Geometry Problems: Circles, Chords, and Quadrilaterals
Geometry Problems and Solutions
Here are the solutions to the geometry questions:
Cyclic Quadrilateral Angle Sum
Question: The sum of either pair of opposite angles of a cyclic quadrilateral is 180°. If ‘n’ represents this sum and ‘m’ is also 180°, find ‘m-n’.
Solution: In a cyclic quadrilateral, the sum of opposite angles is always 180°. Therefore, \(n = 180^\circ\). Given \(m = 180^\circ\), we have:
\[m – n = 180^\circ – 180^\circ = 0\]
Circle’s Largest Chord and Radius
Question: If the largest chord of a circle is \(\pi n\) where \(n\) is a constant, what is the radius of that circle?
Solution: The largest chord of a circle is its diameter. Therefore, \(\pi n\) is the diameter. The radius \(r\) is half of the diameter:
\[r = \frac{\pi n}{2}\]
Angles in the Same Circle Segment
Question: Angles in the same segment of a circle are opposite. (True or False)
Solution: False. Angles in the same segment of a circle are equal, not opposite.
Geometric Proofs
Question: Prove the following statements:
Chords Subtending Equal Angles
Statement: If chords of congruent circles subtend equal angles at their centers, then the chords are equal.
Proof: In congruent circles, if two chords subtend equal angles at the center, the arcs corresponding to these chords are equal. Hence, the chords are equal.
Cyclic Quadrilateral with Diameter Diagonals
Statement: If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Proof: If the diagonals of a cyclic quadrilateral are diameters, each angle subtended by a diameter is a right angle (90°). Therefore, all four angles of the quadrilateral are right angles, making it a rectangle.
Perpendicular from Center to Chord
Statement: The perpendicular from the center of a circle to a chord bisects the chord.
Proof: Let \(O\) be the center of the circle, \(AB\) be the chord, and \(OM\) be the perpendicular from \(O\) to \(AB\). Triangles \(OMA\) and \(OMB\) are congruent by RHS (Right angle-Hypotenuse-Side) congruence. Hence, \(AM = MB\), proving that \(OM\) bisects \(AB\).
Circle Radius from Parallel Chords
Question: Two equal chords with length 48 cm are parallel. The distance from each chord to the center is 7 cm. Find the radius of that circle.
Solution: Let the radius be \(r\). Using the Pythagorean theorem in the right triangle formed by the radius, half the chord, and the perpendicular distance from the center to the chord:
\[r^2 = 24^2 + 7^2\]
\[r^2 = 576 + 49\]
\[r^2 = 625\]
\[r = \sqrt{625}\]
\[r = 25 \text{ cm}\]
City Distances on a Circular Circumference
Question: Three cities are located on the circumference of a circle named Rampur, Janakpur, and Pratap Nagar. The radius of that circle is 20 km. If the distance between Rampur to Janakpur and Janakpur to Pratap Nagar is both 20 km, find the distance between Rampur to Pratap Nagar.
Solution: Since the distance between Rampur to Janakpur and Janakpur to Pratap Nagar is equal to the radius, the three cities (Rampur, Janakpur, Pratap Nagar) form an equilateral triangle with the center of the circle. Therefore, the distance between Rampur to Pratap Nagar is also 20 km.
Arc Defined by Diameter Ends
Question: An arc when its ends are the ends of a diameter is called:
Solution: d) Semicircle
Circles Through Non-Collinear Points
Question: How many circles pass through three given non-collinear points?
Solution: a) 01
Total Length of Major and Minor Arcs
Question: What is the total length of the major arc and minor arc?
Solution: The total length of the major arc and minor arc is equal to the circumference of the circle.