The base of a three-dimensional figure is bound by the circle x² + y² = 9. Vertical cross sections that are perpendicular to the y-axis are isosceles triangles with height equal to 3. Algebraically, find the area of each triangle. 54- O A(Y) = O A(y) = 3 V 9 O A(Y) = 2 19-y² O A(y) = √√√9-y²
The base of a three-dimensional figure is bound by the circle x² + y² = 9. Vertical cross sections that are perpendicular to the y-axis are isosceles triangles with height equal to 3. Algebraically, find the area of each triangle. 54- O A(Y) = O A(y) = 3 V 9 O A(Y) = 2 19-y² O A(y) = √√√9-y²
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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![**Finding the Area of Cross-Sectional Triangles in a Three-Dimensional Figure**
**Problem Statement:**
The base of a three-dimensional figure is bound by the circle \(x^2 + y^2 = 9\). Vertical cross-sections that are perpendicular to the y-axis are isosceles triangles with height equal to 3.
**Objective:**
Algebraically find the area of each triangle.
**Illustration:**
There is a diagram showing a circle on the \(xy\)-plane with the equation \(x^2 + y^2 = 9\). The circle is centered at the origin (0,0) with a radius of 3. The area within this circle is shaded.
**Graphical Details:**
The y-axis is labeled from -5 to 5, and the x-axis is labeled from -5 to 5. The circle intersects these axes at \(x = 3, x = -3, y = 3,\) and \(y = -3\).
Formatted as a multiple-choice question, the problem presents the following options for the area \(A(y)\) of each triangle:
1. \(A(y) = \sqrt{9 - y^2}\)
2. \(A(y) = 3\sqrt{9 - y^2}\)
3. \(A(y) = 2\sqrt{9 - y^2}\)
4. \(A(y) = \frac{1}{2} \sqrt{9 - y^2}\)
The solution involves choosing the correct mathematical expression for the area of the isosceles triangles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F883933e7-6864-477d-8bb2-712659e12b95%2F73d766ad-0b91-40dc-ae14-0218d18abdf1%2F6arsjta_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding the Area of Cross-Sectional Triangles in a Three-Dimensional Figure**
**Problem Statement:**
The base of a three-dimensional figure is bound by the circle \(x^2 + y^2 = 9\). Vertical cross-sections that are perpendicular to the y-axis are isosceles triangles with height equal to 3.
**Objective:**
Algebraically find the area of each triangle.
**Illustration:**
There is a diagram showing a circle on the \(xy\)-plane with the equation \(x^2 + y^2 = 9\). The circle is centered at the origin (0,0) with a radius of 3. The area within this circle is shaded.
**Graphical Details:**
The y-axis is labeled from -5 to 5, and the x-axis is labeled from -5 to 5. The circle intersects these axes at \(x = 3, x = -3, y = 3,\) and \(y = -3\).
Formatted as a multiple-choice question, the problem presents the following options for the area \(A(y)\) of each triangle:
1. \(A(y) = \sqrt{9 - y^2}\)
2. \(A(y) = 3\sqrt{9 - y^2}\)
3. \(A(y) = 2\sqrt{9 - y^2}\)
4. \(A(y) = \frac{1}{2} \sqrt{9 - y^2}\)
The solution involves choosing the correct mathematical expression for the area of the isosceles triangles.
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