Category: CBSE Class 10

RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B
• RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself

MCQ
Question 1.
Solution:
In the given figure,
PT is the tangent and PQ is the chord of the circle with centre O.
∠OPT = 50°

OP is radius and PT is the tangent.
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ + ∠QPT = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 40°
In ∆OPQ,
∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°) = 180° – 80°
= 100° (b)

Question 2.
Solution:
Angle between two radii of a circle = 130°

Then ∠APB = 180° – ∠AOB
= 180°- 130° = 50° (c)

Question 3.
Solution:
In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O
∠APB = 80°

OA is radius of the circle and AP is the tangent
OA ⊥ AP ⇒ ∠OAP = 90°
OP bisects ∠APB,
∠APO = \(\frac { 1 }{ 2 }\) x 80 = 40°
∠POA = 180° – (90° + 40°)
= 180° – 130° = 50° (b)

Question 4.
Solution:
In the given figure, AD and AE are tangents to the circle with centre O.
BC is the tangent at F which meets AD at C and AE at B
AE = 5 cm

AE and AD are the tangents to the circle
AE = AD = 5 cm
Tangents from an external point drawn to the circle are equal
CD = CF and BE = BF
Now, perimeter of ∆ABC = AB + AC + BC
= AB + AC + BF + CF (BE = BF and CF = CD)
= AB + AC + BE + CD
= AB + BE + AC + CD
= AE + AD
= 5 + 5 = 10 cm (b)

Short-Answer Questions
Question 5.
Solution:
In the given figure, a quadrilateral ABCD is circumscribed a circle touching its sides at P, Q, R and S respectively.
AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm

A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD …(i)
Now, AP and AS are tangents to the circle
AP = AS = 5 cm …(ii)
Similarly, CQ = CR = 3 cm
BP = BQ = x – 5 = 4
BQ = BC – CQ = 7 – 3 = 4 cm
x – 5 = 4
⇒ x = 4 + 5 = 9cm

Question 6.
Solution:
In the given figure, PA and PB are the tangents drawn from P to the circle.
OA and OB are the radii of the circle and AP and BP are the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
In quad. AOBP
∠A + ∠B = 90° + 90° = 180°
But these are opposite angles of a quadrilateral
AOBP is a cyclic quadrilateral
A, O, B, P are concyclic

Question 7.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O from an external point P.
∠PBA = 65°,
To find : ∠OAB and ∠APB
In ∆APB
AP = BP (Tangents from P to the circle)
∠PAB = ∠PBA = 65°
∠APB = 180° – (∠PAB + ∠PBA)
= 180° – (65° + 65°) = 180° – 130° = 50°
OA is radius and AP is tangent
OA ⊥ AP
∠OAP = 90°
∠OAB = ∠OAP – ∠PAB = 90° – 65° = 25°
Hence, ∠OAB = 25° and ∠APB = 50°

Question 8.
Solution:
Given : In the figure,
BC and BD are the tangents drawn from B
to the circle with centre O.
∠CBD = 120°
To prove : OB = 2BC
Construction : Join OB.
Proof: OB bisects ∠CBD (OC = OD and BC = BD)

Question 9.
Solution:
(i) A line intersecting a circle in two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinitely many tangents.

Question 10.
Solution:
Given : In a circle, from an external point P, PA and PB are the tangents drawn to the circle with centre O.
To prove : PA = PB
Construction : Join OA, OB and OP.
Proof : OA and OB are the radii of the circle and AP and BP are tangents.

OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
PA = PB (c.p.c.t.)
Hence proved.

Short-Answer Questions
Question 11.
Solution:
Given : In a circle with centre O and AB is its diameter.
From A and B, PQ and RS are the tangents drawn to the circle

To prove : PQ || RS
Proof : OA is radius and PAQ is the tangent
OA ⊥ PQ
∠PAO = 90° …(i)
Similarly, OB is the radius and RBS is tangent
∠OBS = 90° …(ii)
From (i) and (ii)
∠PAO = ∠OBS
But there are alternate angles
PQ || RS

Question 12.
Solution:
Given : In the given figure,
In ∆ABC,
AB = AC.
A circle is inscribed the triangle which touches it at D, E and F
To prove : BE = CE
Proof: AD and AF are the tangents drawn from A to the circle
AD = AF
But, AB = AC
AB – AD = AC -AF
⇒ BD = CF
But BD = BE and CF = CE (tangent drawn to the circle)
But BD = CF
BE = CE
Hence proved.

Question 13.
Solution:
Given : In a circle from an external point P, PA and PB are the tangents to the circle
OP, OA and OB are joined.

To prove: ∠POA = ∠POB
Proof: OA and OB are the radii of the circle and PA and PB are the tangents to the circle
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
∠POA = ∠POB (c.p.c.t.)
Hence proved.

Question 14.
Solution:
Given : A circle with centre O, PA and PB are the tangents drawn from A and B which meets at P.
AB is chord of the circle
To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP

Proof: OA is radius and AP is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
Similarly, OB ⊥ BP ⇒ ∠OBP = 90°
In ∆OAB, OA = OB (radii of the circle)
∠OAB = ∠OBA
⇒ ∠OAP – ∠OAB = ∠OBP – ∠OBA
⇒ ∠PAB = ∠PBA
Hence proved.

Question 15.
Solution:
Given : A parallelogram ABCD is circumscribed a circle.
To prove : ABCD is a rhombus.
Proof: In a parallelogram ABCD.
Opposite sides are equal and parallel.

AB = CD and AD = BC
Tangents drawn from an external point of a circle to the circle are equal.
AP = AS BP = BQ
CQ = CR and DR = DS
Adding, we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC [AB = CD and AD = BC]
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
But AB = CD and BC = AD
AB = BC = CD = AD
Hence || gm ABCD is a rhombus.

Question 16.
Solution:
Given : O is the centre of two concentric circles with radii 5 cm and 3 cm respectively.
AB is the chord of the larger circle which touches the smaller circle at P.

OP and OA are joined.
To find : Length of AB
Proof: OP is the radius of the smaller circle and touches the smaller circle at P
OP ⊥ AB and also bisects AB at P
AP = PB = \(\frac { 1 }{ 2 }\) AB
Now, in right ∆OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
Hence AB = 2 x AP = 2 x 4 = 8 cm

Long-Answer Questions
Question 17.
Solution:
In the figure, quad. ABCD is circumscribed about a circle which touches its sides at P, Q, R and S respectively

To prove : AB + CD = AD + BC
Proof: Tangents drawn from an external point to a circle are equal
AP = AS
BP = BQ
CR = CQ
DR = DS
Adding, we get,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence AB + CD = AD + BC

Question 18.
Solution:
Given : A quad. ABCD circumscribe a circle with centre O and touches at P, Q, R and S respectively
OA, OB, OC and OD are joined forming angles AOB, BOC, COD and DOA
To prove : ∠AOB + ∠COD = 180°
and ∠BOC + ∠AOD = 180°
Construction : Join OP, OQ, OR and OS

Proof: In right ∆AOP and ∆AOS,
Side OP = OS (radii of the same circle)
Hyp. OA = OA (common)
∆AOP = ∆AOS (RHS axiom)
∠1 = ∠2 (c.p.c.t.)
Similarly, we can prove that
∠4 = ∠3
∠5 = ∠6
∠8 = ∠7
Adding, we get
∠1 + ∠4 + ∠5 + ∠8 = ∠2 + ∠3 + ∠6 + ∠7
⇒ (∠1 + ∠8) + (∠4 + ∠5) = (∠2 + ∠3) + (∠6 + ∠7)
⇒ ∠AOB + ∠COD = ∠AOD + ∠BOC
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (angles at a point)
∠AOB + ∠COD = ∠AOD + ∠BOC = 180°
Hence proved

Question 19.
Solution:
Given : From an external point P, PA and PB are the tangents drawn to the circle,
OA and OB are joined.
To prove : ∠APB + ∠AOB = 180°
Construction : Join OP.

Proof : Now, in ∆POA and ∆PBO,
OP = OP (common)
PA = PB (Tangents from P to the circle)
OA = OB (Radii of the same circle)
∆POA = ∆PBO (SSS axiom)
∠APO = ∠BPO (c.p.c.t.)
and ∠AOP = ∠BOP (c.p.c.t.)
OA and OB are the radii and PA and PB are the tangents
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
In ∆POA,
∠OAP = 90°
∠APO + ∠AOP = 90°
Similarly, ∠BPO + ∠BOP = 90°
Adding, we get
(∠APO + ∠BPO) + (∠AOP + ∠BOP) = 90° + 90°
⇒ ∠APB + ∠AOB = 180°.
Hence proved.

Question 20.
Solution:
Given : PQ is chord of a circle with centre O.
TP and TQ are tangents to the circle
Radius of the circle = 10 cm
i.e. OP = OQ = 10 cm and PQ = 16 cm

To find : The length of TP.
OT bisects the chord PQ at M at right angle.
PM = MQ = \(\frac { 16 }{ 2 }\) = 8 cm
In right ∆PMO,
OP² = PM² + MO² (Pythagoras Theorem)
⇒ (10)² = (8)² + MO²
⇒ 100 = 64 + MO²
⇒ MO² = 100 – 64 = 36 = (6)²
MO = 6 cm
Let TP = x and TM = y
In right ∆TPM,
TP² = TM² + PM²
⇒ x² = y² + 8²
⇒ x² = y² + 64 …(i)
and in right ∆TPM
OT² = TP² + OP²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y + 36 = y2 + 64 + 100 {From (i)}
⇒ 12y = 64 + 100 – 36 = 128

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Choose the correct answer in each of the following questions.
Question 1.
Solution:
Distance between P (-6, 8) and origin O (0, 0)

Question 2.
Solution:
Distance of the point (-3, 4) from x-axis = 4 (c)

Question 3.
Solution:
Let point P (x, 0) is on x-axis which is equidistant from points A (-1, 0) and B (5, 0)

Question 4.
Solution:
R (5, 6) is the midpoint of the line segment joining the points A (6, 5) and B (4, y)

Question 5.
Solution:
Point C (k, 4) divides the join of points A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 6.
Solution:
Vertices of ∆ABC are A (0, 4), B (0, 0) and C (3, 0)

Question 7.
Solution:
A (1, 3), B (-1, 2), C (2, 5) and D (x, 4) are the vertices of a ||gm ABCD.
AB = CD

Question 8.
Solution:
Points A (x, 2), B (-3, -4) and C (7, -5) are collinear

Question 9.
Solution:
Area of ∆ABC whose vertices are A (5, 0), B (8, 0) and C (8, 4)

Question 10.
Solution:
Area of ∆ABC with vertices A (a, 0), O (0, 0) and B (0, b)

Question 11.
Solution:
P( \(\frac { a }{ 2 }\), 4) is midpoint of the line segment joining the points A (-6, 5) and B (-2, 3)

Question 12.
Solution:
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3)

Question 13.
Solution:
Let coordinates of P be (x, 7)
P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1

Question 14.
Solution:
Let coordinates of one end of diameter of a circle are A (2, 3) coordinates of centre are (-2, 5)
Let coordinates of other end B of the diameter be (x, y)

Question 15.
Solution:
In the given figure, P (5, -3) and Q (3, y) are the points of trisection of the line segment joining A (7, -2) and B (1, -5)

Question 16.
Solution:
Midpoint of AB is P (0, 4) coordinates of B are (-2, 3)

Question 17.
Solution:
Let point P (x, y) divides AB with vertices A(2, -5) and B (5, 2) in the ratio 2 : 3

Question 18.
Solution:
A (-6, 7) and B (-1, -5) two points, then

Question 19.
Solution:
Let point P (x, 0) on x-axis is equidistant from A (7, 6) and B (-3, 4)

Question 20.
Solution:
Distance of P (3, 4) from the x-axis = 4 units. (b)

Question 21.
Solution:
Let a point P (x, 0) on x-axis divides the line segment.
AB joining the points A (2, -3) and B (5, 6) in the ratio m : n.

Question 22.
Solution:
Let a point A (0, y) on y-axis divides the line segment PQ joining the points P (-4, 2) and Q (8, 3) in the ratio of m : n.

Question 23.
Solution:
P (-1, 1) is the midpoint of line segment joining the points A (-3, b) and B (1, b + 4)

Question 24.
Solution:
Let the point P (x, y) divides the line segment joining the points A (2, -2) and B (3, 7) in the ratio k : 1, then

Question 25.
Solution:
AD is the median of ∆ABC with vertices A (4, 2), B (6, 5) and C (1, 4)

Question 26.
Solution:
A (-1, 0), B (5, -2) and C (8, 2) are the vertices of a ∆ABC then centroid

Question 27.
Solution:
Two vertices A (-1, 4), B (5, 2) of a ∆ABC and its centroid G is (0, -3)
Let (x, y) be the co-ordinates of vertex C, then

Question 28.
Solution:
Points A (-4, 0), B (4, 0) and C (0, 3) are vertices of a ∆ABC.

Question 29.
Solution:
Points P (0, 6), Q (-5, 3) and R (3, 1) are the vertices of a ∆PQR

Question 30.
Solution:
Points A (2, 3), B (5, k) and C (6, 7) are collinear
Area of ∆ABC = 0

Question 31.
Solution:
Points A (1, 2), B (0, 0) and C (a, b) are collinear
Area of ∆ABC = 0

Question 32.
Solution:
A (3, 0), B (7, 0) and C (8, 4)
Area ∆ABC

Question 33.
Solution:
AOBC is a rectangle with vertices A (0, 3), O (0, 0) and B (5, 0) and each diagonal of rectangle are equal.

Question 34.
Solution:
Points are A (4, p) and B (1, 0)
Distance = 5 units

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) Let P (x, y) be the required point which divides the line joining the points A (-1, 7) and B (4, -3) in the ratio 2 : 3.

Question 2.
Solution:
Let (7, -2) and B (1, -5) be the given points and P (x, y) and Q (x’, y’) are the points of trisection.

Question 3.
Solution:
Let coordinates of P be (x, y) which divides the line segment A (-2, -2) and B (2, -4) in

Question 4.
Solution:
Let the coordinates of A be (x, y) which lies on line joining P (6, -6) and Q (-4, -1)

Question 5.
Solution:
Points P, Q, R and S divides a line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts.

Question 6.
Solution:
Points P, Q and R in order divide a line segment joining the points A (1, 6) and B (5, -2) in 4 equal parts.

P divides AB in the ratio of 1 : 3 Let coordinates of P be (x, y), then

Question 7.
Solution:
The line segment joining the point A (3, -4) and B (1, 2) is trisected by the points P (p, -2) and Q(\(\frac { 1 }{ 2 }\), q).

Question 8.
Solution:
Mid point of the line segment joining the points A (3, 0) and B (-5, 4)

Question 9.
Solution:
(2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

Question 10.
Solution:
Mid point of the line segment joining the points A (2a, 4) and B (-2, 3b) is C (1, 2a + 1)

Question 11.
Solution:
The line segment joining the points A (-2, 9) and B (6, 3) is a diameter of a circle with centre C.

Question 12.
Solution:
AB is diameter of a circle with centre C.
Coordinates of C (2, -3) and of B (1, 4)

Question 13.
Solution:
Let P (2, 5) divides the line segment joining the points A (8,2) and B (-6, 9) in the ratio m : n

Question 14.
Solution:

Question 15.
Solution:
Let P (m, 6) divides the join of A (-4, 3) and B (2, 8) in the ratio k : 1
Then coordinates of P will be

Question 16.
Solution:
Let point P (-3, k) divides the join of A (-5, -4) and B (-2, 3) in the ratio m : n, then

Question 17.
Solution:
Let point P on the x-axis divides the line segment joining the points A (2, -3) and B (5, 6) the ratio m : n
Let P is the point on x-axis whose coordinates are (x, 0)

Question 18.
Solution:
Let a point P on y-axis divides the line segment joining the points A (-2, -3) and B (3, 7) in the ratio m : n
Let the coordinates of P be (0, y)

Question 19.
Solution:
Let a point P (x, y) on the given line x – y – 2 = 0 divides the line segment joining the points A (3, -1) and B (8, 9) in the ratio m : n, then

Question 20.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)
Let AD, BE and CF are the medians of sides BC, CA and AB respectively, then
Coordinates of D will be =

Question 21.
Solution:
Centroid of ∆ABC where coordinates of A are (-1, 0), of B are (5, -2) and of C are (8, 2)

Question 22.
Solution:
G (-2, 1) is the centroid of ∆ABC whose vertex A is (1, -6) and B is (-5, 2)
Let the vertex C be (x, y), then

Question 23.
Solution:
O (0, 0) is the centroid of ∆ABC in which B is (-3, 1), C is (0, -2)
Let A be (x, y), then

Question 24.
Solution:
Points are A (3, 1), B (0, -2), C (1, 1) and D (4, 4)

Question 25.
Solution:
Points P (a, -11), Q (5, b), R (2, 15) and S (1, 1) are the vertices of a parallelogram PQRS.
Diagonals of a parallelogram bisect each other.
O is mid point of PR and QS.

Question 26.
Solution:
Three vertices of a parallelogram ABCD are A (1, -2), B (3, 6), C (5, 10).
Let fourth vertices D be (x, y)

Question 27.
Solution:
Let a point P (0, y) on 7-axis, divides the line segment joining the points (-4, 7) and (3, -7) in the ratio m : n

Question 28.
Solution:

Question 29.
Solution:
Let a point P (x, 0) divides the line segment joining the points A (3, -3) and B (-2, 2) in the ratio m : n

Question 30.
Solution:
Base QR of an equilateral APQR lies on x- axis is O (0, 0) is mid point PQR and coordinate of Q are (-4, 0).
Coordinate of R will be (4, 0)

Question 31.
Solution:
Base BC of an equilateral triangle ABC lies on y-axis in such a way that origin O (0, 0) lies is the middle of BC and coordinates of C are (0, -3).
Coordinates of B will be (0, 3)

Question 32.
Solution:
Let the points P (-1, y) lying on the line segment joining points A (-3, 10) and B (6, -8) divides it in the ratio m : n.

Question 33.
Solution:
In rectangle ABCD, A (-1, -1), B (-1, 4), C (5, 4), D (5, -1)
P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Question 34.
Solution:
P is mid point of line segment joining the points A (-10, 4) and B (-2, 0)

Question 35.
Solution:
Let coordinates of P and Q be (0, y) and (x, 0) respectively.
Let M (2, -5) be the mid-point of PQ.
By midpoint formula

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:
Let the other two vertices be (h, k) and (m, n).
Hence, the vertices in order are (3, 2), (-1, 0), (h, k) and (m, n).
It is to be kept in mind that the diagonals of a parallelogram bisect each other.
Hence, the point of intersection (2, -5) is the midpoint of the diagonal whose ends are (3, 2) and (h, k). Then

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -A)

Question 2.
Solution:
Vertices of quadrilateral ABCD are A (3,-1), B (9, -5), C (14, 0) and D (9, 19)
Join diagonal AC.

Question 3.
Solution:
PQRS is a quadrilateral whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2)
Join PR which forms two triangles PQR and PSR.

Question 4.
Solution:
ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4)
Join AC which terms two triangles ABC and ADC

Question 5.
Solution:

Question 6.
Solution:
D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC.
Vertices are A (2, 1), B (4, 3) and C (2, 5)

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)
D is midpoint of BC


Similarly area of ∆ACD = 5 sq. units (10 – 5 = 5 sq. units)
Hence, median divides the triangle into two triangles of equal in area.

Question 8.
Solution:
In ∆ABC, coordinates of A are (1, -4) and let C and E are the midpoints of AB and AC respectively.
Coordinates of F are (2, -1) and of E are (0, -1)
Let coordinates of B be (x1, y1) and of C be (x2, y2)

Question 9.
Solution:
A (6, 1), B (8, 2) and C (9, 4) are the three vertices of a parallelogram ABCD.
E is the midpoint of DC.
Join AE, AC and BD which intersects at O.
O is midpoint of AC

Question 10.
Solution:
(i) Vertices of a ∆ABC are A (1, -3), B (4, p) and C (-9, 7) and area = 15 sq. units

Question 11.
Solution:
Vertices of ∆ABC are A (k + 1, 1), B (4, -3) and C (7, -k) and area = 6 sq. units.

Question 12.
Solution:
Let the vertices of a triangle ABC are A (-2, 5), B (k, -4) and C (2k + 1, 10) and area = 53 sq. units

Question 13.
Solution:
(j) Points are A (2, -2), B (-3, 8) and C (-1, 4)

Question 14.
Solution:
Points are A (x, 2), B (-3, -4) and C (7, -5)

Question 15.
Solution:
Points are given A (-3, 12), B (7, 6) and C (x, 9)

Question 16.
Solution:
Points are given P (1, 4), Q (3, y) and R (-3, 16)

Question 17.
Solution:
The given points are A (-3, 9), B (2, y) and C (4, -5)

Question 18.
Solution:
The points are given A (8, 1), B (3, -2k) and C (k, -5)

Question 19.
Solution:
The points are given A (2, 1), B (x, y) and C (7, 5)

Question 20.
Solution:
The points are given A (x, y), B (-5, 7) and C (-4, 5)

Question 21.
Solution:
Points are given A (a, 0), B (0, b) and C (1,1)
Points are collinear.
Area of ∆ABC = 0

Question 22.
Solution:
The points are given P (-3, 9), Q (a, ti) and R (4, -5)
Points are collinear.
Area of ∆PQR = 0

Question 23.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

Question 24.
Solution:
Let A (a, a²), B (b, b²) and C (0, 0)
For the points A, B and C to collinear area of ∆ABC must be zero.
Now, area of ∆ABC

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) A (9, 3) and B (15, 11)

Question 2.
Solution:
Distance from origin O (0, 0) and the given points (x, y) = √(x² + y²)

Question 3.
Solution:
Points A (x, -1), B (5, 3) and AB = 5 units

Question 4.
Solution:
Points A (2, -3), B (10, y) and AB = 10

Question 5.
Solution:
Points P (x, 4), Q (9, 10) and PQ = 10

Question 6.
Solution:
Point A (x, 2) is equidistant from B (8, -2 and C (2, -2)

Question 7.
Solution:
A (0, 2) is equidistant from B (3, p) and C ip, 5)
Then AB = AC

Question 8.
Solution:
Let point P (x, 0) is on x-axis and P is equidistant from A (2, -5) and B (-2, 9)

Question 9.
Solution:
Let the points on x-axis be P (x,, 0) and Q (x2, 0) and A (11, -8)

or x – 5 = 0, then x = 5
Points are (17, 0) and (5, 0)

Question 10.
Solution:
Let point P (0, y) is on the y-axis, then

Question 11.
Solution:
P (x, y) is equidistant from A (5, 1) and B (-1, 5)

Question 12.
Solution:
P (x, y) is equidistant from A (6, -1) and B (2, 3)

Question 13.
Solution:
Let the coordinates of the points be O (x, y)

Question 14.
Solution:
Points A (4, 3) and B (x, 5) lie on a circle with centre O (2, 3)

Question 15.
Solution:
Point C (-2, 3) is equidistant from points A (3, -1) and B (x, 8)

Question 16.
Solution:
Point P (2, 2) is equidistant from the two points A (-2, k) and B (-2k, -3)

Question 17.
Solution:
(i) Let point P (x, y) is equidistant from A (a + b, b – a) and B (a – b, a + b), then

Question 18.
Solution:
We know that if the sum of any two of these distances is equal to the distance of the third, then the points are collinear.
Now, (i) Let the points are A (1, -1), B (5, 2), C (9, 5)

Question 19.
Solution:
The points are A (7, 10), B (-2, 5) apd C (3, -4)

Question 20.
Solution:
Points are A (3, 0), B (6, 4) and C (-1, 3)

Question 21.
Solution:
Points are A (5, 2), B (2, -2) and C (-2, t)

Question 22.
Solution:
Points are A (2, 4), B (2, 6) and C (2 + √3, 5)

Question 23.
Solution:
Let the points are A (-3, -3), B (3, 3), C (-3√3, 3√3)

Question 24.
Solution:
Points are A (-5, 6), B (3, 0), C (9, 8)

Question 25.
Solution:
Points are O (0, 0), A (3, √3) and B (3, -√3)

Question 26.
Solution:
(i) Points are A (3, 2), B (0, 5), C (-3, 2), D (0, -1)

Question 27.
Solution:
Points are A (-3, 2), B (-5, -5), C (2, -3), D (4, 4)

Question 28.
Solution:
Points are A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)

Question 29.
Solution:
Points are A (6, 1), B (8, 2), C (9, 4) and D (7, 3)

Question 30.
Solution:
Points are A (2, 1), B (5, 2), C (6, 4) and D (3, 3)

Question 31.
Solution:
Points are A (1, 2), B (4, 3), C (6, 6) and D (3, 5)

Question 32.
Solution:
(i) Points are A (-4, -1), B (-2, -4), C (4, 0) and D (2, 3)

Question 33.
Solution:

Question 34.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Very-Short-Answer Questions
Question 1.
Solution:
Points are A (-1, y), B (5, 7) and centre O (2, -3y).
Points A and B lie on the circle with centre O.

or y + 1 = 0, then y = -1
y = -1, 7

Question 2.
Solution:
Point A (0, 2) is equidistant from B (3, p) and also from C (p, 5)

Question 3.
Solution:
Three vertices of a rectangle ABCD are B (4, 0), C (4, 3) and D (0, 3)
Its diagonal are equal.

Question 4.
Solution:
Point P (k – 1, 2) is equidistant from two points A (3, k) and B (k, 5)

Question 5.
Solution:
Let P (x, 2) divides the join of A (12, 5) and B (4, -3) in the ration m : n.

Question 6.
Solution:

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)

Question 8.
Solution:
Point C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 9.
Solution:
Let a point P (x, 0) on x-axis is equidistant from two points A (-1, 0) and B (5, 0)

Question 10.
Solution:
Distance between two points

Question 11.
Solution:
The points (3, a) lies on the line 2x – 3y = 5
It will satisfy it.
2 x 3 – 3 x a = 5
6 – 3a = 5 => 3a = 6 – 5 = 1
a = \(\frac { 1 }{ 3 }\)

Question 12.
Solution:
Points A (4, 3) and B (x, 5) lie on the circle with centre O (2, 3)
OA = OB

Question 13.
Solution:
P (x, y) is equidistant from the point A (7, 1) and B (3, 5)

Question 14.
Solution:
O (0, 0) is the centroid of ∆ABC whose vertices are A (a, b), B (b, c) and C (c, a)

Question 15.
Solution:
Coordinates of centroid

Question 16.
Solution:
Let point P (4, 5) divides the join of A (2, 3) and B (7, 8) in the ratio m : n

Question 17.
Solution:
Points are given A (2, 3), B (4, k) and C (6, -3)
Points are collinear.

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
S_n = 3n² + 6n

Question 5.
Solution:
S_n = 3n² – n
S₁ = 3(1)² – 1 = 3 – 1 = 2
S₂ = 3(2)² – 2 = 12 – 2 = 10
T₂ = 10 – 2 = 8 and
T₁ = 2
(i) First term = 2
(ii) Common difference = 8 – 2 = 6
T_n = a + (n – 1) d = 2 + (n – 1) x 6
= 2 + 6n – 6 = 6n – 4

Question 6.
Solution:

Question 7.
Solution:
Let a be first term and d be the common difference of an AP.
Since, we have,
a_m = a + (m – 1) d = \(\frac { 1 }{ n }\) …(i)

Question 8.
Solution:
AP is 21, 18, 15,…
Here, a = 21,
d = 18 – 21 = -3,
sum = 0
Let number of terms be n, then

Question 9.
Solution:
AP is 9, 17, 25,…
Here, a = 9, d = 17 – 9 = 8
Sum of terms = 636
Let number of terms be n, then

Which is not possible being negative and fraction.
n = 12
Number of terms = 12

Question 10.
Solution:
AP is 63, 60, 57, 54,…
Here, a = 63, d = 60 – 63 = -3 and sum = 693
Let number of terms be n, then

22th term is zero.
There will be no effect on the sum.
Number of terms are 21 or 22.

Question 11.
Solution:

Question 12.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …, 49
Here, a = 1, d = 3 – 1 = 2, l = 49

= \(\frac { 25 }{ 2 }\) x 50 = 25 x 25 = 625

Question 13.
Solution:
Numbers between 200 and 400 which are divisible by 7 will be 203, 210, 217,…, 399
Here, a = 203, d = 7, l = 399

Question 14.
Solution:
First forty positive integers are 0, 1, 2, 3, 4, …
and numbers divisible by 6 will be 6, 12, 18, 24, … to 40 terms
Here, a = 6, d = 12 – 6 = 6, n = 40 .

Question 15.
Solution:
First 15 multiples of 8 are 8, 16, 24, 32, … to 15 terms
Here, a = 8, d = 16 – 8 = 8 , n = 15

Question 16.
Solution:
Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, …, 693
Here, a = 306, d = 9, l = 693

Question 17.
Solution:
Three digit numbers are 100, 101, …, 999
and numbers divisible by 13, will be 104, 117, 130, …, 988
Here, a = 104, d = 13, l = 988
T_n (l) = a + (n – 1) d

Question 18.
Solution:
Even natural numbers are 2, 4, 6, 8, 10, …
Even natural numbers which are divisible by 5 will be
10, 20, 30, 40, … to 100 terms
Here, a = 10, d = 20 – 10 = 10, n = 100

Question 19.
Solution:

Question 20.
Solution:
Let a be the first term and d be the common difference of the AP, then

Question 21.
Solution:
Let a be the first term and d be the common difference, then

Question 22.
Solution:
Let a be the first term and d be the common difference, then

Question 23.
Solution:
In an AP
a = 17, d = 9, l = 350
Let number of terms be n, then

Question 24.
Solution:
Let a be the first term, d be the common difference, then

Question 25.
Solution:
In an AP, let a be the first term and d be the common difference, then

Question 26.
Solution:
Let a be the first term and d be the common difference of an AP, then

Question 27.
Solution:
Let a be first term and d be the common difference, then

Question 28.
Solution:
Let first term be a and d be the common difference in an AP, then

Question 29.
Solution:
Let a be the first term and d be the common difference of an AP, then

Question 30.
Solution:
Let a₁ and a₂ be the first terms of the two APs and d be the common difference, then
a₁ = 3, a₂ = 8

Question 31.
Solution:
Let a be the first term and d be the common difference, then
S₁₀ = -150

Question 32.
Solution:
Let a be the first term and d be the common difference, then

Question 33.
Solution:
Let a be the first term and d be the common difference of an AP, then

Question 34.
Solution:
(i) AP is 5, 12, 19,… to 50 terms
Here, a = 5, d = 12 – 5 = 7, n = 50

Question 35.
Solution:
Let a₁, a₂ be the first term and d₁, d₂ be common difference of the two AP’s respectively.
Given, ratio of sum of first n terms =

Question 36.
Solution:
Let a be the first term and d be the common difference of an AP.

Question 37.
Solution:

Question 38.
Solution:

Question 39.
Solution:
AP is -12, -9, -6,…, 21
Here, a = -12, d = -9 – (-12) = -9 + 12 = 3, l = 21

Question 40.
Solution:
S₁₄ = 1505
Let a be the first term and d be the common difference, then
a = 10

Question 41.
Solution:
Let a be the first term and d be the common difference of an AP, then

Question 42.
Solution:
In the school, there are 12 classes, class 1 to 12 and each class has two sections.
Each class plants double of the class
i.e. class 1 plants two plant, class 2 plants 4 plants, class 3 plants 6 plants, and so on.
So, total plants will be for each class each sections = 2 + 4 + 6 + 8 … + 24
Here, a = 2, d = 2, l = 24, n = 12

Each class has. two sections.
Plants will also be doubt
i.e. Total plants = 156 x 2 = 312

Question 43.
Solution:
In a potato race,
Bucket is at 5 m from first potato and then the distance between the two potatoes is 3m.
There are 10 potatoes.
The player, pick potato and put it in the bucket one by one.
Total distance in m to be covered, for 1st, 2nd, 3rd, … potato.

Question 44.
Solution:
Total number of trees = 25
Distance between them = 5 m in a line.
There is a water tank which is 10 m from the first tree.
A gardener waters these plants separately.
Total distance for going and coming back

Question 45.
Solution:
Total sum = ₹ 700
Number of cash prizes = 7
Each prize in ₹ 20 less than its preceding prize.
Let first prize = ₹ x
Then second prize = ₹ (x – 20)
Third prize = ₹ (x – 40) and so on.

Question 46.
Solution:
Total savings = ₹ 33000
Total period = 10 months
Each month, a man saved ₹ 100 more than its preceding of month.
Let he saves ₹ x in the first month.

Question 47.
Solution:
Total debt to be paid = ₹ 36000
No. of monthly installments = 40
After paying 30 installments, \(\frac { 1 }{ 3 }\) of his debt left
i.e. ₹ 36000 x \(\frac { 1 }{ 3 }\) = ₹ 12000
and amount paid = ₹ 36000 – ₹ 12000 = ₹ 24000
Monthly installments are in AP.
Let first installment = x
and common difference = d

Question 48.
Solution:
A contractor will pay the penalty for not doing the work in time.
For the first day = ₹ 200
For second day = ₹ 250
For third day = ₹ 300 and so on
The work was delayed for 30 days
Total penalty, he paid
200 + 250 + 300 + ….. to 30 terms
Here, a = 200, d = 50, n = 30
Total sum = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
= \(\frac { 30 }{ 2 }\) [2 x 200 + (30 – 1) x 50]
= 15 [400 + 29 x 50]
= 15 [400 + 1450]
= 15 x 1850 = ₹ 27750

Question 49.
Solution:
Child will put 5 rupee on 1st day, 10 rupee (2 x 5 rupee)
on 2nd day, 15 rupee (3 x 5 rupee) on 3rd day etc.
Total saving = 190 coins = 190 x 5 = 950 rupee
The above problem can be written as Arithmetic Progression series
5, 10, 15, 20, ……
with a = 5, d = 5, S_n = 950
Let n be the last day when piggy bank become full.

⇒ n (n + 20) – 19 (n + 20) = 0
⇒ (n + 20) (n – 19) = 0
⇒ n + 20 = 0 or n – 19 = 0
⇒ n = -20 or n = 19
cannot be negative, hence n = 19
She can put money for 19 days.
Total saving is 950 rupees.

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B
• RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself

Very-Short-Answer Questions
Question 1.
Solution:
In the given figure, a circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.
AB = 6 cm, BC = 9 cm, CD = 8 cm.

To find : The length of side AD.
A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD
=> 6 + 8 = 9 + AD
=> 14 = 9 + AD
=> AD = 14 – 9 = 5 cm

Question 2.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O.
∠APB = 50°
To find : Measure of ∠OAB.
Construction : Join OB.

In ∆APB,
PA = PB (tangents of the circle)
∠PAB = ∠PBA
But, ∠PAB + ∠PBA + ∠APB = 180° (Angles of a triangle)
=> ∠PAB + ∠PAB + 50° = 180°
=> 2∠PAB = 180° – 50° = 130°
∠PAB = 65°
But ∠OAP = 90° (OA is radius and PA is tangent)
∠OAB = 90° – 65° = 25°

Question 3.
Solution:
In the given figure, O is the centre of a circle.
PT and PQ are tangents to the circle from an external point P.
R is any point on the circle. RT and RQ are joined.
∠TPQ = 70°
To find : ∠TRQ
Construction : Join TO and QO.

∠TPQ = 70°
∠TOQ = 180° – 70° = 110° (OT and OQ are perpendicular on TP and QP)
Now, ∠TOQ is on the centre and ∠TRQ is on the remaining part of the circle.
∠TRQ = \(\frac { 1 }{ 2 }\) x ∠TOQ = \(\frac { 1 }{ 2 }\) x 110° = 55°

Question 4.
Solution:
In the given figure, common tangents AB and CD to the two circles with centres O1 and O2 intersect each other at E.
To prove : AB = CD.
Proof : EA and EC are tangents to the circle O1
EA = EC …(i)
Similarly, EB and ED are tangents to the circle O2.
EB = ED …(ii)
Adding (i) and (ii),
EA + EB = EC + ED
=> AB = CD
Hence, AB = CD

Question 5.
Solution:
In the given figure, PT is the tangent to the circle with centre O at P.
PQ is a chord of the circle and ∠TPQ = 70°.
To find : The measure of ∠POQ.
PT is tangent and OP is the radius.
∠OPT = 90°
But ∠QPT = 70°
∠OPQ = 90° – 70° = 20°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 20°
and ∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (20° + 20°)
= 180° – 40° = 140°

Short-Answer Questions
Question 6.
Solution:
In the given figure, ∆ABC is circumscribed a circle with centre O and radius 2 cm.
Point D divides BC in such a way that
BD = 4 cm, DC = 3 cm, OD = 2 cm
Area of ∆ABC = 21 cm²
To find : AB and AC.
Construction : Join OA, OB, OC, OE and OF.

BD and BF are tangents to the circle.
BF = BD = 4 cm.
Similarly, CD and CE are tangents.
CE = CD = 3 cm
and AF and AE are tangents
AE = AF = x (suppose)
Now, area of ∆ABC = \(\frac { 1 }{ 2 }\) x Perimeter of ∆ABC x Radius
21 = \(\frac { 1 }{ 2 }\) (AB + BC + CA) x OD
=> 21 x 2 = [4 + 3 + 3+ x + x + 4) x 2
=> 42 = (14 + 2x) x 2
=> 14 + 2x = \(\frac { 42 }{ 2 }\) = 21
=> 2x = 21 – 14 = 7
x = \(\frac { 7 }{ 2 }\) = 3.5
AB = AF + FB = 3.5 + 4 = 7.5 cm
AC = AE + CE = 3.5 + 3 = 6.5 cm

Question 7.
Solution:
Given : Two concentric circles with centre O and radii 5 cm and 3 cm respectively.

AB is chord of larger circle which touches the smaller circle at C.
To find : The length of chord AB.
Construction : Join OA and OC.
AB is tangent and OC is radius of the smaller circle.
OC ⊥ AB and OC bisects AB at C. (AB is chord and OC ⊥ AB)
In right ∆OAC,
OA² = OC² + AC² (Pythagoras Theorem)
=> (5)² = (3)² + AC²
=> 25 = 9 + AC²
=> AC² = 25 – 9 = 16 = (4)²
=> AC = 4
and AB = 2 x AC = 2 x 4 = 8cm

Question 8.
Solution:
Given : AB is the tangent to the circle with centre O at point P.
PL ⊥ AB

To prove : PL passes through O.
Let PQ ⊥PT where Q lies on the circle.
∠QPT = 90°
Let PQ does not pass through the centre O.
Join PO and produce it to meet the circle at L.
PO being the radius of the circle drawn from the point of contact P.
OP ⊥ AB
=> ∠OPB = 90° => ∠LPB = 90°
But, PQ ⊥ AB
∠QPB = 90°
It is possible only if L and Q coincide each other.
Hence, PQ passes through the centre and is perpendicular from the point of contact.

Question 9.
Solution:
In the given figure, two tangents RQ and RP are drawn from the external point R to the circle with centre O.
∠PRQ = 120°
To prove : OR = PR + RQ
Construction : Join OP and OQ.
Also join OR.

Question 10.
Solution:
In the given figure, a circle is inscribed in a ∆ABC touches the sides AB, BC and CA at D, E and F respectively.
AB = 14 cm, BC = 8 cm and CA = 12 cm.

To find : The length of AD, BE and CF.
Let AD = x, BE = y and CF = z
AD and AF are the tangents to the circle from A.
AD = AF = x
Similarly,
BE and BD are tangents
BD = BE = y
and CF and CE are the tangents
CE = CF = z
Now, AB + BC + CA = 14 + 8 + 12 = 34
=> (x + y) + (y + z) + (z + x) = 34
=> 2 (x + y + z) = 34
=> x + y + z = 17 …(i)
But x + y = 14 cm …(ii)
y + z = 8 cm …(iii)
z + x = 12 cm …(iv)
Subtracting (iii), (iv) and (ii) from (i) term by term
x = 17 – 8 = 9 cm
y = 17 – 12 = 5 cm
z = 17 – 14 = 3 cm
Hence, AD = 9 cm, BE = 5 cm and CF = 3 cm.

Question 11.
Solution:
In the given figure, O is the centre of the circle.
PA and PB are the tangents.
To prove : AOBP is a cyclic quadrilateral.
Proof: OA is radius and PA is tangent
OA ⊥ PA
=> ∠OAP = 90° ….. (i)
Similarly, OB is radius and PB is tangent.
OB ⊥ PB
=> ∠OBP = 90° ….. (ii)
Adding (i) and (ii),
∠OAP + ∠OBP = 90° + 90° = 180°
But these are opposite angles of the quadrilateral AOBP.
Quadrilateral AOBP is a cyclic.

Question 12.
Solution:
In two concentric circles with centre O, a chord AB of the laiger circle touches the smaller circle at C.
AB = 8 cm and radius of larger circle = 5 cm
Join OA, OC
To find, the radius of smaller circle,
AB is the tangent and OC is the radius
OC ⊥ AB

AC = CB = \(\frac { 8 }{ 2 }\) = 4 cm
OA = 5 cm
In right ∆OCA,
OA² = OC² + AC² (Pythagoras Theorem)
(5)² = OC² + (4)²
OC² = (5)² – (4)² = 25 – 16 = 9 = (3)²
OC = 3
Radius of smaller circle = 3 cm

Question 13.
Solution:
In the given figure, PQ is a chord of a circle with centre O.
PT is the tangent ∠QPT = 60°.
To find : ∠PRQ.
Construction : Take a point M on the alternate segment.
Join MP and MQ.

∠MPQ = ∠QPT = 60° (Angles in the alternate segment)
∠PMQ + ∠PRQ = 180° (Opposite angles of a cyclic quadrilateral)
=> 60° + ∠PRQ = 180°
∠PRQ = 180° – 60° = 120°
Hence, ∠PRQ = 120°

Question 14.
Solution:
In the given figure,
PA and PB are the two tangents to the circle.
With centre O, OA and AB are joined
∠APB = 60°
To find : The measure of ∠OAB
PA and PB are tangents to the circle from P
PA = PB
∠PAB = ∠PBA
But ∠APB = 60°
∠PAB + ∠PBA = 180° – 60° = 120°
2 ∠PAB = 120°
∠PBA = 60°
OA is radius and PA is tangent.
OA ⊥ PA
∠OAP = 90°
=> ∠OAB + ∠PAB = 90°
=> ∠OAB + 60° = 90°
=> ∠OAB = 90° – 60° = 30°
Hence, ∠OAB = 30°

Question 15.
Solution:
Since, tangents drawn from an external point are equally inclined to the line joining centre to that point.

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

Very-Short and Short-Answer Questions
Question 1.
Solution:
(3y – 1), (3y + 5) and (5y+ 1) are in AP
(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
⇒ 2 (3y + 5) = (5y + 1) + (3y – 1)
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5

Question 2.
Solution:
k, (2k – 1) and (2k + 1) are the three successive terms of an AP.
(2k – 1) – k = (2k + 1) – (2k – 1)
⇒ 2 (2k – 1) = 2k + 1 + k
⇒ 4k – 2 = 3k + 1
⇒ 4k – 3k = 1 + 2
⇒ k = 3
k = 3

Question 3.
Solution:
18, a, (b – 3) are in AP
⇒ a – 18 = b – 3 – a
⇒ a + a – b = -3 + 18
⇒ 2a – b = 15

Question 4.
Solution:
a, 9, b, 25 are in AP.
9 – a = b – 9 = 25 – b
b – 9 = 25 – b
⇒ b + b = 22 + 9 = 34
⇒ 2b = 34
⇒ b= 17
a – b = a – 9
⇒ 9 + 9 = a + b
⇒ a + b = 18
⇒ a + 17 = 18
⇒ a = 18 – 17 = 1
a = 18, b= 17

Question 5.
Solution:
(2n – 1), (3n + 2) and (6n – 1) are in AP
(3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)
⇒ (3n + 2) + (3n + 2) = 6n – 1 + 2n – 1
6n + 4 = 8n – 2
⇒ 8n – 6n = 4 + 2
⇒ 2n = 6
⇒ n = 3
and numbers are
2 x 3 – 1 = 5
3 x 3 + 2 = 11
6 x 3 – 1 = 17
i.e. (5, 11, 17) are required numbers.

Question 6.
Solution:
Three digit numbers are 100 to 990 and numbers which are divisible by 7 will be
105, 112, 119, 126, …, 994
Here, a = 105, d= 7, l = 994
T_n = (l) = a + (n – 1) d
⇒ 994 = 105 + (n – 1) x 7
⇒ 994 – 105 = (n – 1) 7
⇒ (n – 1) x 7 = 889
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128
Required numbers are 128

Question 7.
Solution:
Three digit numbers are 100 to 999
and numbers which are divisible by 9 will be
108, 117, 126, 135, …, 999
Here, a = 108, d= 9, l = 999
T_n (l) = a + (n – 1) d
⇒ 999 = 108 + (n – 1) x 9
⇒ (n – 1) x 9 = 999 – 108 = 891
⇒ n – 1 = 99
⇒ n = 99 + 1 = 100

Question 8.
Solution:
Sum of first m terms of an AP = 2m² + 3m
S_m = 2m² + 3m
S₁ = 2(1)² + 3 x 1 = 2 + 3 = 5
S₂ = 2(2)² + 3 x 2 = 8 + 6=14
S₃ = 2(3)² + 3 x 3 = 18 + 9 = 27
Now, T₂ = S₂ – S₁ = 14 – 5 = 9
Second term = 9

Question 9.
Solution:
AP is a, 3a, 5a, …
Here, a = a, d = 2a

Question 10.
Solution:
AP 2, 7, 12, 17, …… 47
Here, a = 2, d = 7 – 2 = 5, l = 47
nth term from the end = l – (n – 1 )d
5th term from the end = 47 – (5 – 1) x 5 = 47 – 4 x 5 = 47 – 20 = 27

Question 11.
Solution:
AP is 2, 7, 12, 17, …
Here, a = 2, d = 7 – 2 = 5
a_n = a + (n – 1) d = 2 + (n – 1) x 5 = 2 + 5n – 5 = 5n – 3
Now, a₃₀ = 2 + (30 – 1) x 5 = 2 + 29 x 5 = 2 + 145 = 147
and a₂₀ = 2 + (20 – 1) x 5 = 2 + 19 x 5 = 2 + 95 = 97
a₃₀ – a₂₀ = 147 – 97 = 50

Question 12.
Solution:
T_n = 3n + 5
T_n-1 = 3 (n – 1) + 5 = 3n – 3 + 5 = 3n + 2
d = T_n – T_n-1 = (3n + 5) – (3n + 2) = 3n + 5 – 3n – 2 = 3
Common difference = 3

Question 13.
Solution:
T_n = 7 – 4n
T_n-1 = 7 – 4(n – 1) = 7 – 4n + 4 = 11 – 4n
d = T_n – T_n-1 = (7 – 4n) – (11 – 4n) = 7 – 4n – 11 + 4n = -4
d = -4

Question 14.
Solution:
AP is √8, √18, √32, …..
⇒ √(4 x 2) , √(9 x 2) , √(16 x 2), ………

Question 15.
Solution:

Question 16.
Solution:
AP is 21, 18, 15, …n
Here, a = 21, d = 18 – 21 = -3, l = 0
T_n (l) = a + (n – 1) d
0 = 21 + (n – 1) x (-3)
0 = 21 – 3n + 3
⇒ 24 – 3n = 0
⇒ 3n = 24
⇒ n = 8 .
0 is the 8th term.

Question 17.
Solution:
First n natural numbers are 1, 2, 3, 4, 5, …, n
Here, a = 1, d = 1

Question 18.
Solution:
First n even natural numbers are 2, 4, 6, 8, 10, … n
Here, a = 2, d = 4 – 2 = 2

Question 19.
Solution:
In an AP
First term (a) = p
and common difference (d) = q
T₁₀ = a + (n – 1) d = p + (10 – 1) x q = (p + 9q)

Question 20.
Solution:

Question 21.
Solution:
2p + 1, 13, 5p – 3 are in AP, then
13 – (2p + 1) = (5p – 3) – 13
⇒ 13 – 2p – 1 = 5p – 3 – 13
⇒ 12 – 2p = 5p – 16
⇒ 5p + 2p = 12 + 16
⇒ 7p = 28
⇒ p = 4
P = 4

Question 22.
Solution:
(2p – 1), 7, 3p are in AP, then
⇒ 7 – (2p – 1) = 3p – 7
⇒ 7 – 2p + 1 = 3p – 7
⇒ 7 + 1 + 7 = 3p + 2p
⇒ 5p = 15
⇒ p = 3
P = 3

Question 23.
Solution:

Question 24.
Solution:

d = T₂ – T₁ = 14 – 8 = 6
Common difference = 6

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

Question 1.
Solution:
(3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
(4k – 6) – (3k – 2) = (k + 2) – (4k – 6)
⇒ 2(4k – 6) = (k + 2) + (3k – 2)
⇒ 8k – 12 = 4k + 0
⇒ 8k – 4k = 0 + 12
⇒ 4k = 12
k = 3

Question 2.
Solution:
(5x + 2), (4x – 1) and (x + 2) are in AP.
(4x – 1) – (5x + 2) = (x + 2) – (4x – 1)
⇒ 2(4x – 1) = (x + 2) + (5x + 2)
⇒ 8x – 2 = 6x + 2 + 2
⇒ 8x – 2 = 6x + 4
⇒ 8x – 6x = 4 + 2
⇒ 2x = 6
x = 3

Question 3.
Solution:
(3y – 1), (3y + 5) and (5y + 1) are the three consecutive terms of an AP.
(3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)
⇒ 2(3y + 5) = 5y + 1 + 3y – 1
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5

Question 4.
Solution:
(x + 2), 2x, (2x + 3) are three consecutive terms of an AP.
2x – (x + 2) = (2x + 3) – 2x
⇒ 2x – x – 2 = 2x + 3 – 2x
⇒ x – 2 = 3
⇒ x = 2 + 3 = 5
x = 5

Question 5.
Solution:
(a – b)², (a² + b²) and (a + b)² will be in AP.
If (a² + b²) – (a – b)² = (a + b)² – (a² + b²)
If (a² + b²) – (a² + b² – 2ab) = a² + b² + 2ab – a² – b²
2ab = 2ab which is true.
Hence proved.

Question 6.
Solution:
Let the three numbers in AP be
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
and (a – d) x a x (a + d) = 80
a(a² – d²) = 80
⇒ 5(5² – d²) = 80
⇒ 25 – d² = 16
⇒ d² = 25 – 16 = 9 = (±3)²
d = ±3
Now, a = 5, d = +3
Numbers are 5 – 3 = 2
5 and 5 + 3 = 8
= (2, 5, 8) or (8, 5, 2)

Question 7.
Solution:
Let the three numbers in AP be a – d, a and a + d

Question 8.
Solution:
Sum of three numbers = 24
Let the three numbers in AP be a – d, a, a + d

Question 9.
Solution:
Let three consecutive in AP be a – d, a, a + d
a – d + a + a + d = 21
⇒ 3a = 21

Question 10.
Solution:
Sum of angles of a quadrilateral = 360°
Let d= 10
The first number be a, then the four numbers will be
a, a + 10, a + 20, a + 30
a + a + 10 + a + 20 + a + 30 = 360
4a + 60 = 360
4a = 360 – 60 = 300
Angles will be 75°, 85°, 95°, 105°

Question 11.
Solution:
Let the four numbers in AP be a – 3d, a – d, a + d, a + 3d, then
a – 3d + a – d + a + d + a + 3d = 28

Question 12.
Solution:
Let the four parts of 32 be a – 3d, a – d, a + d, a + 3d

Question 13.
Solution:
Let the three terms be a – d, a, a + d
a – d + a + a + d = 48
⇒ 3a = 48
⇒ a = 16
and (a – d) x a = (a + d) + 12
⇒ a(a – d) = 4 (a + d) + 12
⇒ 16 (16 – d) = 4(16 + d) + 12
⇒ 256 – 16d = 64 + 4d + 12 = 4d + 76
⇒ 256 – 76 = 4d + 16d
⇒ 180 = 20d
⇒ d = 9
Numbers are (16 – 9, 16), (16 + 9) or (7, 16, 25)

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B
• RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself

Question 1.
Solution:
PT is the tangent to the circle with centre O and radius OT = 20 cm.
P is a point 29 cm away from O.

OP = 29 cm, OT = 20 cm
OT is radius and PT is the tangent
OT ⊥ PT
Now, in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
⇒ (29)² = (20)² + PT²
⇒ 841 = 400 + PT²
⇒ PT² = 841 – 400
⇒ PT² = 441 = (21)²
⇒ PT = 21
Length of tangent PT = 21 cm

Question 2.
Solution:
P is a point outside the circle with centre O and OP = 25 cm
PT is the tangent to the circle and OT is the radius

OT ⊥ PT
PT = 24 cm
Now, in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
⇒ 25² = OT² + (24)²
⇒ 625 = OT² + 576
⇒ OT² = 625 – 576 = 49 = (7)²
OT = 7 cm
or radius of the circle is 7 cm.

Question 3.
Solution:
Given : Two concentric circles with centre O and radii 6.5 cm and 2.5 cm respectively.
AB is a chord of the larger circle which touches the smaller circle at C.

To find : The length of AB.
Join OC, OA.
AB is tangent to the smaller circle and OC is the radius.
OC ⊥ AB and OC bisects AB at C.
AC = CB
OA = 6.5 cm, OC = 2.5 cm
Now, in right ∆OAC,
OA² = OC² + AC² (Pythagoras Theorem)
6.5² = 2.5² + AC²
⇒ 42.25 = 6.25 + AC²
⇒ AC² = 42.25 – 6.25 = 36 = (6)²
⇒ AC = 6
Length of chord AB = 2 x AC = 2 x 6 = 12 cm

Question 4.
Solution:
Given : In the given figure, a circle is inscribed in a triangle ABC which touches the sides AB, BC, CA at D, E and F respectively.
AB = 12 cm, BC = 8 cm and AC = 10 cm.

To find : Lengths of AD, BE and CF.
AD and AF are tangents to the circle from A.
AD = AF = x.
Similarly, BD and BE are tangents to the circle.
BD = BE = y
and CE and CF are tangents to the circle
CE = CF = z
x + y + 12 …(i)
y + z = 8 …(ii)
z + x = 10 …(iii)
Adding, 2(x + y + z) = 12 + 8 + 10 = 30
x + y + z = 15 …(iv)
Now, subtracting (ii), (iii) and (i) respectively from (iv)
x = 15 – 8 = 7
y = 15 – 10 = 5
z = 15 – 12 = 3
AD = 7 cm, BE = 5 cm and CF = 3 cm

Question 5.
Solution:
Given : In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O.
OA and OB are joined.
To prove : A, O, B and P are concyclic.
Proof : PA is tangent and OA is the radius.
OA ⊥ PA
∠OAP = 90° …(i)
Similarly, OB is the radius and PB is the tangent
OB ⊥ PB
∠OBP = 90° …(ii)
Adding (i) and (ii)
∠OAP + ∠OBP = 90° + 90° = 180°
But these are the opposite angles of the quadrilateral AOBP
Quadrilateral AOBP is a cyclic
Hence A, O, B and P are concylic

Question 6.
Solution:
Given : In the given figure, chord AB of larger circle of the two concentric circles with centre O, touches the smaller circle at C.
To prove : AC = CB.
Construction : Join OC, OA and OB.

Proof: AB is tangent to the smaller circles and OC is the radius.
OC ⊥ AB.
In right ∆OAC and ∆OBC,
Hypotenuse OA = OB (radii of the same circle)
Side OC = OC (common)
∆OAC = ∆OBC (RHS axiom)
AC = CB (c.p.c.t.)

Question 7.
Solution:
Given : In the figure, from an external point P of the circle, PA and PB are tangents to the circle with centre O.
CD is a tangent at E.
PA = 14 cm.
To find : Perimeter of ∆PCD.
Proof: PA and PB are tangents from P to the circle.
PA = PB …(i)
CA and CE are tangents to the circle.
CA = CE …(ii)
Similarly,
DB = DE
Now, perimeter of ∆PCD
= PC + CD + PD
= PC + CE + ED + PD
= PC + CA + BD + PD
= PA + PB [From (i) and (ii)]
= 14 + 14 = 28 cm

Question 8.
Solution:
A circle with centre O, is inscribed in a ∆ABC touching it at P, Q, R respectively
AB = 10 cm, AR = 7 cm and CR = 5 cm

To find, the length of BC
AP and AR are the tangents to the circle
AP = AR = 7 cm
BP = AB – AP = 10 – 7 = 3 cm
BP and BQ are the tangents to the circle
BQ = BP = 3 cm
Similarly, CQ = CR = 5 cm
BC = BQ + CQ = 3 + 5 = 8 cm

Question 9.
Solution:
In the figure, a circle with centre O, touches the sides of a quadrilateral ABCD at P, Q, R and S respectively.

AB = 6 cm, BC = 7 cm and CD = 4 cm
To find, the length of AD.
We know that tangents from an external point to a circle are equal.
AP = AS, BP = BQ
CR = CQ and DR = DS
Now, AP + BP + CR + DR = AS + BQ + CQ + DC
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AD = (AB + CD) – BC = (6 + 4 – 7) = 3 cm (AB = 6 cm, CD = 4 cm and BC = 7 cm)
Hence, AD = 3 cm.

Question 10.
Solution:
In the given figure, an isosceles ∆ABC in which AB = AC, is circumscribed a circle.
The circle touches its sides BC, CA and AB at P, Q and R respectively.
To prove : P bisects the base BC.
i.e. BP = PC
Proof : AR and AQ are tangents to the circle.
AR = AQ But AB = AC
AB – AR = AC – AQ
⇒ BR = CQ …(i)
BR and BP are tangents to the circle.
BR = BP …(ii)
Similarly,
CP and CQ are tangents
CP = CQ …(iii)
BR = CQ (proved)
From (ii) and (iii),
BP = CP or BP = PC
Hence, P is midpoint of BC.

Question 11.
Solution:
In the given figure, O is the centre of two concentric circles with radii 4 cm and 6 cm respectively.
PA and PB are the tangents drawn from P, to the outer circle and inner circle respectively.
PA = 10 cm

To find, the length of PB (upto one place of decimal)
OA and OB are the radii and PA and PB are two tangents to the circles respectively
OA ⊥ PA and OB ⊥ PB
In right ∆OAP,
OP² = OA² + PA² (Pythagoras Theorem)
= (6)² + (10)² = 36 + 100 = 136
Similarly, in right ∆OBP,
OP² = OB² + PB²
136 = (4)² + PB²
⇒ 136 = 16 + PB²
⇒ PB² = 136 – 16 = 120
PB = √120 cm = 2√30 cm = 2 x 5.47 = 10.94 = 10.9 cm

Question 12.
Solution:
In the given figure, ∆ABC circumscribed the circle with centre O.
Radius OD = 3 cm
BD = 6 cm, DC = 9 cm
Area of ∆ABC = 54 cm²
To find : Lengths of AB and AC.
Construction : Join OA, OB, OC, OE and OF.

Proof : OE = OF = OD = 3 cm, radii of the same circle.
BD and BF are tangents to the circle.
BD = BF = 6 cm
Similarly, CD = CE = 9 cm and AE = AF = x (Suppose)

Question 13.
Solution:
In the given figure, PQ is a chord of the circle with centre O.
TP and TQ are tangents, OP and OT are joined.
Radius of the circle is 3 cm and PQ = 4.8 cm.

Question 14.
Solution:
PQ and RS are two parallel tangents which
touches the circle at A and B. O is the centre of the circle.
OA and OB are joined.
To prove : AB passes through the centre O of the circle.
Construction : Draw OC || PQ or RS.

Proof : OA is radius and PQ is tangent.
OA ⊥ PQ ⇒ ∠OAP = 90°
Similarly, OB is radius and RS is the tangent
OB ⊥ RS ⇒ ∠OBR = 90°
PQ || OC
∠AOC + ∠OAP = 180° (Co-interior angles)
∠AOC + 90° = 180°
∠AOC = 180° – 90° = 90°
Similarly, ∠BOC = 90°
∠AOC + ∠BOC = 90° + 90° = 180°
AOB is a straight line.
Hence, AB passes through the centre of the circle.

Question 15.
Solution:
In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD.
The circle touches the siaes of quadrilateral at P, Q, R and S respectively.
AB = 29 cm, AD = 23 cm, ∠B = 90°
DS = 5 cm
To find : The radius of the circle.
Construction : Join OP and OQ.
Proof: Let OP = OQ = r
∠B = 90°
PBQO is a square.
DR and DS are the tangents to the circle.
DR = DS = 5 cm
AR = AD – DR = 23 – 5 = 18 cm
AR and AQ are tangents to the circle.
AQ = AR = 18 cm
QB = AB – AQ = 29 – 18 = 11 cm
PBQO is a square.
OP = OQ = BQ = 11 cm
Hence, radius of the circle (r) = 11 cm

Question 16.
Solution:
In the given figure, TP is the tangent from an external point T and ∠PBT = 30°.
To prove : BA : AT = 2 : 1
Proof: ∠APB = 90° (Angle in a semicircle)
∠PBT = 30° (given)
∠PAB = 90° – 30° = 60°
But, ∠PAT + ∠PAB = 180° (Linear pair)
⇒ ∠PAT + 60° = 180°
⇒ ∠PAT = 180° – 60° = 120°
and ∠APT = ∠PBA = 30° (Angles in the alternate segment)
In ∆PAT,
∠PTA = 180° – (120° + 30°) = 180° – 150° = 30°
PA = AT
In right ∆APB,

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.