Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs. Answer in cm^2
Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs. Answer in cm^2
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**
Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs.
*Answer in cm²*
**Detailed Explanation**
In the problem, we are asked to find the area of the largest rectangle that can fit inside a right triangle with its sides lying along the triangle's legs. The legs of the triangle are 4 cm and 6 cm.
To solve this problem, recognize that the largest rectangle inscribed in a right triangle, with its sides parallel to the legs, takes on a specific configuration:
- The rectangle will be positioned such that one vertex of the rectangle coincides with the right-angle vertex of the triangle.
- The opposite vertex will lie on the hypotenuse.
To determine the dimensions of this rectangle, we use the property of similarity in triangles. If we assume the dimensions of the rectangle are `x` and `y`, where `x` is along the 4 cm leg and `y` is along the 6 cm leg, the rectangle and the triangle will form two smaller, similar triangles. The ratios of the corresponding sides equal those of the original triangle:
\[
\frac{x}{4} = \frac{y}{6}
\]
From this equation, we can express one of the sides in terms of the other:
\[
y = \frac{6}{4}x = \frac{3}{2}x
\]
The area \(A\) of the rectangle is given by the product of its sides:
\[
A = x \cdot y = x \cdot \frac{3}{2}x = \frac{3}{2}x^2
\]
The maximum area occurs when this product is maximized under the constraint defined by the edges of the triangle:
\[
x + \frac{3}{2}x = 4 \\
\frac{5}{2}x = 4 \\
x = \frac{8}{5} = 1.6
\]
\[
y = \frac{3}{2}x = \frac{3}{2} \cdot \frac{8}{5} = \frac{24}{10} = 2.4
\]
Thus, the maximum area of the rectangle is:
\[
A = x \cdot y = 1.6 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6e87594-a0d7-48c0-af39-cad58e8108aa%2F2fc419ec-d7d8-4575-bf68-d906b2f22205%2Fqrj2m1o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs.
*Answer in cm²*
**Detailed Explanation**
In the problem, we are asked to find the area of the largest rectangle that can fit inside a right triangle with its sides lying along the triangle's legs. The legs of the triangle are 4 cm and 6 cm.
To solve this problem, recognize that the largest rectangle inscribed in a right triangle, with its sides parallel to the legs, takes on a specific configuration:
- The rectangle will be positioned such that one vertex of the rectangle coincides with the right-angle vertex of the triangle.
- The opposite vertex will lie on the hypotenuse.
To determine the dimensions of this rectangle, we use the property of similarity in triangles. If we assume the dimensions of the rectangle are `x` and `y`, where `x` is along the 4 cm leg and `y` is along the 6 cm leg, the rectangle and the triangle will form two smaller, similar triangles. The ratios of the corresponding sides equal those of the original triangle:
\[
\frac{x}{4} = \frac{y}{6}
\]
From this equation, we can express one of the sides in terms of the other:
\[
y = \frac{6}{4}x = \frac{3}{2}x
\]
The area \(A\) of the rectangle is given by the product of its sides:
\[
A = x \cdot y = x \cdot \frac{3}{2}x = \frac{3}{2}x^2
\]
The maximum area occurs when this product is maximized under the constraint defined by the edges of the triangle:
\[
x + \frac{3}{2}x = 4 \\
\frac{5}{2}x = 4 \\
x = \frac{8}{5} = 1.6
\]
\[
y = \frac{3}{2}x = \frac{3}{2} \cdot \frac{8}{5} = \frac{24}{10} = 2.4
\]
Thus, the maximum area of the rectangle is:
\[
A = x \cdot y = 1.6 \times
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