Two tangents, PQ and QR, are drawn from an external point Q to a circle of centre O. If OQ = 41 cm, $∠PQR=60^0$ and the radius of the circle is 9 cm, find the perimeter of $ΔPRQ$
120 cm
The question involves two tangents, PQ and QR, drawn from an external point Q to a circle with center O. We are given the following data:
- OQ (distance from the center to the external point Q) = 41 cm
- ∠PQR = 60°
- Radius of the circle = 9 cm
We need to calculate the perimeter of ΔPRQ.
Let's solve this step-by-step:
- From the property of tangents drawn from an external point to a circle, the lengths of tangents from a single external point are equal. Thus, PQ = QR.
- In the triangle ΔOQP and ΔOQR, both are right triangles because the radius is perpendicular to the tangent at the point of contact. Hence, ∠OQP = ∠OQR = 90°.
- We have OQ = 41 cm and OP = OR = 9 cm (radius).
- Using the Pythagorean theorem in ΔOQP:
- OP^2 + PQ^2 = OQ^2
- Substituting the given values: 9^2 + PQ^2 = 41^2
- Calculating further: 81 + PQ^2 = 1681
- Therefore, PQ^2 = 1681 - 81 = 1600
- It follows that PQ = \sqrt{1600} = 40 \text{ cm}
- We have found that PQ = QR = 40 cm.
- Now we calculate the perimeter of ΔPRQ:
- Perimeter = PQ + QR + PR = 40 cm + 40 cm + PR
- Considering ΔPQR where ∠PQR = 60°, we know PR is opposite this angle.
- Since PQ = QR, ΔPQ is an isosceles triangle. Let PR = x:
- We apply the cosine rule in ΔPQR: PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(60°)
- Substitute the values: PR^2 = 40^2 + 40^2 - 2 \times 40 \times 40 \times \frac{1}{2}
- Simplifying: PR^2 = 1600 + 1600 - 1600 = 1600
- This gives us PR = \sqrt{1600} = 40 \text{ cm}
- Finally, the Perimeter of ΔPRQ: 40 + 40 + 40 = 120 \text{ cm}
Therefore, the perimeter of ΔPRQ is 120 cm.
Hence, the correct answer is 120 cm.
Similar Questions
- The adjacent sides of a parallelogram are $4a$ and $3a$. If the angle between them is $60^{\circ}$, then one of the diagonals of the parallelogram will be:
- Every interior angle of a regular octagon is $135^\circ$. Find the exterior angle of the octagon.
- PQ is a diameter of circle whose centre is O. If a point R lies on a circle and $\angle RPO$ is $42^\circ$, then find $\angle RQP$.
- ABC is an equilateral triangle and O is its circumcentre. If the side of triangle is 6 cm, then the $\angle BOC$ is:
- If every interior angle of a regular polygon is $144^\circ$, then the polygon has ______ sides.
- The ratio of the numbers of sides of two regular polygons is $1 : 2$. If each interior angle of the first polygon is $140^\circ$, then the measure of each interior angle of the second polygon is:
- If the sum of the interior angles of a regular polygon is equal to four times the sum of its exterior angles, then what is the number of diagonals in the polygon?
- Find the ratio of the measure of an angle of a regular pentagon to that of a regular octagon.
- What is the measure of each exterior angle of a regular octagon?
- The difference between the interior and exterior angles at a vertex of a regular polygon is $140^\circ$. The number of sides of the polygon is:
Important Questions from Geometry
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.
