KSEEB Solutions

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.
Find the area of the triangle whose vertices are:
i) (2, 3), (-1,0), (2,-4)
ii) (-5, -1), (3, -5), (5, 2)
Solution:
i) Let A (2, 3) = (x₁, y₁)
B (-1, 0) = (x₂, y₂)
C (2, -4) = (x₃, y₃).
Area of the triangle from the given data:

∴ Area of ∆ ABC = 10.5 sq. units

ii) Let A (-5, -1) = (x₁, y₁)
B (3, -5) = (x₂, y₂)
C (5, 2) = (x₃, y₃).
Area of the triangle from the given data :

∴ Area of ∆ ABC =32 sq. units

Question 2.
In each of the following find the value of ‘k’, for which the points are collinear.
i) (7, -2), (5, 1), (3, k)
ii) (8, 1), (k, -4), (2, -5)
Solution:
i) Let A (7,-2)= (x₁, y₁)
B (5, 1) = (x₂, y₂)
C (3, k) = (x₃, y₃).
Points are collinear, therefore the area of the triangle formed by these is zero (0).

-k + 4 = 0
-k = -4
∴ k = 4

ii) Let A (8, 1) = (x₁, y₁)
B (k, -4) = (x₂, y₂)
C (2, -5) = (x₃, y₃).
Area of Triangle ABC = 0
∴ ABC is a straight line.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let the mid-points of ∆ABC are P, Q, R and also the mid-points of AB, BC and AC.

As per Mid-point formula,


\(=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
Area of ∆PQR : Area of ∆ABC
∴ 1 : 4

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)

Solution:
Area of quadrilateral ABCD =?
In the quadrilateral ABCD, diagonal AC is drawn which divides ∆ABC, ∆ACD.
Sum of these triangles is equal to the Area of the quadrilateral.
i) Now, Area of ∆ABC :

= 10.5 sq. units

ii) Area of ∆ACD

∴ Area of quadrilateral:
= Area of ∆ABC + Area of ∆ACD = 10.5 + 17.5 = 28 sq. units.
∴ Area of quadrilateral ABCD = 28 sq.units.

Question 5.
You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Solution:
AD Median is drawn to side BC which is the vertex of ∆ABC.
Median AD divides AABC into two triangles ∆ABD and ∆ADC which are equal in area.
Now Median AD bisects BC.
∴ BD = DC.
i) As per the Mid-Points formula,
Coordinates of D are
= \(\sqrt{(4)^{2},(0)^{2}}\)
= \(\sqrt{16,0}\)
= (4,0)
∴ Coordinates of D are (4, 0)

ii) Now, Area of ∆ABD:

iiii) Now, Area of ADC:

= \(\frac{1}{2}\) × -6
= – 3 sq. units.
∴ Area of ∆ADC = -3 sq. units.
∴ Area of ∆ABC :
= Area of ∆ABD + Area of ∆ADC
= (-3) +(-3)
= -6 sq. units.

iv) Now, Area of ∆ABC : (Direct Method)

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangle Ex 2.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2

Question 1.
In the following fig. (i) and (ii). DE || BC.
Find EC in (i) and AD in (ii).

Solution:
(i) In ∆ABC, DE || BC, EC =?
\(\quad \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{1.5}{3}=\frac{1}{\mathrm{EC}}\)
1.5 EC = 1 × 3
\(\mathrm{EC}=\frac{3}{1.5}\)
\(E C=\frac{30}{15}\)
∴ EC = 2 cm

(i) In ∆ABC, DE || BC, AD =?
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}\)
\(\quad A D=\frac{1.8}{5.4} \times \frac{7.2}{1}\)
\(=\frac{1.8}{54} \times \frac{72}{1}\)
= 0.6 × 4
∴ AD = 2.4 cm

Question 2.
E and F are points on the sides PQ and PR respectively of a PQR For each of the following cases, state whether EF || QR:
(i) PE = 3.9cm. EQ = 3 cm
PF=3.6cm. FR = 2.4 cm.
(ii) PE = 4 cm. QE = 4.5 cm.
PF = 8 cm. RF = 9 cm.
(iii) PQ = 1.28 cm. PR = 2.56 cm.
PE = 0.18 cm. PF = 0.36 cm.
Solution:

In ∆PQR, if EF || QR then,
\(\frac{\mathrm{PE}}{\mathrm{QR}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
\(\frac{3.9}{3}=\frac{3.6}{2.4}\)
1.3 ≠ 1.5
Sides are not dividing In ratio.
∴ EF is not parallel to QR.

(ii)

In ∆PQR, if EF || QR then
\(\frac{\mathrm{PE}}{\mathrm{QR}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
\(\frac{4}{4.5}=\frac{8}{9}\)
\(\frac{8}{9}=\frac{8}{9}\)
Here sides are dividing in ratio.
∴ EF ||QR

(iii)

PE + EQ = PQ
0.18 + EQ = 1.28
∴ EQ=1.28 – 0.18
EQ = 1.1 cm.
Similarly. PF + FR = PR
0.36 + FR = 2.56
FR = 2.56 – 0.36
FR = 2.2cm.
In ∆PQR, if EF || QR then
\(\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
\(\frac{0.18}{1.1}=\frac{0.36}{2.2}\)
\(\frac{1.8}{11}=\frac{3.6}{22}\)
\(\frac{1.8}{11}=\frac{11.8}{11}\)
Here sides are dividing in ratio.
∴ EF || QR

Question 3.
In the following figure. If LM || CB and LN || CD, prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Solution:
Data: LM || CB and LN || CD then prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)
Solution: In ∆ACB, LM || CB
\(\quad \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AM}}{\mathrm{AB}} \quad \ldots \ldots(1)(\text { Theorem } 1)\)
Similarly in ∆ADC, LN || CD
\(\quad \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AN}}{\mathrm{AD}} \quad \ldots \ldots(2) \quad(\ldots \text { Theorem } 1)\)
From equation (1) and (2) we have
\(\frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)
\(\quad \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Question 4.
In the following figure, DE ||AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)

Solution:
Data: In this figure, AE || AC and DF || AE, then we have to prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)
Solution: In ∆ABC, DE || AC.
\(\quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}} \ldots \ldots(1) \quad(\text { Theorem } 1)\)
Similarly, In ∆ABE, DF ||AE.
\(\quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)
from equation (1) and (2). we have
\(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)
\(\quad \frac{\mathrm{BE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)

Question 5.
In the following figure. DE || OQ and DF || OR. Show that EF || QR.

Solution:
Data: In this figure. DE || OQ and DF || OR.
To Prove: EF || QR
Solution: In ∆POQ, DE || OQ.
\(\quad \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}} \quad \ldots \ldots \text { (i) }(\text { Theroem } 1)\)
Similarly. In ∆POR. DF || OR.
\(\quad \frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}} \quad \ldots \ldots \text { (ii) }(\text { Theorem } 1)\)
from equation (1) and (11), we have
\(\frac{P F}{E Q}=\frac{P D}{D O}=\frac{P F}{F R}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
In ∆PQR. if \(\frac{P E}{E Q}=\frac{P F}{F R}\) then EF || QK. (∵ Theorem 1).

Question 6.
In the following figure, A, B an C are points on OP. OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:
Data: In ∆QPR, AB || PQ and AC || PR. A, B, and C are the points on OP. OQ and OR.
To Prove: BC || QR
Solution: In ∆OPQ, AB || PQ.
\(\quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}} \quad \ldots \ldots \text { (i) }(\text { Theorem } 1)\)
Similarly, In ∆OPR. AC|| PR.
\(\quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \quad \ldots \ldots \text { (i) }(\text { Theorem } 1)\)
from equation (i) and (ii), we have
\(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)
\(\quad \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)
∴ BC || QR (∵ Theorem 2)

Question 7.
Using Theorem 2.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:
Data: In ∆ABC. D is the mid-point of AB.
DE Is drawn parallel to BC from D.
To Prove: DE bisects AC side at E.
Solution: In ∆ADE and ∆ABC,
∠D = ∠B (corresponding angles)
∠E = ∠C (corresponding angles)
∴∆ADE || ∆ABC
\(\quad \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\)
\(\frac{1}{2}=\frac{\mathrm{AE}}{\mathrm{AC}}\)
AC = 2AD
∴ AC = AE + EC
∴ E is the mid-point of AC
∴ DE bisects AC at E.

Question 8.
Using Theorem 2.2, prove that the line joining the mid-points of any two sides of a triangle Is parallel to the third side. (Recall that you have done it in Class IX)

Solution:
Solution: In ∆ABC, D and E are mid-points of AB and AC.
∴ AD = DB
AE = EC
AB = 2AD
AC = 2AE
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{2}\)
As per S.S.S. Postulate.
∆ADE ~ ∆ABC
∴ They are equiangular triangles.
∴ ∠A is common.
∠ADE = ∠ABC
∠ADE = ∠ACE
These are pair of corresponding angles
∴ DE || BC

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Solution:
Data : ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
To Prove: \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Proof: In Trapezium ABCD. AB || DC.
∴ In ∆AOB and ∆DOC,
∠OCO = ∠OAB (alternate angles)

∠ODC = ∠OBA (alternate angles)
∠DOC = ∠AOB (vertically opposite angles)
∴ ∆AOB and ∆DOC are equiangular triangles.
∴ ∆AOB ||| ∆DOC
Similar triangles divides sides in ratio.
\(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\) . Show that ABCD is a trapezium.

Solution:
Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
To Prove: ABCD is a trapezium.
Solution: In the qudrilateral ABCD, \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
\(\quad \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
It means sides of ∆AOB and ∆DOC divides proportionately.
∴ ∆AOB ||| ∆DOC.
Similarly. ∆AOD ||| ∆BOC.
Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC
∆ABD = ∆ABC.
Both triangles are on the same base AB and between two pair of lines and equal in area.
∴ AB || DC.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangle Ex 2.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2

In Questions 1 to 3, choose the correct option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
A) 7 cm.
B) 12 cm.
C) 15 cm.
D) 24.5 cm.
Solution:
A) 7 cm.

In ⊥∆OPQ, ∠P = 90°
∴ As per Pythagoras theorem,
OP² + PQ² = OQ²
OP² + (24)² = (25)²
OP² + 576 = 625
OP² = 625 – 576
OP² = 49
∴ OP = 7 cm.

Question 2.
In the following figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
A) 60°
B) 70°
C) 80°
D) 90°
Solution:
B) 70°

∠POQ =110° (given)
OP ⊥PT, OQ ⊥QT ( ∵ Angle between radius and tangent)
∴ ∠OPT = ∠OQT = 90°
Now, OPTQ is a quadrilateral.
∴ ∠O + ∠P + ∠Q + ∠T = 360°
110° + 90° + 90° + ∠T = 360°
290° + ∠T = 360°
∴∠T = 360° – 290°
∴ ∠T = 70°
∴ ∠PTQ = 70°

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
A) 50°
B) 60°
C) 70°
D) 80°
Solution:
A) 50°

OA, OB are radii,
PA, PB are tangents
∴ ∠PAO = 90°, ∠PBO = 90°
In quadrilateral OAPB,
∠BOA = 180° – 80° = 100°
But 

∴∠POA = 50°.

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Data: The tangents drawn at the ends of a diameter of a circle are parallel.

AOB is the p diameter in the circle with centre ‘O’.
Tangent PAQ is at the endpoint A, and
Tangent RBS is A at the endpoint B.
OA is the radius,
PAQ is tangent
∴ ∠PAO = 90°
OB is the radius,
RBS is tangent.
∴ RBO = 90°
Now, ∠PAO + ∠RBO = 90° + 90° = 180°
AB intersects straight lines PAQ and RBS.
Sum of a pair of interior angles is 180°. Hence these lines are parallel to each other.
∴ PQ || RS proved.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution:
Data: Perpendicular at the point of contact to the tangent to a circle passes through the centre.
PQ is the tangent to circle with centre ’O’.
Let the perpendicular drawn at the point P is ∠RPQ.
∠RPQ = 90° …………. (i)
Radius drawn to circle at the point of contact is perpendicular.
∴ ∠OPQ = 90° …………. (ii)
From eqn. (i) and eqn. (ii), we have
∠RPQ = ∠OPQ = 90° .
This is a contradiction, because
∠RPQ is the part of ∠OPQ.
∴ ∠RPQ < ∠OPQ
∴ ∠RPQ ≠∠OPQ.
∴ The perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution:
Data: OB is the radius, BA is the tangent.
∴ ∠OBA = 90°, OA = 5 cm. (data)
Length of tangent, AB = 4 cm.
∴ Radius, OB =?
In ⊥∆OBA, ∠OBA = 90°
∴ OB² + BA² = OA² .
OB² + (4)² = (5)²
OB² + 16 = 25
OB² = 25 – 16
OB² = 9
∴ OB = 3 cm.
∴ Radius of circle, OB = 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:
Radius of big circle,
OA = OB = 5 cm.
Radius of small circle, OP = 3 cm.
Angle between radius and tangent is 90°.
∴ ∠OPA = ∠OPB = 90°
( ∵ Chord AB is tangent to small circle.)
Now, in ⊥∆ OPA, ∠OPA = 90°
OP² + AP² = OA²
(3)² + AP² = (5)²
9 + AP² = 25
∴ AP² = 25 – 9
AP² = 16
∴ AP = 4 cm.
Similarly, in ⊥∆OPB, PB = 4 cm.
∴ Length of chord, AB = AP + PB = 4 + 4
∴ Chord, AB = 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see the fig. given below). Prove that AB + CD = AD + BC.

Solution:
Data: In the quadrilateral, ABCD is drawn to circumscribe a circle.
To Prove: AB + CD = AD + BC
The lengths of tangents drawn from an extremal point to a circle are equal.
∴ Tangents AP and AS are drawn from point A.
∴ AP = AS
Tangents BP and BQ are drawn from point B.
∴ BP = BQ
Tangents CQ, CR are drawn from point C.
∴ CQ = CR.
Tangents DR, DS are drawn from point A.
∴ DR = DS
L.H.S.: AB + CD = AP + PB + CR + RD
= AS + BQ + CQ + DS
= AS + SD + BQ + CQ
∴ AB + CD = AD + BC
∴ L.H.S. = R.H.S.

Question 9.
In the following figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Data: XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B
To Prove: ∠AOB = 90°.
Tangent XY || Tangent X’Y’
∴ ∠PAB + ∠QBA = 180° (∵ interior angles)

∠OAB + ∠OBA = 90°
Now, in AOAB,
∠AOB + ∠OAB + ∠OBA = 180°
∠AOB + 90° = 180°
∴ ∠AOB = 180° – 90°
∴ ∠AOB = 90°.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:

To Prove: ∠APB + ∠AOB = 180°
OA is radius, PA is tangent.
∴∠PAO = 90°
OB is radius, PB is tangent.
∴ ∠PBO = 90°
Now OAPB is a quadrilateral.
∴∠PAO + ∠PBO = 90° + 90° = 180°
Sum of four angles of a quadrilateral is 360°
∴ ∠PAO + ∠PBO + ∠APB + ∠AOB = 360°
180° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 360° – 180°
∴ ∠APB + ∠AOB = 180°
If the sum of two angles is equal to 180°, then they are supplementary angles.
∴ ∠APB and ∠AOB are supplementary angles.

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:

To prove: Parallelogram ABCD is a rhombus.
ABCD is a parallelogram in which ‘O’ is the centre of the interior circle drawn in the parallelogram.
Tangents AC and BD are drawn which intersect at ‘OP’.
Sides of the parallelogram AB. BC. CD and AD touch at the points P, Q, R, and S respectively.
OP, OS joined.
We have OP ⊥ AB, OS ⊥ AD.
In ⊥∆ OPB and ∆OSD,
∠OPB = ∠OSD =90°
OB = OD (tangent is bisected)
OP = OS (Radii)
∴ ∆OPB = ∆OSD
∴ PB = SD → (1)
AP = AS → (2) (∵ tangent of the external point)
By combining eqn. (1) and Eqn. (2),
AP + PB = AS + SC
AB = AD
Similarly, AB = BC = CD = DA
∴ Parallelogram ABCD is a rhombus

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm. respectively, (see the figure given). Find the sides AB and AC.
Solution:

To Prove: Side AB = ? and Side AC = ?
BD = 8 cm. DC = 6 cm.
BE = 8 cm. CF = 6 cm.
AE = AF = x cm.
In ∆ABC,
a = BC = BD + DC =8 + 6 =14 cm.
b = CA = (x + 6) cm.
c = AB = (x + 8) cm.
\(S=\frac{a+b+c}{2} \quad=\frac{14+(x+6)+(x+8)}{2}\)
\(=\frac{2 x+28}{2}\)
= (x + 14) cm.
Area of ∆ABC = \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{(x+14) \times x \times 8 \times 6}\)
\(=\sqrt{48 x \times(x+14)} c m^{2}\) …………….. (i)
Area of ∆ABC
= ∆OBC + ∆OCA + ∆OAB W.
\(=\frac{1}{2} \times 4 \times a+\frac{1}{2} \times 4 \times b+\frac{1}{2} \times 4 \times c\)
– 2(a + b + c)
= 2 × 2S
= 4S
= 4(x +14) cm²………………. (ii)
From eqn. (i) and eqn. (ii),
\(\sqrt{48 x \times(x+14)}=4(x+14)\)
48x × (x + 14) = 16 x (x + 14)2
3x = x + 14
∴ x = 7 cm.
AB = c = x + 8 = 7 + 8 = 15 cm.
AC = b = x + 6 = 7 + 6 = 13 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution:
Data: Circle with centre ‘O’ is circumscribed in a quadrilateral ABCD.
To Prove: ∠AOD + ∠BOC = 180°
Sides of the quadrilateral AB, BC, CD, and DA touches at the points P, Q, R, and S respectively.
OA. OB, OC, OD, and OP, OQ, OR, OS are joined.
OA bisects ∠POS.
∠1 =∠2
∠3 = ∠4
∠5 = ∠6
∠7 = ∠8
2(∠1 + ∠4 + ∠5 + ∠8) = 360°
(∠1 + ∠8) (∠4 + ∠5) = 180°
∴ ∠AOD = ∠BOD = 180°
∴ Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2

Question 1.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution:
Let P(x, y) is the required point.
P(x, y) divides the points A(-1, 7),
B (4, -3) in the ratio m₁ : m₂ = 2 : 3.
As per Section Formula, coordinates of P(x, y) are

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3)

Solution:
Let A(x₁, y₁) = A(4,-1)
B(x₂, y₂) = B(-2, -3).
P and Q are the points of trisection of AB.
AP = PQ = QB
i) P divides AB in the ratio m₁ : m₂ :: 1 : 2.
∴ As per Section Formula, Coordinates of P are :

ii) Q divides AB in the ratio 2 : 1.
∴ As per Section formula,

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in the following figure.

Niharika runs \(\frac{1}{4}\) th the distance of AD on then 2nd line and posts a green flag
Preet runs \(\frac{1}{5}\) th the distance AD on the 5 eighthline and posts a red flag. What is the distance between both the flags ? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
Origin, A(0, 0)
AB along x – axis
AD along y – axis
In the parallel line AD in 2^nd row, Rashmi sees green flag at the distance of \(\frac{1}{4}\) .
∴ AD = \(\frac{1}{4}\) × 100 = 25 m.
∴ Coordinates of green flag are G(2, 25)
Similarly we can find out coordinates of Red flag as R(8, 20).
∴ Distance between G and R is

It means blue flag is in 5th line and parallel to AD at the distance of 22.5 m.

Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
We have to check whether the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Let P(- 1, 6) divides AB in the ratio m₁ : m₂.
By using Section formula,

∴ P(-1, 6) point divides the line segment in the ratio 2 : 7.

Question 5.
Find the ratio in whcih the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:
Let the line segment joining A and B points divides x-axis in the ratio k : 1.
If coordinates of y, P(x, 0) compared,
\(\frac{k \times(5)+1 \times(-5)}{k+1}=0\)
5k – 5 = 0
∴ k = 1
∴ Ratio will be k : 1 = 1 : 1
∴ Point P(x, 0) is the mid-point of line segment AB.
∴ x = \(\frac{1-4}{2}=\frac{-3}{2}\)
∴ Dividing point, P(\(-\frac{3}{2}\), 1)

Question 6.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution:
AC and BD are the diagonals of a parallelogram ABCD and diagonals bisect at O’.
∴ Coordinate of the midpoint of AC = Coordinate of midpoint BD.
As per Mid-point formula,

∴ y = 3
∴ x = 6, y = 3.

Question 7.
Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Solution:
AB is the diameter.
‘O’ is centre of the circle, bisects AB
Let coordinates of A are (x, y)
As per the Mid-point formula,

∴ The coordinates of point A are (3, -10)

Question 8.
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \(=\frac{3}{7}\) AB and P lies on the line segment AB.

Solution:
Let coordinates of P are (x, y).
AP = \(\frac{3}{7}\) AB AP : AB = 3 : 7
As per Section formula,
AP : AB = 3 : 7.
∴ AP + PB = AB
3 + PB = 7
∴ PB = = 7 – 3 = 4
∴ Let AP : PB = 3 : 4 = m₁ : m₂.
Coordinates of P are

Question 9.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution:
Let A (-2, 2) = (x₁, y₁).
B (2, 8) = (x₄, y₄).
P, Q, R points divide equally the side AB.
Here, we have AP = PQ = QR = RB.
∴ AP : PB = 1 : 3
AQ : QB = 2 : 2 = 1 : 1
∴ AR : RB = 3 : 1.
As per Section formula
i) P divides AB in the ratio 1 : 3

ii) Q divides AB in the ratio 2 : 2 means it is mid-point of AB.
∴ As per the Mid-point formula,

Q(x₂, y₂) = (0, 5)

iii) R divides AB in the ratio 3 : 1
m₁ = 3, m₂ = 1

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals]

Solution:
ABCD is a rhombus.
AC and BD are the diagonals and intersects at ‘O’ and bisects perpendicularly.
∴ Let Diagonal BD = d₁ Diagonal AC = d₂,

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1

Question 1.
Check whether the following are quadratic equations:
i) (x + 1)² = 2 (x – 3)
ii) x² – 2x = (-2) (3 – x)
iii) (x – 2) (x + 1) = (x – 1) (x + 3)
iv) (x – 3) (2x + 1) = x (x + 5)
v) (2x – 1) (x – 3) = (x +5) (x – 1)
vi) x² + 3x + 1 = (x – 2)²
vii) (x + 2)³ = 2x (x² – 1)
viii) x³ – 4x² – x + 1 = (x – 2)³
Solution:
i) (x + 1)² = 2(x – 3)
x² + 2x + 1 = 2x – 6
x² + 2x – 2x + 1 – 6 = 0
x² – 5 = 0
It is in the form of ax² + c = 0
Therefore, the given equation is a Quadratic equation.

ii) x² – 2x = (- 2) (3 – x)
x² – 2x = – 6 + 2x
x² – 2x – 2x + 6 = 0
x² – 4x + 6 = 0
It is in the form of ax2 + bx + c = 0
Therefore, the given equation is a Quadratic equation.

iii) (x – 2) (x + 1) = (x – 1) (x + 3)
x² + x – 2x – 2 = x² + 3x – x – 3
x² – x – 2 = x² + 2x – 3
x² – x² – x – 2x – 2 + 3 = 0
– 3x + 1 = 0
It is in not in the form of ax² + bx + c = 0
Therefore, the given equation is not a Quadratic equation.

iv) (x – 3) (2x + 1) = x (x + 5)
2x² + x – 6x – 3 = x2 + 5x
2x² – 5x – 3 = x² + 5x
2x² – x² – 5x – 5x – 3 = 0
x² – 10x – 3 = 0
It is in the form of ax² + bx + c = 0
Therefore, the- given equation is a Quadratic equation.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
2x² – 6x – x + 3 = x² – x + 5x – 5
2x² – 7x + 3 = x² + 4x – 5
2x² – x² – 7x – 4x + 3 + 5 = 0
x² – 11x + 8 = 0
It is in the form of ax² + bx + c = 0
Therefore, the given equation is a Quadratic equation.

vi) x² + 3x + 1 = (x – 2)²
x² + 3x + 1 = x² – 4x + 4
x² – x² + 3x + 4x + 1 – 4 = 0
7x – 3 = 0
It is not in the form of ax² + bx + c = 0
Therefore, the given equation is not a Quadratic equation.

vii) (x + 2)³ = 2x (x² – 1)
x³ + (3)³ + 3 × 2 × x (x + 2) = 2x³ – 2x
x³ + 27 + 6x (x + 2) = 2x³ – 2x
x³ + 27 + 6x² + 12x = 2x³ – 2x
2x³ – x³ – 6x² – 2x – 12x – 27 = 0
x³ – 6x² – 14x – 27 = 0
It is not in the form of ax² + bx + c = 0
Therefore, the given equation is not a Quadratic equation.

viii) x³ – 4x² – x + 1 = (x – 2)³
x³ – 4x² – x + 1 = x³ – (2)³ – 3 × 2 × x(x – 2)
x³ – 4x³ – x + 1 = x³ – 8 – 6x (x – 2)
x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x
x³ – x³ – 4x² + 6x² – x – 12x + 1 + 8 = 0
2x² – 13x + 9 = 0
It is in the form of ax² + bx + c = 0
Therefore, the given equation is a Quadratic equation.

Question 2.
Reprsent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth of rectangular plot (b) be ’x’ m.
Then the length of th plot is one more than twice its breadth,
∴ Length (l)= 2x + 1 m.
But Length × Breadth = Area of rectangle
l × b = A
∴ x × (2x + 1) = 528 sq.m.
2x² + x = 528
∴ 2x² + x – 528 = 0 is the required equation.
Now, we have to find out the value of ‘x’ :
2x² + x – 528 = 0
2x² – 32x + 33x – 528 = 0
2x(x – 16) + 33(x – 16) = 0
(x – 16) (2x + 33) = 0
If x – 16 = 0, then x = 16
If 2x + 33 = 0, then x = -33/2
∴ Breadth (b) = 16 m.
Length (l) = (2x + 1) = 2(16) + 1 = 32 + 1 = 33m
∴ Length (l) = 33 m
Breadth (b) = 16 m.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let one positive integer be x’.
The Next integer is (x + 1)
Their product is 306.
∴ x (x + 1) = 306
x² + x = 306
∴ x² + x – 306= 0. This is required equation.
Now, we have to solve for positive integer.
x² + x – 306 = 0
x² + 18x – 17x – 306 = 0
x(x + 18) – 17(x + 18) = 0
(x + 18) (x – 17) = 0
If x + 18 = 0, then x = -18
If x – 17 = 0, then x = 17
∴ x = 18, OR x = 17.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution:
Let the present age of Rohan be ‘x’, then His mother’s age will be (x + 26)
After 3 years, Age of Rohan is (x + 3). After 3 years his mother’s age will be
= (x + 26 + 3)
= (x + 29)
Then product of their ages is 360.
∴ (x + 3) (x + 29) = 360
x² + 29x + 3x + 87 = 360
x² + 32x + 87 = 360
x² + 32x + 87 – 360 = 0
x² + 32x – 273 = 0.
This is the required equation.
Now, we have to solve for the value of ‘x’:
x² + 32x – 273 = 0
x² + 39x – 7x – 273 = 0
x(x + 39) – 7(x + 39) = 0
(x + 39) (x – 6) = 0
If x + 39 = 0, then x = -39
If x – 6 = 0, then x = 7
Present age of Rohan’s mother
= x + 26
= 7 + 26
= 33 years.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the initial speed of a train be ‘x’ km/h.
Time required to travel x km is 1 hour. Time required to travel 480 km ………?
\(\frac{480}{x}\) hr
If its speed decreases to 8 km/h, then it is (x – 8) km/h.
Time required to cover (x – 8) km is 1 Hr.
Time required to cover 480 km ………..?

∴ x(3x + 456) = 480 (x – 8)
3x² + 456x = 480x + 3840
3x² + 456x – 480 x + 3840 = 0
3x² – 24x + 3840 = 0
∴ x² – 8x + 1280 = 0
This is the required equation.
Now, we have to solve for x :
x² – 8x + 1280 = 0
x² – 40x + 32x + 1280 = 0
x(x – 40) + 32(x + 40) = 0
(x – 40) (x + 32) = 0
If x – 40 = 0, then x = 40
If x + 32 = 0, then x = -32
∴ Average speed of train is 40 km/hr.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Question 1.
Aftab tells his daughter, “Seven years ago, I was, seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Solution:
Let the present age of father be x’
Let the age of daughter be ‘y’.
Before seven years, age of father is x – 7
Before seven years, age of daughter is y – 7.
Father was 7 times as old as daughter.
∴ x – 7 = 7 (y – 7)
x – 7 = 7y – 49
x – 7y = -49 + 7
x – 7y = – 42
∴ x – 7y + 42 = 0 …………….. (i)
After three years, age of father is x + 3
After three years, age of daughter is y + 3
Three years from now will three times as old as father.
∴ x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y + 3 – 9 = 0
3 – 3y – 6 = 0 ………….. (ii)
∴ Algebraically linear equations :
x – 7y + 42 = 0
x – 3y – 6 =0
If we represent this linear equations through graph, we have
i) x – 7y + 42 = 0
-7y = -x – 42
7y = x + 42
\(\quad y=\frac{x+42}{7}\)

x	0	7
\(y=\frac{x+42}{7}\)	6	7

ii) x – 3y – 6 = 0
-3y = -x + 6
3y = x – 6
\(\quad y=\frac{x-6}{3}\)

x	0	+6
\(y=\frac{x-6}{3}\)	-2	0

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebriacally and geometrically.
Solution:
Let the cost of each bat be Rs. ‘x’
Let the cost of each ball be Rs. ‘y’.
3x + 6y = 3900
∴ x + 2y = 1300 ………. (i)
Cost of another bat and ball be x + 3y = 1300 ………….. (ii)
∴Algebriacally
x + 2y = 1300
x + 3y = 1300
And to represent through graph :
i) x + 2y = 1300
2y = 1300 – x
\(\quad x=\frac{1300-x}{2}\)

x	100	200
\(y=\frac{1300-x}{2}\)	600	500

ii) x + 3y = 1300
3y = 1300 – x
\(\quad y=\frac{1300-x}{3}\)

x	100	400
\(y=\frac{1300-x}{3}\)	400	900

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Solution:
Algebraically,
If Cost of each kg. of apple be Rs. ‘x’,
Cost of grapes be Rs. ‘y’. Then
2x + y = 160
4x + 2y = 300
To represent geometrically,
i) 2x + y = 160
y = -2x + 160

x	20	80
y = 2x + 160	120	0

ii) 4x + 2y = 300
2x + y= 150
y = 150 – 2x

x	40	60
y = 150 – 2x	70	30

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1

Question 1.
Find the distance between the following pairs of points:
i) (2, 3), (4, 1)
ii) (-5, 7), (-1, 3)
iii) (a, b), (-a,-b)
Solution:
i) If A(x₁, y₁) = A(2, 3)
B(x₂, y₂) = B(4, 1), then

ii) Let P(x₁, y₁) = P(-5, 7)
Q(x₂, y₂) = Q(-1, 3)

iii) A (a, b), B (-a, -b)

Question 2.
Find the distance between the points (0, 0) and (36, 15). can you now find the distance between the two towns A and B discussed in Section 7.2?
Solution:
The distance between the points A(0, 0) and B(36, 15):

∴ AB = 39
∴ The distance between the two towns A and B discussed in section 7.2 is

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Whether the points A (1, 5), B (2, 3) and C (-2, -11) are collinear?

Here, AB + BC = AC, then it is straight line.
But \(\sqrt{5}+\sqrt{221} \neq \sqrt{261}\)
∴ These points are non-collinear.

Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:

Here, AB = BC.
∴ If Two sides of the triangle ABC are equal to each other then it is an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points, A, B, C, and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct.

Solution:
A(3, 4), B (6, 7), C (9, 4), D (6, 1)


Four sides of quadrilateral ABCD are equal to each other.
∵ AB = BC = CD = DA = \(3 \sqrt{2}\)
Diagonal AC = Diagonal BD
∴ ABCD is a square.
∴ Champa is correct among the two.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons 1 for your answer :
i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
i) A (1, -2), B (1, 0), C (-1, 2), D (-3, 0)

In the quadrilateral ABCD,
diagonal AC = diagonal BD = 4 cm., and
AB = BC = CD = DA = \(\sqrt{8}\) units.
i. e., Here all four sides are equal to each other then ABCD is a square.

ii) P (-3, 5), Q (3, 1), R (0, 3), S (-1, -4)
As per Distance formula,

Here all four sides of quadrilateral PQRS are not equal to each other. Hence PQRS is a quadrilateral.

iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)
As per the Distance formula,

In the quadrilateral ABCD
AB = DC = \(\sqrt{10}\)
BC = AD = \(\sqrt{18}\)
Diagonal AC ≠ Diagonal BD

Adjacent sides are equal to each other and diagonals are not equal to each other. This is a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

Solution:

we have PA = PB
∴ PA² = PB²
x² – 4x + 29 = x² + 4x + 85
-4x – 4x = 85 – 29
-8x = 56
8x = -56
∴ x = \(\frac{-56}{8}\)
∴ x = – 7
∴ Coordinates of P are (x, 0), it means (-7, 0).
∴ Required point (x, 0) = (-7, 0)
∴ x = -7.

Question 8.
Find the values of ‘y’ for which the distance between the points P(2, -3) and Q (10, y) is 10 units.
Solution:
If the distance between P and Q, PQ = 10 units, then y =?

By Squaring on both sides,
y² + 6y + 73 = 100
y² + 6y + 73 – 100 = 0
y² + 6y – 27 = 0
y² + 9y – 3y – 27 = 0

y (y + 9) – 3(y + 9) = 0
(y + 9) (y – 3) = 0
If y + 9 = 0 then OR y – 3 = 0 then
y = -9 y = 0
∴ y = -9 OR +3.

Question 9.
If Q (0, 1) is equidistant from P (5, -3) and R(x, 6), find the value of x. Also, find the distances QR and PR.

Solution:
Point Q is equidistant from P and R, then PQ = QR, x = ?

Question 10.
Find a relation between ‘x’ and ‘y’ such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

Solution:
A (3, 6) = A (x₁, y₁) AP = PB.
P(x, y)
Let B(-3, 4) = B(x₂, y₂).

Here, AP = PB
∴ AP² = PB²
x² – 6x + y² – 12y + 45 = x² + y² + 6x – 8y + 25
-6x – 6x – 12y + 8y = 25 – 45
– 12x – 4y = -20
12x + 4y = 20
∴ 3x + y = 5
∴ 3x + y – 5 = 0

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter Chapter 7 Coordinate Geometry Ex 7.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter Chapter 7 Coordinate Geometry Exercise 7.1, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.1

Question 1.
How many tangents can a circle have?
Solution:
There is only one tangent at a point of the circle.

Question 2.
Fill in the blanks :

1. A tangent to a circle intersects it in __________ point (s).
2. A line intersecting a circle in two points is called a ____________
3. A circle can have _____________ parallel tangents at the most
4. The common point of a tangent to a circle and the circle is called _____________ .

Answers:

1. one
2. tangent
3. two
4. point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ =12 cm, Length PQ is.:
a) 12 cm.
b) 13 cm.
c) 8.5 cm.
d) 19 cm.
Solution:
d) \(\sqrt{119} \mathrm{cm}\).

A tangent PQ at a point P of a circle.
OP is the radius drawn from centre of the circle to the tangent.
∴ OP ⊥ PQ.
Now, in ⊥ ∆OPQ, ∠P = 90°
∴ As per Pythagoras theorem,
OP² + PQ² = OQ²
(5)² + PQ² = (12)²
25 + PQ² = 144
PQ² = 144 – 25
PQ² = 119
∴ √PQ²  = √119
∴ PQ = √119 cm.

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:

CD is the tangent parallel to AB.
AB is secant, it meets RS at P.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.1

Question 1.
Fill In the b1ank using the correct word given in brackets:
i) All circles are _______ . (congruent, similar)
Solution:
congruent

ii) All squares are _______. (similar, congruent)
Solution:
similar

iii) All ______ triangles are similar. (isosceles, equilateral)
Solution:
equilateral

iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are _______ and
(b) their corresponding sides are _______. (equal, proportional).
Solution:
(a) equal
(b) proportional

Question 2.
Give two different examples of pair of
i) similar figures
Solution:
coin, wheel of a cart.

ii) non-similar figures.
Solution:
A square Rhombus

Question 3.
State whether the following quadrilaterals are similar or not:

Solution:
Here, two diagrams PQRS and ABCD are rhombus and square whose sides are in proportional. But the corresponding angle is not equal. Hence these are not similar.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² – 8u
(v) t² – 15
(vi) 3x² – x – 4
Solution:

(iv) 4u² – 8u
= 4u² – 8u + 0
= 4u (u – 2)
If 4u = 0, then u = 0
If u – 2 = 0, then u = 2
∴ Zeroes are 0 and 2

(v) t² – 15
= t² + 0 – 15

(vi) 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)
If 3x – 4 = 0, then x = \(\frac{4}{3}\)
If x + 1 = 0, then x = -1
∴ Zeroes are \(\frac{4}{3}\) and -1.

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), 1
(ii) \(\sqrt{2}, \frac{1}{3}\)
(iii) 0, \(\sqrt{5}\)
(iv) 1, 1
(v) \(-\frac{1}{4}, \frac{1}{4}\)
(vi) 4, 1
Solution:
(i) \(\frac{1}{4}\), 1
Here m + n = \(\frac{1}{4}\), mn = -1
∴ Quadratic Equation

(ii) \(\sqrt{2}, \frac{1}{3}\)
Here m + n = \(\sqrt{2}\), mn = \(\frac{1}{3}\)
∴ Quadratic Equation

(iii) 0, \(\sqrt{5}\)
Here m + n = 0, mn = \(\sqrt{5}\)
∴ Quadratic Equation

(iv) Standard form of quadratic polynomial
sum and product of its zeroes is
K(x² – sum of the zeroes) x + product of zeroes.
= K(x² – 1x + 1)
Taking K = 1
= x² – x + 1

(v) \(-\frac{1}{4}, \frac{1}{4}\)
Here m + n = \(-\frac{1}{4}\), mn = \(\frac{1}{4}\)
∴ Quadratic Equation

(vi) Standard form of quadratic polynomial sum and product of its zeroes is
= K[x² – (sum of the zeroes) x + Product of zeroes]
= K(x² – 4x + 1)
Taking K = 1
= 1(x² – 4x + 1)
= x² – 4x + 1

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