Example Question - circle geometry
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Tangent and Secant Angle Relationships in a Circle
<p>Consider triangle AEB and AED:</p> <p>\angle AEB = \angle AED = 90^{\circ} \text{ (angle formed by a tangent and a chord is a right angle)}</p> <p>Consider triangle ABE and the exterior angle \angle EAX:</p> <p>\angle EAB + \angle ABE = \angle EAX</p> <p>\angle ABE = \frac{\angle BCD}{2} \text{ (angle at the center is twice the angle at the circumference)}</p> <p>Now consider the quadrilateral BCED:</p> <p>\angle BCD + \angle BED = 180^{\circ} \text{ (opposite angles of a cyclic quadrilateral sum to 180 degrees)}</p> <p>\angle BED = 180^{\circ} - \angle BCD</p> <p>Solve for \angle BCD using the fact that \angle AED = 90^{\circ}:</p> <p>\angle AED = \angle BED = 180^{\circ} - \angle BCD</p> <p>90^{\circ} = 180^{\circ} - \angle BCD</p> <p>\angle BCD = 90^{\circ}</p> <p>So, \angle ABE = \frac{\angle BCD}{2} = \frac{90^{\circ}}{2} = 45^{\circ}</p> <p>Now, solve for \angle EAX:</p> <p>\angle EAX = \angle EAB + \angle ABE</p> <p>\angle EAX = 80^{\circ} + 45^{\circ}</p> <p>\angle EAX = 125^{\circ}</p> <p>Consider the exterior angle \angle AXF for triangle AXF:</p> <p>\angle AFX + \angle FAX = \angle EAX</p> <p>\angle AFX = \angle EAX - \angle FAX</p> <p>\angle AFX = 125^{\circ} - 80^{\circ}</p> <p>\angle AFX = 45^{\circ}</p>
Calculating the Angle of a Tangent-Secant Triangle in a Circle Geometry Problem
<p>Let \( \angle BAX = y \) and \( \angle BFX = z \).</p> <p>Since AE is tangent to the circle at B, \( \angle AEB = 90^\circ \).</p> <p>By the alternate segment theorem, \( \angle ABX = \angle AEB = 90^\circ \).</p> <p>Consider triangle ABX: \( x + y + 90^\circ = 180^\circ \)</p> <p>\( x + y = 90^\circ \) ...(1)</p> <p>Since AE and AX are tangent to the circle at E and X, \( \angle EAX = 80^\circ \).</p> <p>In triangle AEX: \( y + 80^\circ + 90^\circ = 180^\circ \)</p> <p>\( y = 180^\circ - 80^\circ - 90^\circ \)</p> <p>\( y = 10^\circ \) ...(2)</p> <p>Using (1) and (2), we find \( x \):</p> <p>\( x + 10^\circ = 90^\circ \)</p> <p>\( x = 90^\circ - 10^\circ \)</p> <p>\( x = 80^\circ \) ...(3)</p> <p>Since BF is tangent to the circle at F and BC is a secant, \( \angle BFX = \angle BCX \) (angles in the alternate segment).</p> <p>Triangle BCF is isosceles (BF = BC as radii of the same circle), so \( \angle BCF = \angle BFC \).</p> <p>Consider the sum of angles in triangle BCF: \( z + z + x = 180^\circ \)</p> <p>\( 2z = 180^\circ - x \)</p> <p>Using the value of \( x \) from (3): \( 2z = 180^\circ - 80^\circ \)</p> <p>\( 2z = 100^\circ \)</p> <p>\( z = 50^\circ \)</p> <p>Finally, \( \angle AFX = x - z \)</p> <p>\( \angle AFX = 80^\circ - 50^\circ \)</p> <p>\( \angle AFX = 30^\circ \)</p>
Circle Geometry Problems
<p>Dado que la imagen es un poco borrosa y no se pueden discernir todos los detalles con precisión, resolveré la pregunta número 4, que parece estar más clara y requiere el cálculo del área de un sector y la longitud de un arco dado un ángulo central y el radio de un círculo.</p> <p>El área \( A \) de un sector de un círculo con radio \( r \) y ángulo central \( \theta \) (en radianes) es:</p> <p>\[ A = \frac{1}{2} r^2 \theta \]</p> <p>La longitud \( L \) de un arco con radio \( r \) y ángulo central \( \theta \) (en radianes) es:</p> <p>\[ L = r \theta \]</p> <p>En la pregunta 4, dan \( r = 3 \, cm \) y \( \theta = 60^\circ \). Primero convertimos el ángulo a radianes, recordando que \( 180^\circ \) equivalen a \( \pi \) radianes:</p> <p>\[ \theta = 60^\circ \cdot \frac{\pi \text{ rad}}{180^\circ} = \frac{\pi}{3} \text{ rad} \]</p> <p>Sustituimos \( r \) y \( \theta \) en las fórmulas para área y longitud de arco:</p> <p>\[ A = \frac{1}{2} (3\, cm)^2 \cdot \frac{\pi}{3} = \frac{9\pi}{6}\, cm^2 = \frac{3\pi}{2}\, cm^2 \]</p> <p>\[ L = 3\, cm \cdot \frac{\pi}{3} = \pi\, cm \]</p> <p>Por lo tanto, el área del sector es \( \frac{3\pi}{2}\, cm^2 \) y la longitud del arco es \( \pi\, cm \).</p>
Completing a Logical Argument Involving Circle Geometry
Premis/Premise 1: <p>\text{Jika jejari sebuah bulatan ialah 7 cm, maka lilitan bulatan itu ialah 14x.}</p> <p>\text{If the radius of the circle is 7 cm, then the circumference of the circle is 14x.}</p> Premis/Premise 2: <p>\text{Untuk mencari lilitan sebuah bulatan, kita gunakan rumus } C = 2\pi r \text{.}</p> <p>\text{To find the circumference of a circle, we use the formula } C = 2\pi r \text{.}</p> Kesimpulan/Conclusion: <p>\text{Jejari bulatan itu bukan 7 cm.}</p> <p>\text{The radius of the circle is not 7 cm.}</p> Maka, Premis 2 harus menyatakan hubungan antara jejari dan lilitan dengan menggunakan formula yang betul supaya argumen logik ini sah. Premis 2 adalah: <p>\text{Oleh itu, jika jejari bulatan itu ialah 7 cm, kita boleh gunakan rumus } C = 2\pi r \text{ untuk mencari lilitannya, yang mana } C = 2\pi(7) = 14\pi \text{ cm, dan bukannya 14x.}</p> <p>\text{Therefore, if the radius of the circle is 7 cm, we can use the formula } C = 2\pi r \text{ to find its circumference, which is } C = 2\pi(7) = 14\pi \text{ cm, not 14x.}</p>
Finding Inscribed Angle Measure in a Circle
The image depicts a circle with a central angle ∠XZY measuring 144° and an inscribed angle ∠YZV that intercepts the same arc XY as the central angle. According to the inscribed angle theorem, the measure of an inscribed angle is half the measure of the central angle that intercepts the same arc. Therefore, to find the measure of angle ∠YZV, we simply need to take half of the measure of ∠XZY. ∠XZY = 144° ∠YZV = 1/2 × ∠XZY ∠YZV = 1/2 × 144° ∠YZV = 72° So, the measure of ∠YZV is 72 degrees.
Calculate Circumference of Circle from Given Area
The question in the image asks for the circumference of a circle whose area is 49 m². To find the circumference, we first need to determine the radius of the circle. The formula for the area of a circle is: \[ A = \pi r^2 \] where \( A \) is the area and \( r \) is the radius. Given that the area \( A \) is 49 m², we can solve for \( r \) as follows: \[ 49 = \pi r^2 \] Divide both sides by \( \pi \) to get: \[ \frac{49}{\pi} = r^2 \] Take the square root of both sides to solve for \( r \): \[ r = \sqrt{\frac{49}{\pi}} = \frac{\sqrt{49}}{\sqrt{\pi}} = \frac{7}{\sqrt{\pi}} \] Now we have the radius. The circumference \( C \) of a circle is given by the formula: \[ C = 2\pi r \] We can substitute \( r = \frac{7}{\sqrt{\pi}} \) into this formula to find the circumference: \[ C = 2\pi\left(\frac{7}{\sqrt{\pi}}\right) \] Now let's simplify: \[ C = \frac{14\pi}{\sqrt{\pi}} \] Multiplying the top and bottom by \( \sqrt{\pi} \) to rationalize the denominator, we get: \[ C = \frac{14\pi\sqrt{\pi}}{\pi} = 14\sqrt{\pi} \] Finally, you can leave the answer in terms of \( \pi \) as the question requests. The circumference \( C \) is: \[ C = 14\sqrt{\pi} \: \text{meters} \]
Calculating Circumference of a Circle from Area
The question asks for the circumference of a circle given its area. The area (A) of a circle is related to its radius (r) by the formula: \[ A = \pi r^2 \] Given that the area is 36π cm², we can set up the equation: \[ 36\pi = \pi r^2 \] We can solve for r by dividing both sides of the equation by π: \[ 36 = r^2 \] Taking the square root of both sides gives us: \[ r = \sqrt{36} \] \[ r = 6 \text{ cm} \] The circumference (C) of a circle is related to the radius by the formula: \[ C = 2\pi r \] Substituting the radius we found into the circumference formula gives us: \[ C = 2\pi \times 6 \] \[ C = 12\pi \] So the circumference of the circle is 12π centimeters.
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