AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = 5 sqrt2 cm, then prove that riangle AOB is an isosceles triangle. Find the radius of the circle.

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Asked on April 24, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-24T06:01:36.899Z

You're on a roll — Here are the solutions to the questions: IV. Answer the following questions: 24. AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = 5sqrt(2) cm, then prove that AOB is an isosceles triangle. Find the radius of the circle. Step 1: Prove AOB is an isosceles triangle. A tangent to a circle is perpendicular to the radius at the point of contact. Therefore, OAB = 90^. In AOB, the sum of angles is 180^. We are given OBA = 45^ (from the diagram). AOB + OBA + OAB = 180^ AOB + 45^ + 90^ = 180^ AOB + 135^ = 180^ AOB = 180^ - 135^ AOB = 45^ Since OBA = 45^ and AOB = 45^, two angles of AOB are equal. Therefore, the sides opposite to these equal angles must be equal, which means OA = AB. Hence, AOB is an isosceles triangle. Step 2: Find the radius of the circle. In the right-angled triangle AOB, we can use trigonometric ratios. We know OBA = 45^ and OB = 5sqrt(2) cm. The radius is OA. ( OBA) = OppositeHypotenuse = (OA)/(OB) (45^) = (OA)/(5sqrt(2) cm) (1)/(sqrt(2)) = (OA)/(5sqrt(2) cm) OA = 5sqrt(2)sqrt(2) cm OA = 5 cm The radius of the circle is 5 cm. 25. Find the zeroes of the quadratic polynomial p(x) = x^2 - 2x - 8 and verify the relationship between the zeroes and the coefficients. Step 1: Find the zeroes of the polynomial. To find the zeroes, set p(x) = 0: x^2 - 2x - 8 = 0 Factor the quadratic expression. We need two numbers that multiply to -8 and add to -2. These numbers are -4 and 2. (x-4)(x+2) = 0 Set each factor to zero: x-4 = 0 x = 4 x+2 = 0 x = -2 The zeroes of the polynomial are 4 and -2. Step 2: Verify the relationship between the zeroes and the coefficients. For a quadratic polynomial ax^2 + bx + c, the sum of zeroes is -(b)/(a) and the product of zeroes is (c)/(a). For p(x) = x^2 - 2x - 8, we have a=1, b=-2, c=-8. Let the zeroes be = 4 and = -2. Sum of zeroes: + = 4 + (-2) = 2 From coefficients: -(b)/(a) = -((-2))/(1) = 2 The sum of zeroes matches the formula. Product of zeroes: = 4 × (-2) = -8 From coefficients: (c)/(a) = (-8)/(1) = -8 The product of zeroes matches the formula. The relationship between the zeroes and coefficients is verified. 26. Prove that "The tangent at any point of a circle is perpendicular to the radius through the point of contact". Step 1: Set up the proof. Let a circle have center O and radius r. Let AB be a tangent to the circle at point P. We need to prove that the radius OP is perpendicular to the tangent AB, i.e., OP AB. Step 2: Consider a point on the tangent. Take any point Q on the line AB, other than P. Since AB is a tangent, it touches the circle at only one point, P. This means that any other point Q on the line AB must lie outside the circle. Step 3: Compare distances. If Q lies outside the circle, then the distance OQ must be greater than the radius OP. OQ > OP Step 4: Conclude the proof. This implies that OP is the shortest distance from the center O to any point on the line AB. In geometry, the shortest distance from a point to a line is always the perpendicular distance. Therefore, OP AB. Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact. 27. ABC PQR. In ABC, A = 55^, C = 35^ and the sides of ABC are in the ratio 3:4:5. If the perimeter of PQR is 60 cm, then find the measure of sides and angles of PQR. Step 1: Find the angles of ABC. The sum of angles in a triangle is 180^. A + B + C = 180^ 55^ + B + 35^ = 180^ 90^ + B = 180^ B = 180^ - 90^ B = 90^ Step 2: Find the angles of PQR. Since ABC PQR, their corresponding angles are equal. P = A = 55^ Q = B = 90^ R = C = 35^ The angles of PQR are P = 55^, Q = 90^, R = 35^. Step 3: Find the sides of PQR. The sides of ABC are in the ratio 3:4:5. Let the sides be 3k, 4k, 5k. Since ABC PQR, the sides of PQR will also be in the same ratio 3:4:5. Let the sides of PQR be 3x, 4x, 5x. The perimeter of PQR is given as 60 cm. Perimeter of PQR = 3x + 4x + 5x 60 cm = 12x x = \

Accepted Answer · 2026-04-24T06:01:36.899Z

You're on a roll — Here are the solutions to the questions: IV. Answer the following questions: 24. AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = 5sqrt(2) cm, then prove that AOB is an isosceles triangle. Find the radius of the circle. Step 1: Prove AOB is an isosceles triangle. A tangent to a circle is perpendicular to the radius at the point of contact. Therefore, OAB = 90^. In AOB, the sum of angles is 180^. We are given OBA = 45^ (from the diagram). AOB + OBA + OAB = 180^ AOB + 45^ + 90^ = 180^ AOB + 135^ = 180^ AOB = 180^ - 135^ AOB = 45^ Since OBA = 45^ and AOB = 45^, two angles of AOB are equal. Therefore, the sides opposite to these equal angles must be equal, which means OA = AB. Hence, AOB is an isosceles triangle. Step 2: Find the radius of the circle. In the right-angled triangle AOB, we can use trigonometric ratios. We know OBA = 45^ and OB = 5sqrt(2) cm. The radius is OA. ( OBA) = OppositeHypotenuse = (OA)/(OB) (45^) = (OA)/(5sqrt(2) cm) (1)/(sqrt(2)) = (OA)/(5sqrt(2) cm) OA = 5sqrt(2)sqrt(2) cm OA = 5 cm The radius of the circle is 5 cm. 25. Find the zeroes of the quadratic polynomial p(x) = x^2 - 2x - 8 and verify the relationship between the zeroes and the coefficients. Step 1: Find the zeroes of the polynomial. To find the zeroes, set p(x) = 0: x^2 - 2x - 8 = 0 Factor the quadratic expression. We need two numbers that multiply to -8 and add to -2. These numbers are -4 and 2. (x-4)(x+2) = 0 Set each factor to zero: x-4 = 0 x = 4 x+2 = 0 x = -2 The zeroes of the polynomial are 4 and -2. Step 2: Verify the relationship between the zeroes and the coefficients. For a quadratic polynomial ax^2 + bx + c, the sum of zeroes is -(b)/(a) and the product of zeroes is (c)/(a). For p(x) = x^2 - 2x - 8, we have a=1, b=-2, c=-8. Let the zeroes be = 4 and = -2. Sum of zeroes: + = 4 + (-2) = 2 From coefficients: -(b)/(a) = -((-2))/(1) = 2 The sum of zeroes matches the formula. Product of zeroes: = 4 × (-2) = -8 From coefficients: (c)/(a) = (-8)/(1) = -8 The product of zeroes matches the formula. The relationship between the zeroes and coefficients is verified. 26. Prove that "The tangent at any point of a circle is perpendicular to the radius through the point of contact". Step 1: Set up the proof. Let a circle have center O and radius r. Let AB be a tangent to the circle at point P. We need to prove that the radius OP is perpendicular to the tangent AB, i.e., OP AB. Step 2: Consider a point on the tangent. Take any point Q on the line AB, other than P. Since AB is a tangent, it touches the circle at only one point, P. This means that any other point Q on the line AB must lie outside the circle. Step 3: Compare distances. If Q lies outside the circle, then the distance OQ must be greater than the radius OP. OQ > OP Step 4: Conclude the proof. This implies that OP is the shortest distance from the center O to any point on the line AB. In geometry, the shortest distance from a point to a line is always the perpendicular distance. Therefore, OP AB. Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact. 27. ABC PQR. In ABC, A = 55^, C = 35^ and the sides of ABC are in the ratio 3:4:5. If the perimeter of PQR is 60 cm, then find the measure of sides and angles of PQR. Step 1: Find the angles of ABC. The sum of angles in a triangle is 180^. A + B + C = 180^ 55^ + B + 35^ = 180^ 90^ + B = 180^ B = 180^ - 90^ B = 90^ Step 2: Find the angles of PQR. Since ABC PQR, their corresponding angles are equal. P = A = 55^ Q = B = 90^ R = C = 35^ The angles of PQR are P = 55^, Q = 90^, R = 35^. Step 3: Find the sides of PQR. The sides of ABC are in the ratio 3:4:5. Let the sides be 3k, 4k, 5k. Since ABC PQR, the sides of PQR will also be in the same ratio 3:4:5. Let the sides of PQR be 3x, 4x, 5x. The perimeter of PQR is given as 60 cm. Perimeter of PQR = 3x + 4x + 5x 60 cm = 12x x = \

AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = 5 sqrt2 cm, then prove that riangle AOB is an isosceles triangle. Find the radius of the circle.

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AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = 5 sqrt2 cm, then prove that riangle AOB is an isosceles triangle. Find the radius of the circle.

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