Quantitative Aptitude — Circles

Formula for circumference of a circle: C = 2 * pi * r Given C = 88 cm. 88 = 2 * (22/7) * r 88 = (44/7) * r r = 88 * (7/44) = 2 * 7 = 14 cm Formula for area of a circle: A = pi * r^2 A = (22/7) * 14 * 14 A = 22 * 2 * 14 = 616 sq cm

According to the circle theorem, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Therefore, angle at circumference = (Angle at center) / 2 Angle at circumference = 70 degrees / 2 = 35 degrees.

Let T be the point where the tangent from P touches the circle. The radius OT is perpendicular to the tangent PT at the point of contact. So, triangle OTP is a right-angled triangle with the right angle at T. Given: Radius OT = 5 cm, Distance OP = 13 cm (hypotenuse). Using Pythagoras theorem: PT^2 + OT^2 = OP^2 PT^2 + 5^2 = 13^2 PT^2 + 25 = 169 PT^2 = 169 - 25 PT^2 = 144 PT = sqrt(144) = 12 cm.

Let the chord be AB and the center of the circle be O. Let M be the midpoint of the chord AB. The perpendicular from the center to a chord bisects the chord. So, AM = MB = (Length of chord) / 2 = 24 cm / 2 = 12 cm. OA is the radius of the circle, so OA = 13 cm. Triangle OMA is a right-angled triangle with the right angle at M. Using Pythagoras theorem: OM^2 + AM^2 = OA^2 OM^2 + 12^2 = 13^2 OM^2 + 144 = 169 OM^2 = 169 - 144 OM^2 = 25 OM = sqrt(25) = 5 cm. The distance of the chord from the center is 5 cm.

According to the property of a cyclic quadrilateral, the sum of opposite angles is 180 degrees. So, Angle A + Angle C = 180 degrees. Given Angle A = 80 degrees. 80 degrees + Angle C = 180 degrees Angle C = 180 degrees - 80 degrees Angle C = 100 degrees.

When two circles touch each other externally, the distance between their centers is equal to the sum of their radii. Let R1 be the radius of the first circle and R2 be the radius of the second circle. R1 = 9 cm R2 = 4 cm Distance between centers = R1 + R2 = 9 cm + 4 cm = 13 cm.

According to the circle theorem, the angle subtended by a diameter at any point on the circumference is a right angle (90 degrees). Since AB is the diameter, angle ACB = 90 degrees. Now, consider triangle ABC. The sum of angles in a triangle is 180 degrees. Angle CAB + Angle ABC + Angle ACB = 180 degrees Given Angle CAB = 40 degrees. 40 degrees + Angle ABC + 90 degrees = 180 degrees Angle ABC + 130 degrees = 180 degrees Angle ABC = 180 degrees - 130 degrees Angle ABC = 50 degrees.

When tangents PA and PB are drawn from an external point P to a circle with center O, the quadrilateral OAPB is formed. We know that the radius is perpendicular to the tangent at the point of contact. So, Angle OAP = 90 degrees and Angle OBP = 90 degrees. The sum of angles in a quadrilateral is 360 degrees. Angle AOB + Angle OBP + Angle APB + Angle OAP = 360 degrees Angle AOB + 90 degrees + 70 degrees + 90 degrees = 360 degrees Angle AOB + 250 degrees = 360 degrees Angle AOB = 360 degrees - 250 degrees Angle AOB = 110 degrees. Alternatively, for tangents from an external point, the angle between the tangents and the angle subtended by the points of contact at the center are supplementary. Angle AOB + Angle APB = 180 degrees. Angle AOB = 180 degrees - 70 degrees = 110 degrees.

Given radius (r) = 10 cm. Angle subtended at the center (theta) = 90 degrees. 1. Area of the minor sector: Formula: Area = (theta/360) * pi * r^2 Area of minor sector = (90/360) * (22/7) * 10 * 10 = (1/4) * (22/7) * 100 = 2200 / 28 = 550 / 7 sq cm. 2. Area of the triangle formed by the chord and two radii: Since the angle at the center is 90 degrees, the triangle is a right-angled triangle. Formula: Area = (1/2) * base * height = (1/2) * r * r Area of triangle = (1/2) * 10 * 10 = 50 sq cm. 3. Area of the minor segment: Area of minor segment = Area of minor sector - Area of triangle = (550/7) - 50 = (550 - 350) / 7 = 200 / 7 sq cm. 4. Total area of the circle: Formula: Area = pi * r^2 Total area = (22/7) * 10 * 10 = 2200 / 7 sq cm. 5. Area of the major segment: Area of major segment = Total area of circle - Area of minor segment = (2200/7) - (200/7) = 2000 / 7 sq cm. 2000 / 7 approximately equals 285.71 sq cm.

Let the two circles have centers O1 and O2 and radii R1 = 5 cm and R2 = 3 cm respectively. They intersect at points A and B. AB is the common chord. The distance between their centers O1O2 = 4 cm. The line joining the centers of two intersecting circles is the perpendicular bisector of their common chord. Let M be the point where O1O2 intersects AB. So, O1M is perpendicular to AB, and AM = MB = AB/2. Consider the right-angled triangle O1MA: Hypotenuse O1A = R1 = 5 cm. Let O1M = x cm. Then AM^2 = O1A^2 - O1M^2 = 5^2 - x^2 = 25 - x^2 -- (Equation 1) Consider the right-angled triangle O2MA: Hypotenuse O2A = R2 = 3 cm. O2M = O1O2 - O1M = (4 - x) cm. Then AM^2 = O2A^2 - O2M^2 = 3^2 - (4 - x)^2 = 9 - (16 - 8x + x^2) -- (Equation 2) Equating Equation 1 and Equation 2 (since AM is the same): 25 - x^2 = 9 - (16 - 8x + x^2) 25 - x^2 = 9 - 16 + 8x - x^2 25 = -7 + 8x 25 + 7 = 8x 32 = 8x x = 32 / 8 = 4 cm. Now substitute x = 4 into Equation 1 to find AM: AM^2 = 25 - 4^2 AM^2 = 25 - 16 AM^2 = 9 AM = sqrt(9) = 3 cm. The length of the common chord AB = 2 * AM = 2 * 3 = 6 cm.