Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 8 Tangents Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.
SSLC Maths Chapter 8 Tangents Important Questions and Answers
Tangents Class 10 Extra Questions Kerala Syllabus
Tangents Class 10 Kerala Syllabus Extra Questions
Question 1.
Choose the correct answer from the options given below.
(i) The distance between two parallel tangents on a circle is 8 cm. What is the radius of the circle?
(a) 4 cm
(b) 2 cm
(c) 5 cm
(d) 8 cm
Answer:
(a) 4 cm
Since the tangent is perpendicular to the radius, the distance between the parallel tangents is the diameter of the circle.
(ii) The length of the tangent to a circle from an outer point and the radius of the circle are equal to 12 cm. What is the distance from the center to the outer point?
(a) 12 cm
(b) 12√3 cm
(c) 12√2 cm
(d) 6 cm
Answer:
(c) 12√2 cm
Use the property of a 45° – 45° – 90° triangle.
(iii) PA, PB are the tangents from P to the circle. If ∠APB = 40°, then what is the measure of ∠AOB?
(a) 140°
(b) 120°
(c) 150°
(d) 110°
Answer:
(a) 140°
Tangents from the outer point to the circle and radii form a cyclic quadrilateral.
(iv) PA, PB are the tangents from P to the circle. If ∠APB = 40°, then what is the measure of angle ACB?
(a) 50°
(b) 60°
(c) 70°
(d) 40°
Answer:
(c) 70°
\(\frac {1}{2}\) × ∠AOB = ∠ACB
(v) P is a point at a distance of 13 cm from the centre of a circle of radius 5 cm. What is the length of the tangents drawn from P to the circle?
(a) 12 cm
(b) 16 cm
(c) 18 cm
(d) 610 cm
Answer:
(a) 12 cm
(vi) In the figure AB = AC = 12 cm. If AP = 4 cm, then what is the length of BC?
(a) 14 cm
(b) 18 cm
(c) 10 cm
(d) 16 cm
Answer:
(d) 16 cm
BP = CQ = 8
BR = CR = 8
BC = 16 cm
(vii) In the circumcircle of triangle PQR, the tangent at P makes 40° with PQ. What is x?
(a) 40°
(b) 50°
(c) 80°
(d) 20°
Answer:
(a) 40°
(viii) The vertices of a square are on a circle. BP is the tangent to the circle at P. What is the measure of ∠PBC?
(a) 50°
(b) 40°
(c) 45°
(d) 30°
Ans:
(c) 45°
(ix) What is the radius of the incircle of an equilateral triangle with a side of 6 cm?
(a) √3
(b) √2
(c) 1
(d) 3
Answer:
(a) √3
(x) The number of tangents that can be drawn from a point outside a circle to the circle.
(a) 3
(b) 1
(c) 2
(d) 4
Answer:
(c) 2
(xi) From a point, the angle between the tangents drawn to a circle is 90°. Find the angle between the lines joining this point to the points of contact.
(a) 90°
(b) 60°
(c) 100°
(d) 80°
Answer:
(a) 90°
Question 2.
Read the two statements given below.
Statement 1: If the angle between the tangents from an outer point is 90 + x, then the angle between the radii to the point where the tangents touch the circle is 90 – x.
Statement 2: Tangents from the outer point and radii to the points where they touch the circle make a cyclic quadrilateral.
Choose the correct answer from those given below.
(a) Statement 1 is true, and Statement 2 is false.
(b) Statement 1 is false, and Statement 2 is true.
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both statements are true, but Statement 2 is not the correct reason for Statement 1.
Answer:
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
Question 3.
Read the two statements given below.
Statement 1: If the area and perimeter of a triangle are numerically equal, that is, both are the same number, then its inradius is 2 units.
Statement 2: If A is the area, s is half of the perimeter, then the inradius r = \(\frac {A}{s}\)
Choose the correct answer from those given below.
(a) Statement 1 is true, and Statement 2 is false.
(b) Statement 1 is false, and Statement 2 is true.
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both statements are true, but Statement 2 is not the correct reason for Statement 1.
Answer:
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
Question 4.
In the figure, PA is the tangent from P to A on the circle, and OA is the radius. If OP = 18 and ∠OPA = 40° then
[sin 40° = 0.64, cos 40° = 0.76, tan 40° = 0.83]
(a) What is the length of the tangent?
(b) What is the radius of the circle?
Answer:
(a) cos 40° = \(\frac{A P}{O P}\)
⇒ 0.76 = \(\frac{A P}{18}\)
⇒ AP = 0.76 × 18 = 13.68 cm
(b) sin 40° = \(\frac{O A}{O P}=\frac{O A}{18}\)
⇒ OA = 11.52 cm
Question 5.
Angles of triangle OAP are in the ratio 1 : 2 : 3, O is the center of the circle, and PA is the tangent from P. ∠P is the smallest angle, and OP = 12 cm.
(a) What is the ratio of the sides of this triangle?
(b) Find the other two sides of the triangle?
Answer:
(a) Angles are x, 2x, 3x
6x = 180°
x = 30°
Angles are 30°, 60°, 90.
Ratio of the sides is 1 : √3 : 2
(b) OA = 6, PA = 6√3
Question 6.
In the figure, O is the centre of the circle, PA is the tangent, and OP = 24 cm.
(a) What are the angles of a triangle?
(b) What is the radius of the circle?
(c) Find the length of the tangent.
Answer:
(a) ∠POA = 180 – 120 = 60°
∠PAO = 90°
∠OPA = 30°
(b) 12 cm
(c) 12√3 cm
Question 7.
In the figure, a circle touches the sides of a triangle. If AP = 1, BQ = 2 and CR = 3 then
(a) What is the perimeter of the triangle?
(b) What is the area of the triangle?
Answer:
(a) AR = 1, CQ = 3, BP = 2
Perimeter of triangle ABC = 12 cm
(b) The perpendicular sides of the right-angled triangle ABC are 3 and 4 cm.
Area = \(\frac {1}{2}\) × 3 × 4 = 6 sq.cm
Question 8.
In the figure, PA = 12 cm, O is the center of the circle, radius of the circle is 5 cm.
(a) What is the length PB?
(b) Find the area of APBO.
Answer:
(a) PB = 12
(b) Area of APBO = 2 × \(\frac {1}{2}\) × 12 × 5 = 60 sq. unit
Question 9.
In the figure, PA and QB are parallel tangents. Another line PQ touches the circle at R and cuts the parallel tangents.
(a) Draw a rough diagram and join OA, OR, and OB.
(b) Name the equal triangles in the figure.
(c) Find the measure of ∠POQ.
Answer:
(a) Draw the diagram yourself.
(b) PA = PR, OA = OR and OP is common.
Triangle PAO and triangle PRO are equal.
Similarly, ∆QRO and ∆QBO are equal.
(c) If ∠POA = ∠POR = x and ∠QOR = ∠QOB = y then 2x + 2y = 180
⇒ x + y = 90
∴ ∠POQ = 90°
Question 10.
In the figure, AB is a common tangent to the circle. The line PC joins the common point of the circle to the point P. It is also a tangent to the circle.
(a) Name the lines of equal length shown in the figure.
(b) If ∠PAC = x and ∠PBC = y then what is ∠ACB?
(c) Prove that ∆ABC is a right triangle.
Answer:
(a) PA = PC, PB = PC
(b) ∠ACB = x + y
(c) In triangle ABC,
2x + 2y = 180
⇒ x + y = 90
Triangle ABC is a right triangle.
Question 11.
In triangle ABC. O is the centre of the circumcircle. ∠BOC = 140°.
(a) What is the measure of ∠BAC?
(b) If PC is the tangent at C, then what is the measure of ∠BCP?
Answer:
(a) ∠BAC = 70°
(b) 70°
Question 12.
ABCDE is a regular pentagon. AD and BD are its diagonals. The tangents intersect at a point P.
(a) What is the angle measure of ∠E and ∠C?
(b) Find ∠ADE, ∠BDC?
(c) What is ∠PAB, ∠PBA?
(d) What is the angle measure of APB?
Answer:
(a) ∠E = ∠C = \(\frac {540}{5}\) = 108°
(b) ∠ADE = 36°
∠BDC = 36°
(c) ∠ADB = 108 – 72 = 36°
∠PAB = ∠PBA = 36°
(d) ∠P = 180 – 72 = 108°
Question 13.
In the figure, the circle touches the sides of the quadrilateral.
If AB = 12, CD = 8, AD = 7.
(a) What is the relation between the length of its sides?
(b) Find BC?
Answer:
(a) AB + CD = AD + BC
(b) 12 + 8 = 7 + BC
⇒ BC = 13
Question 14.
A circle touches two sides of the triangle ABC. The sides AB and AC are equal.
(a) Which lengths shown in the figure are equal?
(b) Prove that BR = CR.
Answer:
(a) AP = AQ, BP = BR, CQ = CR
(b) AB = AC
AB – PA = AC – AQ
⇒ BP = CQ
BP = CQ
⇒ BR = CR
Question 15.
In the figure, O is the center of the triangle ABC.
∠OBC = 20° and ∠OCB = 30°
(a) What is ∠BOC?
(b) Find ∠A.
Answer:
(a) ∠BOC = 180 – (20 + 30) = 130°
(b) This is the incircle of the triangle.
To draw it, the angles must be bisected.
Given that ∠B = 40° and ∠C = 60°
Therefore ∠A = 80°
Question 16.
AD and BC are the common tangents to two circles. The centres of the circles are M and N, respectively. ∠APB = 40°.
(a) Which lines shown in the figure are of equal length?
(b) Prove that AD = BC.
(c) Calculate ∠APB and ∠CPD.
Answer:
(a) PA = PB, PD = PC
(b) By adding the equalities,
PA + PD = PB + PC
AD = BC
(c) ∠AMB = 180 – 40 = 140°
∠CND = 180 – 40 = 140°
Question 17.
In the first figure, the line PA is the tangent from point P to the circle.
In the second figure, OP is the diameter of the semicircle, and A is a point on the semicircle.
(a) In both figures, what is the value of ∠OAP?
(b) Draw a circle with radius 3 cm. Mark a point 7 cm away from the centre. Draw tangents from point P to the circle.
(c) Measure the length of the tangent.
Answer:
(a) 90°
(b) Draw a circle with centre O.
Mark a point P at a distance of 7 cm from the centre.
Draw the diameter OP.
The circle is intersected at points A and B.
Draw the lines PA and PB; these are tangents.
(c) Measure and write the length of the tangents.