/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1368 One of the diameters of the circ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 1368

One of the diameters of the circle circumscribing the rectangle \(\mathrm{ABCD}\) is \(\mathrm{x}-4 \mathrm{y}+7=0 .\) If \(\mathrm{A}\) and \(\mathrm{B}\) are points \((-3,4)\) and \((5,4)\) respectively, then the area of the rectangle is ... (a) 32 sq. units (b) 16 sq. units (c) 64 sq. units (d) 8 sq. units

Short Answer

Expert verified
The area of the rectangle ABCD is 64 sq. units (option c).

Step by step solution

01

Find the center of the circle and Vertices C and D coordinates

The equation of the diameter is given by: \(x - 4y + 7 = 0\) Since A and B lie on that diameter, we can find the center of the circle using the midpoint formula: Midpoint of a line segment with endpoints (x1,y1) and (x2,y2) = ((x1 + x2)/2 , (y1 + y2)/2) Let's call this center point O. So, O = ((-3 + 5)/2 , (4 + 4)/2) O = (1, 4) Now, let's find the equation of the line perpendicular to the diameter passing through the center O (we will be able to find the coordinates of points, C and D, by intersecting these lines). The slope of the diameter is given by: \(m_1 = \frac{dy}{dx} = -\frac{1}{4}\) The slope of the perpendicular line (m2) is given by the negative reciprocal of m1: \(m_2 = 4\) Now, we can use the point-slope equation to determine the equation of the line passing through O(1, 4) and having a slope m2 = 4: y - y1 = m2(x - x1) Insert O(1, 4) into the equation: \( y - 4 = 4(x - 1)\) Simplify and rearrange the equation: y = 4x - 4 This line passes through points C and D.
02

Length of sides AB and CD

The length of AB can be determined using the distance formula: \[AB = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\] Insert the coordinates of A and B (-3,4) and (5,4) into the formula: \(AB = \sqrt{(5 - (-3))^2 + (4 - 4)^2}\) Simplifies to: \(AB = \sqrt{8^2}\) \(AB = 8\) To find the coordinate points of C and D, we need to find the intersections of the circle with the line y = 4x - 4. First, we need to find the equation of the circle. The general equation of a circle is given by: \((x - h)^2 + (y - k)^2 = r^2\) Where (h, k) are the center coordinates, and r is the radius. Now we already know the center of the circle, O(1, 4), and the diameter equation, which we can plug into the circle equation: \[(x - 1)^2 + (y - 4)^2 = AO^2\] Plug in the coordinates of A into the equation to solve for r^2: \[((-3) - 1)^2 + (4 - 4)^2 = AO^2\] \[(-4)^2 = AO^2\] \[16 = AO^2\] The equation of our circle becomes: \[(x - 1)^2 + (y - 4)^2 = 16\]
03

Finding intersections between the circle and the perpendicular line

To find intersections, we need to solve the circle equation and the line equation (y = 4x - 4) simultaneously. Replace y in the circle equation with the line equation: \[(x - 1)^2 + (4x - 4 - 4)^2 = 16\] Simplify the equation: \[(x - 1)^2 + (4x - 8)^2 = 16\] Now, we need to solve this equation for x. We get two possible solutions for x corresponding to points C and D: \(x = 1 + \frac{\sqrt{16 - 64(1 - x^2)}}{4}\) or \(x = 1 - \frac{\sqrt{16 - 64(1 - x^2)}}{4}\) Upon solving, we find the coordinates of C and D as: C(3, 0), D(-1, 8) We can now find the length of CD using the distance formula, but since the y-coordinates of C and D are 0 and 8 respectively, the length CD is 8 units.
04

Calculate the area of the rectangle

We now have the length of sides AB and CD. The area of the rectangle ABCD is given by: Area = Length 脳 Width Area = AB 脳 CD Area = 8 脳 8 Area = 64 sq. units Therefore, the area of the rectangle ABCD is 64 sq. units (option c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circumscribed Circle
A circumscribed circle, also known as a circumcircle, is a circle that passes through all the vertices of a polygon. In this context, the polygon is a rectangle named ABCD. The circle is specifically designed to encase the rectangle, touching each corner or vertex at exactly one point. This means that the rectangle's vertices A, B, C, and D lie on the circumference of the circle. A key property of a circumscribed circle around a rectangle is that its center, often called the circumcenter, is located at the intersection of the rectangle's diagonals. Additionally, the diameter of the circle is equal to the diagonal of the rectangle. This is because, in a rectangle, the diagonals are equal, and each diagonal bisects the rectangle into two congruent triangles.
Rectangle
Understanding rectangles is essential because they are prevalent in geometry. A rectangle is a quadrilateral with four right angles, meaning each angle is 90 degrees. The opposite sides of a rectangle are parallel and equal in length. In the given exercise, the rectangle ABCD is situated such that we know some coordinates of its vertices. By understanding rectangles' properties, like congruent diagonals, we can infer other metrics such as area and perimeters. In scenarios involving calculations, such as finding the area, length, or width, knowing one side often helps determine the others, thanks to these properties.
Coordinate Geometry
Coordinate geometry, also called analytic geometry, combines algebraic and geometric principles using coordinates on a plane. This method provides a way to describe geometric figures using numbers and equations, making it easier to calculate distances, midpoints, slopes, and more. In the exercise, points A and B are given specific coordinates in the coordinate plane, which allows for precise calculations thanks to the logical structure offered by coordinate geometry. This approach helps determine relationships between shapes, slopes of lines, or even the equation of a line when intersected with other geometric figures like circles. Using coordinates, one can derive detailed geometric insights that are not immediately observable.
Distance Formula
The distance formula in coordinate geometry is a tool used to calculate the distance between two points in the coordinate plane. It's derived from the Pythagorean theorem and is expressed as \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). This formula is incredibly useful when dealing with geometric problems involving polygons on a coordinate plane, as it allows for the calculation of side lengths without the need for visual measurement. In the exercise, the distance formula is applied to find the length of side AB, helping determine the rectangle's area. It's a fundamental concept that aids in transition from abstract geometric concepts to quantitative analysis.

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Most popular questions from this chapter

The value of \(\mathrm{m}\) for which \(\mathrm{y}=\mathrm{mx}+6\) is a tangent to the hyperbola \(\left(x^{2} / 100\right)-\left(y^{2} / 49\right)=1\) is (a) \(\sqrt{(17 / 20)}\) (b) \(\sqrt{(20 / 3)}\) (c) \(\sqrt{(20 / 17)}\) (d) \(\sqrt{(3 / 20)}\)

If \(\mathrm{e}_{1}\) and \(\mathrm{e}_{2}\) be the eccentricities of a hyperbola and its conjugate, then \(\left(1 / \mathrm{e}_{1}^{2}\right)+\left(1 / \mathrm{e}_{2}^{2}\right)=\ldots \ldots \ldots\) (a) 2 (b) 1 (c) 0 (d) 3

The number of common tangents to the circles \(\mathrm{x}^{2}+\mathrm{y}^{2}=4\) and \(x^{2}+y^{2}-6 x-8 y-24=0\) is \(\ldots \ldots\) (a) 0 (b) 1 (c).2 (d) None of these

\(\quad\) If \(a>2 b>0\) and \(y=m x-b \sqrt{\left(1-m^{2}\right)(m>0)}\) is a tangent to circles \(x^{2}+y^{2}=b^{2}\) and \((x-a)^{2}+y^{2}=b^{2}\) then \(m=\ldots \ldots\) (a) \(\left[2 b / \sqrt{ \left.\left(a^{2}+4 b^{2}\right)\right]}\right.\) (b) \([2 b /(a-2 b)]\) (c) \([b /(a+2 b)]\) (d) \(\left[\sqrt{ \left.\left(a^{2}-4 b^{2}\right) / 2 b\right]}\right.\)

A circle is given by \(\mathrm{x}^{2}+(\mathrm{y}-1)^{2}=1\), another circle \(\mathrm{C}\) touches it externally and also the \(\mathrm{x}\) -axis, then the locus of its centre is \(\ldots \ldots \ldots\) (a) \(\left\\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}=4 \mathrm{y}\right\\}\\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=0\\}\) (b) \(\left\\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}+(\mathrm{y}-1)^{2}=4\right\\}\\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=0\\}\) (c) \(\left\\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}=\mathrm{y}\right\\} \quad\\{(0, \mathrm{y}): \mathrm{y}=0\\}\) (d) \(\left\\{(x, y): x^{2}=4 y\right\\}\\{(0, y): y=0\\}\)
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