Substituting the lengths of the diagonals AB = √14 and CD = 4 into the area formula, we get Area = (1)/(2) × √14 × 4 = 2√14.

Let the line $x + y = 1$ meet the circle $x^2 + y^2 = 4$ at the points $A$ and $B$ . If the line perpendicular to $AB$ and passing through the midpoint of the chord $AB$ intersects the circle at $C$ and $D$ , then the area of the quadrilateral $ADBC$ is equal to:

Answer: $2\sqrt{14}$

Step-by-step solution

Step 1: Find the midpoint of chord AB

The given circle is $x^2 + y^2 = 4$ , which means its center is at the origin $(0{,}0)$ and its radius is $r=2$ . The line $x+y=1$ is a chord of this circle. The line perpendicular to the chord $AB$ and passing through its midpoint $M$ must pass through the center of the circle. This is a fundamental property of circles: the perpendicular bisector of any chord passes through the center. Since the line $x+y=1$ has a slope of $-1$ , the perpendicular line passing through the center $(0{,}0)$ will have a slope of $1$ . The equation of this perpendicular line is $y=x$ . The midpoint $M$ of the chord $AB$ is the intersection of the line $x+y=1$ and the line $y=x$ . Substituting $y=x$ into $x+y=1$ gives $x+x=1$ , so $2x=1$ , which means $x=1/2$ . Therefore, $y=1/2$ , and the midpoint $M$ is $(1/2, 1/2)$ .

Step 2: Calculate the length of chord AB

The distance from the center $(0{,}0)$ to the midpoint $M(1/2, 1/2)$ is $OM = \sqrt{(1/2)^2 + (1/2)^2} = \sqrt{1/4 + 1/4} = \sqrt{1/2} = 1/\sqrt{2}$ . We can use the Pythagorean theorem in triangle $OMA$ , where $OA$ is the radius $r=2$ . So, $AM^2 = OA^2 - OM^2 = 2^2 - (1/\sqrt{2})^2 = 4 - 1/2 = 7/2$ . Thus, $AM = \sqrt{7/2}$ . The length of the chord $AB$ is $2 \times AM = 2 \times \sqrt{7/2} = \sqrt{4 \times 7/2} = \sqrt{14}$ .

Step 3: Identify the quadrilateral ADBC

The line perpendicular to $AB$ passing through its midpoint $M$ also passes through the center of the circle. This line intersects the circle at points $C$ and $D$ . Since $AB$ is a chord and $CD$ is a diameter (because it passes through the center and is perpendicular to $AB$ ), the quadrilateral $ADBC$ has diagonals $AB$ and $CD$ that are perpendicular bisectors of each other. This means $ADBC$ is a rhombus. Furthermore, since $CD$ is a diameter, $C$ and $D$ are diametrically opposite points. The line $CD$ is the line $y=x$ . The points $C$ and $D$ are the intersections of $y=x$ and $x^2+y^2=4$ . Substituting $y=x$ gives $x^2+x^2=4$ , so $2x^2=4$ , $x^2=2$ , $x = \pm \sqrt{2}$ . So $C = (\sqrt{2}, \sqrt{2})$ and $D = (-\sqrt{2}, -\sqrt{2})$ . The length of the diagonal $CD$ is $2r = 4$ .

Step 4: Calculate the area of the rhombus

The area of a rhombus is given by half the product of its diagonals. Here, the diagonals are $AB$ and $CD$ . We found $AB = \sqrt{14}$ and $CD = 4$ .

Step 5: Substitute values and find the area

Substituting the lengths of the diagonals $AB = \sqrt{14}$ and $CD = 4$ into the area formula, we get $Area = \frac{1}{2} \times \sqrt{14} \times 4 = 2\sqrt{14}$ .

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Let the line x+y=1x + y = 1x+y=1 meet the circle x2+y2=4x^2 + y^2 = 4x2+y2=4 at the points AAA and BBB. If the line perpendicular to ABABAB and passing through the midpoint of the chord ABABAB intersects the circle at CCC and DDD, then the area of the quadrilateral ADBCADBCADBC is equal to: