Assignments for Class 10 Mathematics Circles have been developed for Standard 10 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 10 Mathematics Circles from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 10 Mathematics Circles. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics Circles book and get good marks in class 10 exams.
The perimeter of the circle is referred to as the circumference of the circle.
- A secant divides a circle in two parts and a tangent touches a circle.
- There is one and only one tangent at a point of the circle.
There can be maximum two number of tangents which can be drawn to the circle from any given external point.
The length of two tangents from an external point to a circle is equal.
The tangent to a circle is perpendicular to the radius through the point of contact.
Question. Circles are drawn from the three vertices of a ΔABC (as shown); taken as centre to touch each other externally. If the sides of the triangle are 4 cm, 6 cm and 8 cm, find the radii of the circles.
Solution. 3 cm, 5 cm and 1 cm
Question. The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of a bigger circle. BD is a tangent to the smaller circle, touching it at D. Find the length AD.
Solution. 19 cm
Question. Find the length of the tangent drawn from a point whose distance from the centre is 25 cm when the radius of the circle is 24 cm.
Solution. 7 cm
Question. In the given figure, the four sides AB, BC, CD and DA of a quadrilateral ABCD touches a circle at the points P, Q, R and S respectively. If AB = 6 cm, BC = 7 cm and CD = 4 cm, find AD.
Solution. 3 cm
Question. A point P is 13 cm from the centre of the circle. The length of the tangent drawn from P to the circle is 12 cm. Find the radius of the circle.
Solution. 5 cm
Question. A point P is 29 cm from the centre of the circle. Find the length of the tangent drawn from P to the circle if the radius of the circle is 20 cm.
Solution. 21 cm
Question. P and Q are centres of circles of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm which touches the above circles externally. Given that ΔPRQ = 90°, write an equation in x and solve it.
Solution. x2 + 11x – 102 = 0; x = 6
Question. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. (see figure). Find the length of TP.
Solution. TP = 20/3 cm
Question. A circle is inscribed in a ΔABC having sides 8 cm, 10 cm and 12 cm as shown. Find the length of AD, BE and CF.
Solution. AD = 3 cm, BE = 7 cm and CF= 5 cm
Question. In the given fig. P, Q and R are the points of contact. If AB = 4 cm, BP = 2 cm then the perimeter of ΔABC is
Solution. 12 cm
Question. The distance between two tangent parallel to each other to a circle is 12cm. The radius of circle is
Solution. 6 cm
Question. In the given fig. O is the centre of the circle and PQ is tangent then ∠POQ + QPO is equal to
Question. If PQ is a tangent to a circle of radius 5cm and PQ = 12 cm, Q is point of contact, then OP is
Solution. √119 cm
Question. How many tangents can a circle have?
Question. In the given fig. QS is a tangent to the circle, OS = 8 cm, OQ = 6 cm then the length of QS is
Solution. √28 cm
Question. In the given fig PQ is tangent to outer circle and PR is tangent to inner circle. If PQ = 4 cm, OQ = 3 cm and OR = 2 cm then the length of PR is
Solution. √21 cm
Question. How many tangents can be drawn from a given exterior point to a circle?
Solution. Only 2 Tangents
Question. The chord of a circle of radius 10cm subtends a right angle at its centre. Find the length of the chord.
Question. In the given fig. PQ and PR are tangents to the circle, QOP= 70°, then QPR is equal to
Question. If AB, AC, PQ are tangent in the given figure, and AB = 5 cm, find the perimeter of ΔAPQ.
Solution. 10 cm
Question. ABCD is a quadrilateral such that ∠A = 90°. A circle with centre O, touches the sides AB, BC, CD and DA at E, F, G, H respectively. If CD = 35 cm, AD = 20 cm and CF = 25 cm. Find the radius of the circle.
Solution. 10 cm
Question. ABC is a right angle triangle, right angled at A, AB = 6 cm and AC = 8 cm. A circle with centre O is inscribed inside the triangle. Calculate the value of ‘r’, the radius of the inscribed circle.
Solution. r = 2 cm
Question. ABCD is a quadrilateral such that ∠A = 90°. A circle with center O, touches the sides AB, BC, CD and DA at E, F, G, H respectively. If CD = 30 cm, AD = 15 cm and CF = 20 cm, find the radius of the circle.
Solution. 5 cm
Question. PQ and RS are two parallel tangents to a circle with centre O and another tangent XY, with point of contact C intersects PQ at A and RS at B. Prove that ΔAOB = 90°.
Solution. Join O to A, O to B and O to C.
Since tangent is perpendicular to the radius through the point of contact.
Question. PQR is a right-angled triangle right angled at Q with PQ = 16 cm, QR = 12 cm. A circle with centre at O and radius r is inscribed in ΔPQR. Find the value of r.
Solution. Let the circle touch the sides QR, RP and PQ of ΔPQR at the points A, B and C respectively. Join O to A and O to C.
Then, OC = OA = r cm.
In ΔPQR, using pythagoras theorem,
PR2 = PQ2 + QR2
= 162 + 122 = 256 + 144 = 400
∴ PR = 20 cm.
Since OC is the radius and PQ is the tangent at C, we have ΔOCQ = 90°. Similarly, ΔOAQ = 90°.
Also, ΔAQC = 90° [∴ given]
Hence, the remaining angle AOC of quadrilateral AQCO is also 90°.
Again, QA = QC [∴ lengths of the tangents from the external point to a circle are equal]
∴ The quadrilateral AQCO is a square.
∴ QC = QA = r cm
Hence, PC = (16 – r) cm and RA = (12 – r) cm.
Now, PB = PC (∴ same as above)
Similarly, RB = RA = (12 – r) cm
⇒ PB + RB = (16 – r + 12 – r) cm = (28 – 2r) cm
⇒ PR = (28 – 2r) cm
⇒ 20 cm = (28 – 2r) cm
⇒ 2r = 28 – 20 = 8
⇒ r = 4
Hence, the required length of the radius of the circle is 4 cm. Ans.
Question. The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm. Determine the other two sides of the triangle.
Solution. Since lengths of tangent from an external point to the circle are equal.
∴ BP = BR = 8 cm
CP = CQ = 6 cm
and, AR = AQ = x cm (say)
Also, IP = IQ = IR = 4 cm
Now, 2s = a + b + c
⇒ 2s = 8 + 6 + 6 + x + x + 8
⇒ 2s = 28 + 2x
⇒ s = 14 + x
⇒ s – a = 14 + x – 14 = x, s – b = 14 + x – 6 – x = 8 and
s – c = 14 + x – 8 – x = 6.
Clearly, ar (ΔABC) = ar (ΔIBC) + ar (ΔICA) + ar (ΔIAB)
Squaring both sides, we get
3x (14 + x) = (14 + x)2
⇒ 3x = 14 + x [∴ 14 + x ≠ 0]
⇒ 2x = 14 ⇒ x = 7
⇒ AB = (8 + 7) cm = 15 cm and AC = (6 + 7) cm = 13 cm. Ans.
Question. A point P is 26 cm away from the centre O of a circle and the length PQ of the tangent segment drawn from P to the circle is 10 cm. Find the radius of the circle.
Solution. Let PQ be a tangent to the circle from point P and OQ be the radius at the point of contact.
Question. A circle is touching the side BC of ΔABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ = 1/2 (Perimeter of ΔABC).
Solution. Since tangents from an exterior point to a circle are equal in length.
∴ BP = BQ …(i)
CP = CR …(ii)
and, AQ = AR …(iii)
from eqn. (iii), we have AQ = AR
⇒ AB + BQ = AC + CR
⇒ AB + BP = AC + CP …(iv)
[∴ using (i) and (ii)]
Now, Perimeter of ΔABC = AB + BC + AC
= AB + (BP + PC) + AC
= (AB + BP) + (AC + PC)
= 2 (AB + BP) [using (iv)]
= 2 AQ [using (i)]
AQ = 1/2 (Perimeter of ΔABC).
Assignments for Class 10 Mathematics Circles as per CBSE NCERT pattern
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