38. A rectangle is inscribed in a semicircle with diameter 10 centimeters as shown. Express the area of the rectangle as a function of the height of the rectangle. (A) A(h) = 2h√√25-h² (C) A(h) = 2h√√5-n² (B)_A(h)=h√√10-h² (D) A(h) = 2h√√10-h² (E) A(h) = h√25-h²
38. A rectangle is inscribed in a semicircle with diameter 10 centimeters as shown. Express the area of the rectangle as a function of the height of the rectangle. (A) A(h) = 2h√√25-h² (C) A(h) = 2h√√5-n² (B)_A(h)=h√√10-h² (D) A(h) = 2h√√10-h² (E) A(h) = h√25-h²
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement:
A rectangle is inscribed in a semicircle with a diameter of 10 centimeters as shown in the diagram. Express the area of the rectangle as a function of the height (h) of the rectangle.
### Diagram Explanation:
The diagram shows a semicircle with diameter \( d = 10 \) cm. A rectangle is inscribed inside the semicircle. The height of the rectangle is labeled as \( h \). The base of the rectangle extends to the diameter line of the semicircle.
### Options:
(A) \( A(h) = 2h \sqrt{25 - h^2} \)
(B) \( A(h) = h \sqrt{10 - h^2} \)
(C) \( A(h) = 2h \sqrt{5 - h^2} \)
(D) \( A(h) = 2h \sqrt{10 - h^2} \)
(E) \( A(h) = h \sqrt{25 - h^2} \)
### Conceptual Breakdown:
To solve this problem, we need to:
1. Identify the relationship between the height \( h \) of the rectangle and the total area of the rectangle.
2. Use the Pythagorean theorem to express the width of the rectangle in terms of its height \( h \).
3. Formulate the area of the rectangle \( A \) as a function of \( h \).
**Steps:**
1. **Radius of Semicircle:**
Since the diameter \( d = 10 \) cm, the radius \( r = \frac{d}{2} = 5 \) cm.
2. **Width in terms of height using Pythagorean theorem:**
The width of the rectangle \( w \) is the horizontal leg of a right triangle whose hypotenuse is the radius \( r \) and the other leg is the height \( h \). Therefore, \( w^2 + h^2 = r^2 \).
Given \( r = 5 \), we have:
\[
w^2 + h^2 = 5^2
\]
\[
w^2 + h^2 = 25
\]
\[
w = \sqrt{25 - h^2}
\]
3. **Area of the rectangle as a function of height:**
The area \( A \) of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc716be-e066-4234-bd1a-1a86837575c3%2Faa61ed4d-916b-47c9-bd36-3fd56deee9a4%2Fpjdxvfk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A rectangle is inscribed in a semicircle with a diameter of 10 centimeters as shown in the diagram. Express the area of the rectangle as a function of the height (h) of the rectangle.
### Diagram Explanation:
The diagram shows a semicircle with diameter \( d = 10 \) cm. A rectangle is inscribed inside the semicircle. The height of the rectangle is labeled as \( h \). The base of the rectangle extends to the diameter line of the semicircle.
### Options:
(A) \( A(h) = 2h \sqrt{25 - h^2} \)
(B) \( A(h) = h \sqrt{10 - h^2} \)
(C) \( A(h) = 2h \sqrt{5 - h^2} \)
(D) \( A(h) = 2h \sqrt{10 - h^2} \)
(E) \( A(h) = h \sqrt{25 - h^2} \)
### Conceptual Breakdown:
To solve this problem, we need to:
1. Identify the relationship between the height \( h \) of the rectangle and the total area of the rectangle.
2. Use the Pythagorean theorem to express the width of the rectangle in terms of its height \( h \).
3. Formulate the area of the rectangle \( A \) as a function of \( h \).
**Steps:**
1. **Radius of Semicircle:**
Since the diameter \( d = 10 \) cm, the radius \( r = \frac{d}{2} = 5 \) cm.
2. **Width in terms of height using Pythagorean theorem:**
The width of the rectangle \( w \) is the horizontal leg of a right triangle whose hypotenuse is the radius \( r \) and the other leg is the height \( h \). Therefore, \( w^2 + h^2 = r^2 \).
Given \( r = 5 \), we have:
\[
w^2 + h^2 = 5^2
\]
\[
w^2 + h^2 = 25
\]
\[
w = \sqrt{25 - h^2}
\]
3. **Area of the rectangle as a function of height:**
The area \( A \) of
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