ML Aggarwal ICSE Solutions for Class 10 Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration 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 18 Mensuration Chapter Test.

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Question 1.
A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m², find the cost of the tin for making the container.
Solution:
Height of container opened at the top (h) = 1 m = 100 cm
and diameter = 70 cm
∴Radius (r) = \(\\ \frac { 70 }{ 2 } \) = 35 cm
∴Total surface area = 2πrh + πr²

Question 2.
A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Solution:
Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm
Volume = 30 × 14 × 14 = 5880 cm³

Question 3.
Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.
Solution:
Slant height of a cone (l) = 17 cm
Diameter of base = 30 cm
Radius (r) = \(\\ \frac { 30 }{ 2 } \) = 15 cm

Question 4.
Find the volume of a cone given that its height is 8 cm and the area of base 156 cm².
Solution:
Height of a cone = 8 cm
Area of base = 156 cm
.’. Volume = \(\\ \frac { 1 }{ 3 } \) × area of base × height

Question 5.
The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Solution:
Circumference of the edge of bowl = 132 cm
Radius of a hemispherical bowl

Question 6.
The volume of a hemisphere is \(2425 \frac { 1 }{ 2 } \) cm². Find the curved surface area.
Solution:
Volume of a hemisphere = \(2425 \frac { 1 }{ 2 } \) cm³
= \(\\ \frac { 4851 }{ 2 } \) cm³
Let radius = r, then

Question 7.
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy
Solution:
A wooden solid toy is of a shape of a right circular cone
mounted on a hemisphere.
Radius of hemisphere (r) = 4.2 cm
Total height = 10.2 cm

Question 8.
A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.
Solution:
Diameter of cylindrical part = 0.5 cm
Total length of the capsule = 2 cm

Question 9.
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Solution:
Radius of cylinder = \(\\ \frac { 7 }{ 2 } \)cm
and height of cylinder = 19 – 2 × \(\\ \frac { 7 }{ 2 } \) cm
= 19 – 7 = 12 cm
and radius of hemisphere = \(\\ \frac { 7 }{ 2 } \) cm

Question 10.
The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).
Solution:
Let radius of cone (r) = 5x
then height (h) = 12x

Question 11.
A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Solution:
Let height of cone and cylinder = h
Diameter of the base of cone = 3x
Diameter of base of cylinder = 2x

Question 12.
A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Solution:
Radius of the cone (r) = 9 cm
Height of the cone (h) = 10 cm
Volume of water filled in cone

Question 13.
An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Solution:
Radius of the base of cone = 8 cm

Question 14.
A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.
Solution:

Question 15.
The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their
(i) curved surfaces
(ii) volumes.
Solution:
Radius of the base of a cone (r) = 3 cm
Height (h) = 4 cm

Question 16.
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Solution:
Radius of base of a cone (r) = 2. 1 cm
and height (h) = 8.4 cm

Question 17.
How many lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.
Solution:
Dimensions of a solid rectangular lead piece
= 66 cm × 42 cm × 21 cm
.’. Volume = 66 × 42 × 21 cm³
Diameter of a lead shot = 4.2 cm

Question 18.
Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Solution:
Radius of a cylinder (r) = 3 cm
Height (h) = 5 cm

Question 19.
A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Solution:
Radius of hemisphere = 8 cm
Volume = \(\frac { 2 }{ 3 } \pi { r }^{ 3 }{ cm }^{ 3 }\)

Question 20.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.
Solution:
Radius of hemispherical bowl = 6 cm
.’. Volume of the water in the bowl

Question 21.
The diameter of a metallic sphere is 42 cm. It is metled and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.
Solution:
Diameter of sphere = 42 cm
Radius of sphere =\(\\ \frac { 42 }{ 2 } \) = 21 cm

Question 22.
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Radius of sphere = \(\\ \frac { 6 }{ 2 } \) = 3 cm

Question 23.
A solid sphere of radius 6 cm is metled into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Solution:
Radius of solid sphere = 6 cm
Volume of solid sphere = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

Question 24.
A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical vessel, full of water, in such a Way that the whole solid is submerged in water. If the radius of the cylindrical vessel is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical vessel.
Solution:
Radius of hemisphere (r) = 3.5 cm
Height of cone (h₁) = 4 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test, drop a comment below and we will get back to you at the earliest.

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.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

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.

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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.

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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.