ML Aggarwal Solutions

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Constructions Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Constructions Chapter Test.

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Question 1.
Draw a circle of radius 3 cm. Mark its centre as C and mark a point P such that CP = 7 cm. Using ruler and compasses only, Construct two tangents from P to the circle.
Solution:

Steps of Construction :

1. Draw a circle with centre C and radius 3 cm.
2. Mark a point P such that CP = 7 cm.
3. With CP as diameter, draw a circle intersecting the given circle at T and S.
4. Join PT and PS.
5. Draw a tangent at Q to the circle given. Which intersects PT at D.
6. Draw the angle bisector of ∠PDQ intersecting CP at E.
7. With centre E and radius EQ, draw a circle.
   It will touch the tangent T and PS and the given circle at Q.
   This is the required circle.

Question 2.
Draw a line AQ = 7 cm. Mark a point P on AQ such that AP = 4 cm. Using ruler and compasses only, construct :
(i) a circle with AP as diameter.
(ii) two tangents to the above circle from the point Q.
Solution:

Steps of construction :

1. Draw a line segment AQ = 7 cm.
2. From AQ,cut off AP = 4cm
3. With AP as diameter draw a circle with centre O.
4. Draw bisector of OQ which intersect OQ at M.
5. With centre M and draw a circle with radius MQ
   which intersects the first circle at T and S.
6. Join QT and QS.
   QT and QS are the tangents to the first circle.

Question 3.
Using ruler and compasses only, construct a triangle ABC having given c = 6 cm, b = 1 cm and ∠A = 30°. Measure side a. Draw carefully the circumcircle of the triangle.
Solution:

Steps of Construction :

1. Draw a line segment AC = 7 cm.
2. At C, draw a ray CX making an angle of 30°
3. With centre A and radius 6 cm draw an arc
   which intersects the ray CX at B.
4. Join BA.
5. Draw perpendicular bisectors of AB and AC intersecting each other at O.
6. With centre O and radius OA or OB or OC,
   draw a circle which will pass through A, B and C.
   This is the required circumcircle of ∆ABC

Question 4.
Using ruler and compasses only, construct an equilateral triangle of height 4 cm and draw its circumcircle.
Solution:

Steps of Construction :

1. Draw a line XY and take a point D on it.
2. At D, draw perpendicular and cut off DA = 4 cm.
3. From A, draw rays making an angle of 30°
   on each side of AD meeting the line XY at B and C.
4. Now draw perpendicular bisector of AC intersecting AD at O.
5. With centre O and radius OA or OB or OC
   draw a circle which will pass through A, B and C.
   This is the required circumcircle of ∆ABC.

Question 5.
Using ruler and compasses only :
(i) Construct a triangle ABC with the following data: BC = 7 cm, AB = 5 cm and ∠ABC = 45°.
(ii) Draw the inscribed circle to ∆ABC drawn in part (i).
Solution:
Steps of construction :

1. Draw a line segment BC = 7 cm.
2. At B, draw a ray BX making an angle of 45° and cut off BA = 5 cm.
3. Join AC.
4. Draw the angle bisectors of ∠B and ∠C intersecting each other at I.
5. From I, draw a perpendicular ID on BC.
6. With centre, I and radius ID, draw a circle
   which touches the sides of ∆ABC at D, E and F respectively.
   This is the required inscribed circle.

Question 6.
Draw a triangle ABC, given that BC = 4cm, ∠C = 75° and that radius of circumcircle of ∆ABC is 3 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC = 4 cm
2. Draw the perpendicular bisector of BC.
3. From B draw an arc of 3 cm radius which intersects the perpendicular bisector at O.
4. Draw a ray CX making art angle of 75°
5. With centre O and radius 3 cm draw a circle which intersects the ray CX at A.
6. Join AB.
   ∆ABC is the required triangle

Question 7.
Draw a regular hexagon of side 3.5 cm construct its circumcircle and measure its radius.
Solution:
Steps of construction:

1. Draw a regular hexagon ABCDEF whose each side is 3.5 cm.
2. Draw the perpendicular bisector of AB and BC
   which intersect each other at O.
3. Join OA and OB.
4. With centre O and radius OA or OB, draw a circle
   which passes through A, B, C, D, E and P.
   Then this is the required circumcircle.

Question 8.
Construct a triangle ABC with the following data: AB = 5 cm, BC = 6 cm and ∠ABC = 90°.
(i) Find a point P which is equidistant from B and C and is 5 cm from A. How many such points are there ?
(ii) Construct a circle touching the sides AB and BC, and whose centre is equidistant from B and C.
Solution:
Steps of Construction :

1. Draw a line segment BC = 6 cm.
2. At B, draw a ray BX making an angle of 90° and cut off BA = 5 cm.
3. Join AC.
4. Draw the perpendicular bisector of BC.
5. From A with 5 cm radius draw arc which intersects the perpendicular bisector of BC at P and P’.
   There are two points.
6. Draw the angle bisectors of ∠B and ∠C intersecting at 0.
7. From O, draw OD ⊥ BC.
8. With centre O and radius OD, draw a circle which will touch the sides AB and BC.
   This is the required circle.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Constructions Chapter Test are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Question 1.
(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD

Solution:
(a) ∆ABC is an equilateral triangle.

Question 2.
(a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find
(i) ∠BDC (ii) ∠CAE.

(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Solution:
(a) Join DB, CA and CB.
∠ADC = 118° (given)
and ∠ADB = 90°

Question 3.
(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).
(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.

Solution:
(a) Given : O is the centre of the circle.
To Prove : ∠AOC = 2 (∠ACB + ∠BAC).
Proof: In ∆ABC,
∠ACB + ∠BAC + ∠ABC = 180°
(Angles of a triangle)

Question 4.
(a) In the figure (i) given below, AB is diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :
(i)∠ADC (ii) ∠DAC.
(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If ∠APB = 70° and ∠BCD = 60°, find :
(i) ∠AOB (ii) ∠ACB
(iii) ∠ABD (iv) ∠ADB.

Solution:
(a) AB is diameter and DC || AB,
∠CAB = 25°, join AD,BD

Question 5.
(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.
(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.

Solution:
(a) Given : ABDC is a cyclic quadrilateral AB = CD.
To Prove: AD = BC.

Question 6.
A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.
Solution:
Join OT, OP = 13 cm and TP = 12 cm

Question 7.
Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Given : Two circles with centre O and O’ touch each other internally at P.

Question 8.
From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP.
(ii) OP is the perpendicular bisector of the chord AB.
Solution:
Given : From a point P, outside the circle with centre O. PA and PB are the tangents to the circle, OA, OB and OP are joined.

Question 9.
(a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:
AP : BQ = PC : CQ.

(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.

Solution:
(a) Given : Two circles with centres A and B and a transverse common tangent to these circles meet AB at C.

Question 10.
In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

Solution:
In the given figure, two chords with centre A and B touch externally. PM is a tangent to the circle with centre A and QN is tangent to the circle with centre B. PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm. We have to find AB.

Question 11.
Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.
Solution:
∵ AB and CD are two chords of a circle which intersect each other at P, outside the circle.

Question 12.
(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle

(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also find the length of the tangent drawn from P to the circle. .

Solution:
Given : (a) AB is chord of a circle with centre O and PT is tangent and CD is the diameter of the circle which meet at P.
AP = 16 cm, AB = 12 cm, OP = 2 cm
∴PB = PA – AB = 16 – 12 = 4 cm
∵ABP is a secant and PT is tangent.
∴PT² = PA x PB .

Question 13.
In the figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and.the radius of the circle is 6 cm, compute the length of AB. Also find the length of tangent drawn from X to the circle.

Solution:
Chord AB and diameter PQ meet at X on producing outside the circle

Question 14.
(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.
(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:
(i)∠BEC (ii) ∠ACB
(iii) ∠BCD (iv) ∠CED.

Solution:
(a) (i) Given : ABCD is a cyclic quadrilateral.
Ext. ∠PBC = ∠ADC
=> 40° = ∠ADC

Question 15.
(a) In the figure (i) given below, APC, AQB and BPD are straight lines.
(i) Prove that ∠ADB + ∠ACB = 180°.
(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as diameter

(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of ∆ABC cut the circles at E and D respectively. Prove that the points C, E, P and D are concyclic.

Solution:
(a) Given : In the figure, APC, AQB and BPD are straight lines.

Question 16.
(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find
(i) ∠BDC (ii) ∠BEC
(iii) ∠AEC (iv) ∠AOB.
Hence Prove that AOAB is equilateral.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that
(i) ∠CED = 3 ∠CBD (ii) CD = DA.

Solution:
(a) In ∆BCD, BC = CD
∠CBD = ∠CDB
But ∠BCD + ∠CBD + ∠CDB = 180°
(∵ Angles of a triangle)

Question 17.
(a) In the adjoining figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find
(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :
(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.

Solution:
(a) AB and XY are diameters of a circle with centre O.
∠APX = 30°.
To find :
(i) ∠AOX (ii) ∠APY
(iii) ∠BPY (iv) ∠OAX

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Question 1.
Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Solution:
(i) Draw a line segment AB = 8 cm.

(ii) Draw the perpendicular bisector of AB intersecting AB at D.
∴ Every point P on it will be equidistant from A and B.
(iii) Take a point P on the perpendicular bisector.
(iv) Join PA and PB.
Proof : In ∆PAD and ∆PBD
PD = PD (common)
AD = BD (D is mid-point of AB)
∠PDA = ∠PDB (each 90°)
∴ ∆ PAD ≅ ∆ PBD (SAS axiom of congruency)
∴PA = PB (c.p.c.t.)
Similarly we can prove any other point on the perpendicular bisector of AB is equidistant from A and B.
Hence Proved.

Question 2.
A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Solution:
The point P is moving in the space and it is at a constant distance from a fixed point C.
∴ Its locus is a sphere.

Question 3.
Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².
Solution:
Base of ∆PAB = 7 cm
and its area = 14 cm²

Now draw a line XY parallel to AB at a distance of 4 cm.
Now take any point P on XY
Join PA and PB
area of ∆PAB = 14 cm.
Hence locus of P is the line XY which is parallel to AB at distance of 4 cm.

Question 4.
Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS ? Compute the area of the quadrilateral PQRS.
Solution:
Steps of Construction :

(i) Take a line AB = 12 cm
(ii) Take M, the mid point of AB.
(iii) Draw straight lines CD and EF parallel to AB at a distance of 3 cm.
(iv) With centre M and radius 5 cm, draw areas which intersects CD at P and Q and EF at R and S.
(v) Join QR and PS.
PQRS is a rectangle where length
PQ = 8 cm.
Area of rectangle PQRS = PQ x RS = 8 x 6 = 48 cm²

Question 5.
AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
Solution:
(i) AB and CD are the intersecting lines which intersect each other at O.

(ii) Draw a line EF parallel to AB and GH parallel to CD intersecting each other at P
P is the required point.

Question 6.
Two straight lines PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flag staff X, which is equidistant from P and S, and is also equidistant from the road.
Solution:
1 cm = 100 cm
800 m = 8 cm.
Steps of Construction :
(i) Draw the lines PQ and PK intersecting each other at P making an angle of 75°.

(ii) Take a point S on PQ such that PS = 8 cm.
(iii) Draw the perpendicular bisector of PS.
(iv) Draw the angle bisector of ∠KPS intersecting the perpendicular bisector at X. X is the required point which is equidistant from P and S and also from PQ and PK.

Question 7.
Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Solution:
Steps of Construction :

(i) Take PR = 8 cm and draw the perpendicular bisector of PR intersecting it at O.
(ii) From O, out. off OS = OQ = 3 cm
(iii) Join PQ, QR, RS and SP.
PQRS is a rhombus. Whose diagonal are PR and QS.
(iv) PR is the bisector of ∠SPQ.
(v) Draw the perpendicular bisector of SR intersecting PR at X
∴ X is equidistant from PQ and PS and also from S and R.
On measuring length of XR = 3.2 cm (approx)

Question 8.
Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm. the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Solution:
Steps of Construction :

(i) Take AB = 5.1 cm
(ii) At A, with readius \(\\ \frac { 5.6 }{ 2 } \) = 2.8 cm and at B with radius \(\\ \frac { 7.0 }{ 2 } \) = 3.5 cm, draw two arcs
intersecting each other at O.
(iii) Join AO and produce it to C such that OC = AD = 2.8 cm and join BO and produce it to D such that BO = OD = 3.5 cm
(iv) Join BC, CD, DA
ABCD is a parallelogram.
(v) Draw the angle bisector of ∠ABC intersecting CD at P. P is the required point which is equidistant from AB and BC.

Question 9.
By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4cm and ∠DAB = 75°. C is equidistant from the sides AB and AD, also C is equidistant from the points A and B.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 75° and cut off AD = 4 cm.
(iii) Draw the bisector of ∠DAB.
(iv) Draw perpendicular bisector of AB intersecting the angle bisector at C.
(v) Join CB and CD.
ABCD is the required quadrilateral.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Question 1.
Find the equation of a line whose inclination is 60° and y-intercept is – 4.
Solution:
Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
=> y – √3x – 4 Ans.

Question 2.
Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Solution:
Slope of the line 3y + 2x = 12
=> 3y = 12 – 2x
=> 3y = – 2x + 12

Question 3.
If the equation of a line is y – √3x + 1, find its inclination.
Solution:
In the line
y = √3 x + 1
Slope = √3 => tan θ = √3
θ = 60° (∵ tan 60° = √3)

Question 4.
If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.
Solution:
The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points – 4 = 2 m + c …(i)
and 1 = – 3 m + c …(ii)
Subtracting we get,

Question 5.
If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.
Solution:
Let the points to be A (1, 4), B (3, – 2) and C (p, – 5) are collinear and let B (3, – 2)
divides AC in the ratio of m1 : m2

Question 6.
Find the inclination of the line joining the points P (4, 0) and Q (7, 3).
Solution:
Slope of the line joining the points P (4, 0) and Q (7, 3)

Question 7.
Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to \(– \frac { 3 }{ 7 } \)
Solution:
Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5
Adding we get,

Question 8.
If the point A is reflected in the y-axis, the co-ordinates of its image A₁, are (4, – 3),
(i) Find the co-ordinates of A
(ii) Find the co-ordinates of A₂, A₃ the images of the points A, A₁, Respectively under reflection in the line x = – 2
Solution:
(i) ∵ A is reflected in the y-axis and its image is A₁ (4, – 3)
Co-ordinates of A will be ( – 4, – 3)

Question 9.
If the lines \(\frac { x }{ 3 } +\frac { y }{ 4 } =7 \) and 3x + ky = 11 are perpendicular to each other, find the value of k.
Solution:
Given
Equation of lines are

Question 10.
Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).
Solution:
The equation of the line is x – 2y + 8 = 0
=> 2y = x + 8

Question 11.
Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.
Solution:
Equations of the line is
3y = 5 – x
=> 3y = – x + 5

Question 12.
Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.
Solution:
Let slope of the line joining the points A (1, 2) and B (6, 7) be m₁

Question 13.
The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.
Solution:
Slope of line AC (m₁)

Question 14.
A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:

(i) the co-ordinates of A and B
(ii) the equation of the line AB
Solution:
A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y) , and P (2, 1) divides BA in the ratio 3 : 1.

Question 15.
A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.
Solution:
Let the line make intercept a and b with the x-axis and y-axis respectively then the line passes through

Question 16.
If the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.
Solution:
Co-ordinates of A are (3, – 2).

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Question 1.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B
Solution:
Base BC of an equilateral ∆ABC lies on y-axis co-ordinates of point C are (0, – 3), origin (0, 0) is the mid-point of BC.

Question 2.
A and B have co-ordinates (4, 3) and (0, 1), Find
(i) the image A’ of A under reflection in the y – axis.
(ii) the image of B’ of B under reflection in the lineAA’.
(iii) the length of A’B’.
Solution:

(i) Co-ordinates of A’, the image of A (4, 3) reflected in y-axis will be ( – 4, 3).
(ii) Co-ordinates of B’ the image of B (0, 1) reflected in the line AA’ will be (0, 5).
(iii) Length A’B’

Question 3.
Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.
Solution:
Let R be the point whose co-ordinates are (x, y) which divides PQ in the ratio of 3:1.

Question 4.
Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Co-ordinates of A (3, 1) and B ( – 2, 5)
P lies on AB such that

Question 5.
P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Solution:
Given
AP = PQ = QB

Question 6.
The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).
Solution:
Let A (2, – 2), B (8, – 2) and centre of the circle be
O (α + 2, α – 5)

Question 7.
The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:
Given
(3, 5) is the mid-point of A (2, p) and B (q, 4)

Question 8.
The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.
Solution:
Let AB be the diameter where co-ordinates of A are (3, 0) and of B are ( – 5, 6).
Co-ordinates of its origin O will be

Question 9.
In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?
Solution:
Let the point ( – 4, 6) divides the line segment joining the points
A ( – 6, 10) and B (3, – 8), in the ratio m : n

Question 10.
Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.
Solution:
Let P ( – 3, p) divides AB in the ratio of m1 : m2 coordinates of
A ( – 5, – 4) and B ( – 2, 3)

Question 11.
In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.
Solution:
Let the point P which is on x-axis, divides the line segment joining the points A (4, 2) and B (3, – 5) in the ratio of m1 : m2.
and let co-ordinates of P be (x, 0)

Question 12.
If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.
Solution:
Let co-ordinates of A be ( – 4, 3) and of B (6, 3) and of P be (2, y)
Let the ratio in which the P divides AB be m1 : m2

Question 13.
Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.
Solution:
Points are given A (2, – 2), B (3, 7)
and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2 at P and let co-ordinates of

Question 14.
The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.
(i) Write the coordinates of A’ and A”.
(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.
Solution:
A’ is the reflection of A(2, – 3) in the x-axis
(i) ∴ Co-ordinates of A’ will be (2, 3)
Draw a line x = 4 which is parallel to y-axis
A” is the reflection of A’ (2, 3)
∴Co-ordinates OA” will be (6, 3)
(ii) Join AA” which intersects x-axis at P whose
co-ordinate are (4, 0)
Let P divide AA” in the ratio in m1 : m2

Hence P(4, 0) divides AA” in the ratio 1 : 1

Question 15.
ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.
Solution:
Let the co-ordinates of C be (x, y) and other three vertices of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)
∴ ABCD is a parallelogram
Its diagonals bisect each other.
Let AC and BD intersect each other at O.
∴O is mid-points of BD
∴ Co-ordinates of O will be

Question 16.
ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.
Solution:
ABCD is a || gm , m which co-ordinates of A are (2, – 3) and B (-1, -1)
Its diagonals AC and BD bisect each other at M (1, – 4)
∴ M is mid point of AC and BD Let co-ordinates of C be (x1, y1) and of D be (x2, y2) when M is mid point of AC then

Question 17.
In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.

Solution:
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y) and P (3, 1) divides it in the ratio of 2 : 3.

Question 18.
Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.
Find : (i) the co-ordinates of Q.
(ii)the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.

Solution:
(i) Let co-ordinates of Q be (x’, y’) and of R (x”,y”)
Point P (1, 2) divides OQ in the ratio of 2 : 3

Question 19.
If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Solution:
A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of ∆ABC
D, E and F are the midpoints of sides BC, CA and AB respectively and G is the centroid of the ∆ABC

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test are helpful to complete your math homework.

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