in above equation,

(x+y)2=h2+4A (5)

From equation (3), x+y=Ph

Substitute the value of x+y in equation (4),

(Ph)2=h2+4AP2+h22Ph=h2+4A2Ph=P24Ah=P24A2P

Substitute A=6h in above equation,

h=P24(6h)2P2Ph=P224h2Ph+24h=P2h(2P+24)=P2

On further simplification, the value of h=P22P+24

Therefore, the length of the hypotenuse is h=P22P+24.

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