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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20E
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G

Question 1.
Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.
Solution:
Slant height (L) = 17 cm
Radius (r) = 8cm
But l²= r² + h²
⇒  h² = l²-r² = 17² – 8²
⇒ h² = 289 – 64 = 225 = (15)²
∴   h=15 cm.

Question 2.
The curved suface area of a cone is 12320 cm². If the radius of its base is 56 cm, find its height.
Solution:
Curved surface area = 12320 cm²
Radius of base (r) = 56 cm.
Let slant height = l.

Question 3.
The circumference of the base of a 12 m high conical tent is 66m. Find the volume of the air contained in it.
Solution:
Circumference of conical tent = 66 m
and height (h) = 12 m.

Question 4.
The radius and the height of a right circular cone are in the ratio 5 :12 and its volume is 2512 cubic Cm. Find the radius and slant height of the cone. (Take π = 3.14)
Solution:
The ratio between radius and height = 5 : 12
Volume =2512 cm³
Let radius (r) = 5x and
height (h) = 12x
and slant height = l

Question 5.
Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.
Solution:
Let radius of cone y = r
∴ radius of cone x = 3r
Let volume of cone y = V
Then volume of x = 2V
Let h₁ be the height of x and h₂ be the height of y.

Question 6.
The diameters of two cones are equal, if their slant heights are in the ratio of 5:4, find the ratio of their curved surface area.
Solution:
Let radius of each one = r
and ratio between their slant heights =5:4
Let slant height of the first cone = 5x
and slant height of second = 4 x
∴  Curved surface area of the first cone
= πr = πr x 5x = 5πrx.
and curved suface area of second cone
= πr x 4x = 4πrx
∴ Ratio between them = 5 πrx : 4 πrx
= 5:4

Question 7.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Solution:
Let the slant height of first cone = l
then slant height of the second cone = 2l
and let r₁  be the radius of the first cone and r₂ be the radius of the second cone.
Then curved surface area of the first cone = πr₁l
and that of second cone = πr₂2l= 2πr₂l.
According to the condition,

Question 8.
A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5m. Find its volume. How much cloth is required to just cover the heap?
Solution:

Question 9.
Find what length of canvas, 1.5m in width, is required to make a conical tent 48 m in diameter and 7m in height Given that 10% of the canvas is used in folds and stritchings. Also, find the cost of the canvas at the rate of ₹24 per metre.
Solution:

Question 10.
A solid cone of height 8 cm and base radius 6 cm is melted and recast into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.
Solution:
Height of solid cone (h) = 8 cm.
Radius (r) = 6 cm.

Question 11.
The total surface area of a right circular cone of slant height 13 cm is 90π cm². Calculate:
(i) its radius in cm
(ii) its volume in cm³. [Take π = 3.14]
Solution:
Total surface area of cone = 90π cm²
slaint height (l) = 13 cm
Let r be its radius, then
Total surface area = πrl + πr² = πr (l + r)

Question 12.
The area of the base of a conical solid is 38.5 cm² and its volume is 154 cm³. Find curved surface area of the solid.
Solution:
Area of base of a solid cone = 38.5 cm²
and volume  = 154 cm³

Question 13.
A vessel, in the form of an invested cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result, one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged ?
Solution:
Diameter of the base of cone = 25.2 cm

Question 14.
The volume of a conical tent is 1232 m³ and the area of the base floor is 154 m². Calculate the:
(i) radius of the floor.
(ii) height of the tent.
(iii) length of the canvas required to cover this conical tent if its width is 2 m. (2008)
Solution:
Volume of conical tent = 1232 m³
Area of base floor = 154 m²
(i)  Let r be the radius of the floor

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20B are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19

Question 1.
Draw a circle of radius 3 cm. Mark a point P at a distance of 5cm from the centre of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.
Solution:

Steps of Construction:
(i) Draw a circle with centre O and radius 3 cm.
(ii) From O, take a point P such that OP = 5 cm.
(iii) Draw the bisector of OP which intersects OP at M.
(iv) With centre M, and radius OM. draw’ a circle which intersects the given circle at A and B.
(v) Join AP and BP.
AP and BP are the required tangents.
On measuring them, AP = BP = 4 cm.

Question 2.
Draw a circle of diameter 9 cm. Mark a point at a distance of 7.5 cm from the centre of the circle. Draw tangents to the given circle from this exterior point. Measure the length of each tangent.
Solution:

Steps of Construction:
(i) Draw a line segment AB = 9 cm.
(ii) Draw a circle with centre O and AB as diameter.
(iii) Take a point P from the centre at a distance of 7.5 cm.
(iv) Draw an other circle OP as diameter which intersects the given circle at T and S.
(v) Join TP and SP.
TP and SP are are required tangents.
On measuring their lengths, TP = SP = 6 cm.

Question 3.
Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45°.
Solution:
Steps of Construction:
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw two arcs making an angle of 180° – 45° = 135°
so that ∠AOB = 135°.

(iii) At A and B, draw two rays making an angle of 90° at each point which meet each other at P, out side the circle.
Then AP and BP are the required tangents which make an angle of 45° at P.

Question 4.
Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60°.
Solution:

Steps of Construction:
(i) Draw a circle with centre O and radius 4.5 cm.
(ii) Draw two arcs making an angle of 180° – 60° = 120° i.e. ∠AOB = 120°.
(iii) At A and B draw rays making an angle of 90° at each point which meet each other at P outside the circle.
AP and BP are the required tangents which makes an angle of 60° at P.

Question 5.
Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.
Solution:
Steps of Construction:
(i) Draw a line segment BC = 4.5 cm.
(ii) With centres B and C, draw two arcs of radius 4.5 cm. which intersect each other at A.
(iii) Join AB and AC,
(iv) Draw the perpendicular bisectors of AB and BC intersecting each other at O.
(v) With centre O, and radius OA or OB or OC draw a circle which will passes through A, B and C.
This is the required circumcircle of ∆ ABC.
Measuring OA = 2.6 cm

Question 6.
Construct triangle ABC, having given = 7 cm, AB – AC = 1 cm and ∠ABC = 45°.
(ii) Inscribe a circle in the ∆ ABC constructed in (i) above,
Solution:

Steps of Construction:
(i) Draw a line segment BC = 7 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = AB – AC = 1 cm.
(iii) Join EC and draw the perpendicular bisector of EC intersecting BX at A.
(iv) Join AC
∆ ABC is the required triangle.
(v) Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.
(vi) From O, draw perpendicular OL to BC.
(vii) O as centre and OL as radius draw circle which touches the sides of the A ABC. This is the required in-circle of ∆ ABC.
On measuring radius OL = 1.8 cm (approx.).

Question 7.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.
Solution:

Steps of Construction:
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and C, draw two arcs of 5 cm radius each which intersect each other at A.
(iii) Join AB and AC.
(iv) Draw angle bisectors of ∠B and ∠C intersecting each other at O.
(v) From O, draw OL ⊥ BC.
(vi) Now with centre O and radius OL, draw a circle which will touch the sides of the ∆ ABC. Measuring OL =1.4 cm. (approx.).

Question 8.
Using ruler and compasses only,
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, BC = 6.2 cm and ∠CAB = 60°.
(ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre O.
(iii) Draw a perpendicular from O to AB which meets AB in D.
(iv) Prove that AD = BD.
Solution:
Steps of Construction:
(i) Draw a line segment AB = 6 cm.
(ii) At A, draw a ray making an angle of 60° with BC.
(iii) B as centre and 6.2 cm as radius draw an arc which intersect the AX rays at C.
(iv) Join CB.
∆ ABC is the required triangle.
(v) Draw the perpendicular bisectors of AB and AC intersecting each other at O.
(vi) With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
(vii) From O, draw OD ⊥ AB.
Proof: In right ∆ OAD and ∆ ODB
Hyp, OA = OB (radii of the saine circle)
Side OD = OD (Common)
∴ OAD ≅ OBD (R.H.S.)
∴ AD = BD (C.P.C.T.)

Question 9.
Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC and measure its radius.
Solution:

Steps of Construction:
(i) Draw a line segment BC = 4 cm.
(ii) At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.
(iii) From E, draw another perpendicular line EY.
(iv) From C, draw a ray making an angle of 45° with CB, which intersects EY at A.
(v) JoinAB.
∆ ABC is the required triangle.
(vi) Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
(vii) With centre O, and radius OB, draw a circle which passes through A, B and C.
Measuring the radius OB = OC = OA = 2 cm

Question 10.
Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.
(i) What do you call the point O ?
(ii) What is the relation between the distances OA, OB and OC?
(iii) Does the perpendicular bisector of BC pass through O ?
Solution:

(i) Perpendicular bisectors of sides AB and AC intersect each other at O.
(ii) O is called the circum centre of circumcircle of ∆ ABC.
(iii) OA, OB and OC are the radii of the circumcircle.
(iv) Yes, the perpendicular bisector of BC will also pass through O.

Question 11.
The bisectors of angles A and B of a scalene triangle ABC meet at O.
(i) What is the point O called ?
(ii) OR ancLOQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ ?
(iii) What is the relation between angle ACO and angle BCO ?
Solution:

(i) ∆ ABC is a scalene triangle.
(ii) Angle bisectors of ∠A and ∠B intersect each other at O. O is called the incentre of the incircle of ∆ ABC.
(iii) Through O, draw perpendiculars to AB and AC which meet AB and AC at R and Q respectively.
(iv) OR and OQ are the radii of the in circle and OR =OQ.
(v) OC is the bisector of ∠C
∴∠ACO = ∠BCO

Question 12.
(i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
(ii) Find its incentre and mark it I.
(iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle. What is the length of the radius of this circle.
Solution:

Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) With centre B and radius 8 cm draw ah arc.
(iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
∆ ABC is the given triangle.
(v) Draw the angle bisectors of ∠B and ∠A intersecting each other at I.
Then I is the incentre of incircle of ∆ ABC.
(vi) Through I, draw ID ⊥ AB.
(vii) Now from D, cut off DP = DQ = \(\frac { 2 }{ 2 }\) = 1 cm.
(viii) With centre I, and radius IP or IQ, draw a circle which will intersect each side of ∆ ABC cuting chords of 2 cm each.

Question 13.
Construct an equilateral triangle ABC with side 6cm. Draw a circle circumscribing the triangle ABC.
Solution:

Steps of construction:
(i) Draw a line segment BC = 6cm.
(ii) With centre B and C, draw arcs with radius 6 cm each which intersect each other at A.
(iii) Join AB and AC,
then ∆ABC is the equilateral triangle.
(iv) Draw the perpendicular bisectors of BC and AB which intersect each other at O.
(v) Join OB and OC and OA.
(vi) 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 circle.

Question 14.
Construct a circle, inscribing an equilateral triangle with side 5.6 cm.
Solution:

Steps of construction:
(i) Draw a line segment BC = 5.6 cm
(ii) With centre B and C,
draw two arcs of radius 5.6cm each which intersect each other at A.
(iii) Join AB and AC, then ∆ABC is an equilateral triangle.
(iv) Draw the angle bisectors of ∠B and ∠C which intersect each other at I.
(v) From I, draw ID ⊥ BC.
(vi) With centre I and radius ID, draw circle which touches the sides of the ∆ABC. This is the required circle.

Question 15.
Draw a circle circumscribing a regular hexagon of side 5cm.
Solution:

Steps of construction:
(i) Draw a regular hexagon ABCDEF whose each side is 5cm.
(ii) Join its diagonals AD, BE and CF intersecting each other at O.
(iii) With centre O and radius OA, draw a circle which will pass through the vertices of the hexagon A, B, C, D, E and F. This is the required circle.

Question 16.
Draw an inscribing circle of a regular hexagon of side 5.8 cm.
Solution:
Steps of construction:
(i) Draw a line segment AB = 5.8cm.

(ii) At A and B, draw rays making an angle of 120° each and cut off AF = BC = 5.8 cm.
(iii) Again at F and C, draw rays making an angle of 120° each and cut off FE = CD = 5.8 cm.
(iv) JoinDE. Then ABCDEF is the regular hexagon.
(v) Draw the bisectors of ∠A and ∠B intersecting each other at O.
(vi) From O, draw OL J. AB.
(vii) With centre O and radius OL, draw a circle which touches the sides of the hexagon. This is the required incircle of the hexagon.

Question 17.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Solution:
Steps of construction:
(i) Draw a circle of radius 4 cm with centre O.
(ii) Since regular hexagon \(\frac { { 360 }^{ \circ } }{ 6 }\) = 60°, draw radii
OA and OB, such that ∠AOB = 60°.
(iii) Cut off arcs BC, CD, DE, EF and each equal to arc AB on given circle.
(iv) Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.

Question 18.
Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent (2011).
Solution:

Steps of construction:
(i) Draw a line segment OP = 6 cm
(ii) With centre O and radius 3.5 cm, draw a circle
(iii) Draw the mid point of OP.
(iv) With centre M and diameter OP, draw a circle which intersect the circle at T and S.
(v) Join PT and PS.
PT and PS are the required tangent on measuring the length of PT = PS = 4.8 cm

Question 19.
Construct a triangle ABC in which base BC=5.5 cm,AB = 6cmand ∠ABC = 120°.
(i) Construct a circle circumscribing the triangle ABC.
(ii) Draw- a cyclic quadrilateral ABCD so that D is equidistant from B and C.
Solution:

Steps of construction:
(i) Draw BC = 6 cm. x
(ii) At B, draw ∠XBC= 120°.
(iii) From BX, cut off AB = 6 cm.
(iv) Join AC to get ∆ ABC.
(v) Draw the perpendicular bisector of BC and AB. These bisectors meet at O. With O as centre and radius equal tb OA, draw a circle, which passes through A, B and C. This is the required circumcircle of ∆ABC.
(vi) Produce the perpendicular bisector of BC so that it meets the circle at D. Join CD and AD to _ get the required cyclic quadrilateral ABCD.

Question 20.
Using a ruler and compasses only :
(i) Construct a triangle ABC with the following data : AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP
Solution:
Steps of construction:
(i) Draw AB = 3.5

(ii) At B, draw ∠ABX = 120°.
(iii) With B as center draw an arc of radii 6 cm at C.
(iv) Join A and C.
(v) Draw the perpendicular bisector of line BC and draw a circle with BC as diameter.
(vi) Draw angle bisector of ∠B.
Meets the circle at P
∴ P is the required point ∠BCP = 30°

Question 21.
Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. (2014)
Solution:
Steps of construction:
(i) Draw a line segment BC = 6.5 cm.
(ii) From B, draw an arc of radius of 5.5 cm and from C, another arc of 5 cm radius which intersect each other at A.
(iii) Join AB and AC.
∆ABC is required triangle.
(iv) Draw the angle bisectors of ∠B and ∠C which intersect each other at O.
(v) Through O, draw OL ⊥ BC.
(vi) With centre O and radius OL, draw a circle which touches the sides of ∆ABC.
(vii) On measuring, OL = r = 1.5 cm.

Question 22.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC. (2015)
Solution:
Steps of construction:
(i) Draw a line segment AB = 5.5 cm.
(ii) At A, draw a ray AX making an angle of 105°.
(iii) Cut off AC from AX =6 cm.
(iv) JoinCB.
∆ABC is required triangle.
(v) Draw angle bisector CX of ∠C.
CX is the locus of points equidistant from BA and BC.
(vi) Draw the perpendicular bisector of BC which is the locus of points equidistant from the points B and C.
These two loci intersect each other at P.
Join PC and on measuring it, it is 4.8 cm (approx).

Question 23.
Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. (2016)
Solution:
Steps of construction :
(i) Draw AF measuring 5 cm using a ruler.
(ii) With A as the centre and radius equal to AF, draw an arc above AF.
(iii) With F as the centre, and same radius cut the previous arc at O.
(iv) With O as the centre, and same radius draw a circle passing through A and F.
(v) With A as the centre and same radius, draw an arc to cut the circle above AF at B.
(vi) With B as the centre and same radius, draw an arc to cut the circle at C.
(vii) Repeat this process to get remaining vertices of the hexagon at D and E
(viii) Join consecutive arcs on the circle to form the hexagon.
(ix) Draw the perpendicular bisectors of AF, EF and DE.
(x) Extend the bisectors of AF, EF and DE to meet CD, BC and AB at X, L and O respectively.
(xi) Join AD, CF and EB.
(xii) These are the 6 lines of symmetry of the regular hexagon.

Question 24.
Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents. (2016)
Solution:
Steps of construction :
(i) Draw AB = 5 cm using a ruler.
(ii) With A as the centre cut an arc of 3 cm on AB to obtain C.
(iii) With A as the centre and radius 2.5 cm, draw an arc above AB.
(iv) With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
(v) With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed
(vi) Join OB.
(vii) Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
(viii) With M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
(ix) Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.
QB = PB = 3 cm
That is, length of the tangents i.e. 3.2 cm.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20E
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G

Question 1.

The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find :
(i) the volume
(ii) the total surface area.
Solution:
Height of cylinder (h) =  20cm
and radius of its base (r) = 7 cm
(î) Volurne=πr²h

Question 2.
The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold ?
Solution:
Inner radius of a pipe (r) = 2.1 cm
and length of pipe (h) = 12 m = 1200 cm
∴ Volume of water in it = πr²h

Question 3.
A cylinder of circumference 8 cm and length 21 cm roils without sliding for 4 \(\frac { 1 }{ 2 }\) seconds at the rate of 9 complete rounds per second. Find:
(i) the distance travelled by the cylinder in 4 \(\frac { 1 }{ 2 }\) seconds, and
(ii) the area covered by the cylinder in 4 \(\frac { 1 }{ 2 }\) seconds.
Solution:
Circumference of a cylinder = 8 cm

Length of cylinder (h) = 21 cm
It takes 9 complete rounds per second
∴ Curved surface area = 2πrh

Question 4.
How many cubic metres of earth must be dug out to make a well 28 m deep and 2.8 m in diameter ? Also, find the cost of plastering its inner surface at ₹4.50 per sq. metre.
Solution:

Question 5.
What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long ?
Solution:
Diameter of solid cylinder = 2 cm
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm
Let length (h) = x cm
∴ Volume = πr²h = π x 1 x 1 x x
= πx cm³       …(i)
External diameter of hollow cylinder = 20cm
∴ External radius =  \(\frac { 20 }{ 2 }\) = 10 cm
Thickness of cylinder = 0.25 cm
∴ Innerradius= 10-0.25 = 9.75 cm
Length = 15 cm
∴ Volume = π(R² – r²) x h
= π(R + r)(R-r) x h
= π(10 + 9.75)(10-9.75) x 15 cm³
= πx 19.75 x 0.25 x 15 cm³    ………(ii)
Comparing (i) and (ii), we get
∴ π x 19.75 x 0.25 x 15 = π x
x = 19.75 x 0.25 x 15 cm
= 74.0625 = 74.06 cm

Question 6.
A cylinder has a diameter of 20 cm. The area of the curved surface is 100 cm² (sq. cm). Find:
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place.
Solution:

Question 7.
A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm³ of the metal weighs 7.7 g.
Solution:

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in the. level of the water when the solid is submerged.
Solution:
Diameter of base of a cylindrical container =42 cm
∴ Radius = \(\frac { 42 }{ 2 }\) = 21 cm
Size of rectangular solid = 22cmx 14cmx 10.5 cm
∴ Volume of solid = 3234 cm³
∴ Height of water level raise in the container

Question 9.
A cylindrical container with infernal radius of its base 10 cm’, contains water up”to a height of 7 cm. Find the area of the wet surface of the cylinder.
Solution:
Internal radius of cylindrical container (r) = 10cm
Water upto height (h) = 7 cm
∴ Area of wet surface by the water of the container = 2πrh
= π x 19.75 x 0.25 x 15 cm³                       …(ii)
Comparing (i) and (ii), we get
∴ π x 19.75 x 0.25 x 15 = πx
x = 19.75×0.25x 15 cm
= 74.0625
= 74.06 cm

Question 10.
Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.
Solution:
Length of open pipe (h) = 50 cm
External diameter=20 cm
and internal diameter = 6 cm
∴ External radius (R) = \(\frac { 20 }{ 2 }\) = 10 cm
and internal radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm
∴ Total surface area of the open pipe

Question 11.
The height and the radius of the base of a cylinder are in the ratio 3:1. If its volume is 1029 πcm³; find its total surface area.
Solution:
Ratio in height and radius of cylinder = 3:1
Let height = 3x
and radius = x cm
∴ Volume = πr²h = π x 3x x x² = 3πx³
∴ 3πt³= 1029π

∴ x = 7
∴ Height = 7 x 3=21 cm
and radius = 7 cm
Now total surface area = 2πr² + 2 πrh
= 2 πr (r + h)
= 2 x \(\frac { 22 }{ 7 }\) x 7(7 + 21)
=44 x 28 = 1232 cm²

Question 12.
The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.
Solution:
Let radius of the cylinder (r) = 100 cm
and height (h) = 100 cm
∴ Volume = πr²h =π( 100)² x 100 cm³ = 1000000π cm³
Now radius (R) = 100 + 20 = 120 cm
and new height (h) = 100 – 20 = 80 cm
∴ Volume = π(120)² x 80
= π x 14400 x 80cm³= 152000π cm³
Percentage change (increase) in the volume

Question 13.
The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its:
(i) volume
(ii) curved surface area
Solution:
Let radius of the cylinder = 100 cm
and height = 100 cm
∴ Volume = πr²h
= πx 100 x 100 x 100 cm³
= 1000000πcm³
and Curved surface area = 2πrh
= 2 x π x 100 x 100 cm²
=20000π cm²
Decreased radius = 100 – 20 = 80 cm
and increased height = 100 + 10 = 110 cm
∴ Increased volume = π x 80 x 80 x 110 cm³ = 704000π cm³
and increased curved surface =2 x π x 80 x 110 cm² = 17600πcm²
⇒  Decrease in volume = 1000000π – 704000π = 296000π cm³

Question 14.
Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of ₹56 per m.
Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.
Solution:
Diameter of the hollow closed cylinder =20 cm
∴ Radius (r) = \(\frac { 20 }{ 2 }\)
and height (h) = 35 cm

Question 15.
3080 cm³ of water is required to fill a cylindrical vessel completely and 2310 cm³ of water is required to fill it upto 5 cm below the top. Find:
(i) radius of the vessel.
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.
Solution:
Volume of water to fill a cylindrical vessel = 3080 cm³
Volume of water to fill it upto 5 cm below = 2310 cm²
Volume of water for 5 cm height =3080- 2310 = 770cm³

Question 16.
Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it:
(i) shorter side.                 
(ii) longer side.
Solution:
Length of rectangular sheet = 44 cm
and breadth = 33 cm

(i) Folding along shorter side i.e., 33cm
∴ Circumference of cylinder = 33cm

Question 17.
A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.
Solution:
Side of a cube = 11 cm
∴ Volume = (Side)³ = 11 x 11 x 11 cm³ = 1331 cm³
Diameter of cylinder = 28 cm

Question 18.
A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.
Solution:
Diameter of circular tank = 2m
Width of embankment = 2 m
Height = 1.6 m

Question 19.
The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm² and the volume of material in it is 1056 cin³. Find its internal and external radii. Given that the height of the cylinder is 21 cm.
Solution:
Sum of outer and inner surface area of a hollow cylinder = 1056 cm²
Volume of metal used =1056 cm³
Height of cylinder (h) = 21 cm
Let outer radius = R
and inner radius = r
Height = 21 cm
We are given
Outer surface + Inner surface = 1056 cm²
Volume = πR²h – πr²h =1056 cm³
Now, 2πRh + 2πrh =1056

Question 20.
The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm². If its height is 28 cm and the volume of material in it is 704 cm³; find its external curved surface area.
Solution:
Difference in outer and inner curved surface of a hollow cylinder = 352 cm²
Height (h) = 28 cm
Volume of material used = 704 cm³
Let outer radius = R
and inner radius = r
∴ 2πRh-2πrh = 352
2πh(R-r) = 352

Question 21.
The sum of the height and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm²; find the volume of the cylinder.
Solution:

Question 22.
The total surface area of a solid cylinder is 616 cm². If the ratio between its curved surface area and total surface area is 1:2; find the volume of the cylinder.
Solution:
Total surface area of a cylinder = 616 cm²

Question 23.
A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.
Solution:
Height of cylinder (6) = 24 cm
Radius (r)= \(\frac { 40 }{ 2 }\) = 20 cm
∴ Volume of water filled in it = πr²h
= π x 20 x 20 x 24 cm³
= 9600π cm³
Radius of small cylindrical bottle = \(\frac { 8 }{ 2 }\) = 4 cm
and height (6) = 10 cm
∴  Volume of one small bottle = πr²h
π x 4 x 4 x 10 cm³ = 160π cm³
∴ Number of small bottles

Question 24.
Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are melted and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.
Solution:
Diameter of first cylinder = 60 cm
∴ Radius (R)= \(\frac { 60 }{ 2 }\) =30 cm
and height (h) = 30 cm
Radius of second cylinder = 30 cm
and height = 60 cm
Volume of first cylinder = πR²h
= π30 x 30 x 30 cm³ = 27000π cm³
and volume of second cylinder = π x 30 x 30 x 60 cm³ = 54000πcm³
Total volume of both cylinders
= 27000π+ 54000π cm³
= 81000π cm³
Volume of third cylinder = 81000π cm³
Height of third cylinder = 10 cm

Question 25.
The total surface area of a hollow cylinder, which is open from both the sides, is 3575 cm²; area of its base ring is 357.5 cm² and its height is 14 cm. Find the thickness of the cylinder.
Solution:
Total surface area of an opened hollow cylinder = 3575 cm²
Area of ring of its base = 357.5 cm²
Height = 14 cm
Let R and r be the outer and inner radius of the ring.
∴ π(R² – r²) = 357.5

Question 26.
The given figure shows a solid formed of a solid cube of side 40 cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown.
Find the volume and the total surface area of the whole solid [Take π=3.14]

Solution:
Side of the cube = 40 cm
Radius of cylinder = 20 cm
Height (h) = 50 cm
Volume of cube = (40)³ = 64000 cm³
and volume of cylinder = πr²h
= 3.14 x 20 x 20 x 50 cm³
=314×200 = 62800 cm³
∴ Total volume = 64000 + 62800 =126800cm³
Total surface area = (6a² + 2πrh)
= 6 x 40 x 40 + 2 x 3.14 x 20 x 50 =9600+6280
= 15880 cm²

Question 27.
Two right circular solid cylinders have radii in the ratio 3: 5 and heights in the ratio 2:3. Find the ratio between their:
(i) curved surface areas.
(ii) volumes.
Solution:
The ratio is the radii of two right circular solid cylinder = 3:5
and ratio in their heights = 2:3
(i) Let radius of first cylinder = 3x
and height = 2y
∴ Curved surface area = 2πrh
= 2π(3x) (2y) = 12 πxy
and radius of second cylinder = 5x
and height = 3y
∴  Curved surface = 2πrh
= 2π x 5x 3y
= 30πxy
∴ Ratio in their curved surface
= 12πxy : 30πxy
= 2 :5
(ii) Volume of first cylinder = πr²h
= π(3x)² x 2y = 18πx²y
and volume of second cylinder = π(5x)² x 3y
= 75πx²y
∴ Ratio = 18πx²y : 75πx²y
= 6:25

Question 28.
A closed cylindrical tank, made of thin iron sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m², is used in making this tank, if \(\frac { 1 }{ 15 }\) of the sheet actually used was wasted in making the tank ?
Solution:
Diameter of a closed cylindrical tank=8.4 cm
∴ Radius (r) = \(\frac { 8.4 }{ 2 }\) = 4.2 m
and height (h) = 5.4 m
∴ Total surface area = 2πr(r + h)
= 2 x \(\frac { 22 }{ 7 }\) x 4.2(4.2+ 5.4) m²
=26.4 x 9.6 m²=253.44 m²
Area of sheet used in wastage
= \(\frac { 1 }{ 15 }\) of 253.44= 16.896 m²
Total sheet required = 16.896 +253.44 m²
= 270.336 m²
=270.34 m² (approx)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20A are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

Question 1.
The radius of a circle is 8cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10cm from its centre.
Solution:

OP = 10 cm,
raduis OT = 8 cm
∵ OT ⊥ PT
∴ In right ΔOTP,
OP² = OT²+PT²
⇒  (10)² =(8)²+PT²
⇒  100 = 64+PT²
⇒ PT² = 100-64 = 36
∴ PT = \( \sqrt{36} \) = 6 cm

Question 2.
In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.


Solution:

∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent)
⇒  OB² = OA² – AB² ⇒ r² = (r + 7.5)² – 15²
⇒ r² = r² + 56.25 + 15r – 225 168.75
⇒   15r= 168.75
⇒  r =\(\frac { 168.75 }{ 75 }\) ⇒ r=11.25
Hence, radius of the circle = 11.25 cm

Question 3.
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Solution:
Given: Two circles with centre O and O’ touches at P externally. Q is a point on the common tangent through P.
QA and QB are tangents from Q to the circles respectively.
To Prove: QA=QB.
Proof: From Q, QA and QP are the tangents to the circle with centre O
∴  QA=QP ….(i)
Similarly, QP and QB are the tangents to the circle with centre O’
∴ QP=QB   ….(ii)
From (i) and (ii)
QA=QB                           Q.ED.

Question 4.
Two circles touch each other internally. Show 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. Q is a point on the common tangent through P. QP an QB are tangents from Q to the circles respectively.

Question 5.
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

Solution:
Given: Two concentric circles with radius 5 cm and 3 cm with centre O. PQ is the chord of the outer circle which touches the inner circle at L. Join OL and OP.
OL=3 cm, OP = 5 cm

Question 6.
Three circles touch each other externally. A triangle is formed when the centres of these circles are j oined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution:
Three circles touches each other externally
Δ ABC is formed by joining the centres A, B and C of the circles.
AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centres A, B and C be r₁, r₂, r₃ respectively

Question 7.
If the sides of a quadrilateral ABCD touch a circle, prove that: AB + CD = BC + AD.

Solution:
Given: A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q,R and S respectively.

To Prove: AB + CD = BC + AD
Proof: Since AP and AS are the tangents to the circle from external point A.
∴ AP = AS     ….(i)
Similarly, we can prove that,
BP=BQ               ….(ii)
CR=CQ             …(iii)
DR=DS          ….(iv)
Adding, we get:
AP + BP + CR + DR = AS + BQ + CQ + DS
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence AB + CD = BC + AD.     Q.E.D.

Question 8.
If the sides of a parallelogram touch a circle (refer figure of Q/7) prove that the parallelogram is a rhombus.
Solution:
Given : The sides AB, BC, CD and DA of ||gm ABCD touches the circle at P, Q, R and S respectively.

To Prove : ABCD is a rhombus.
Proof : From A, AP and AS are the tangents to the circle.
∴ AP = AS    ….(i)

Question 9.
From the given figure, prove that:
AP + BQ + CR = BP + CQ + AR.

Solution:
Given: In the figure, sides of Δ ABC touch a circle at P, Q, R.
To Prove:
(i) AP + BQ + CR = BP + CQ + AR
(ii) AP + BQ + CR = \(\frac { 1 }{ 2 }\) Perimeter of Δ ABC.
Proof :
∵  From B, BQ and BP are the tangents to the circle.
∴ BQ = BP   ….(i)
Similarly we can prove that
AP= AR    ……… (ii)
and CR = CQ  …..(iii)
Adding we get
AP + BQ + CR = BP + CQ + AR ….(iv)
Adding AP + BQ + CR both sides,
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR+CR.
⇒  2 (AP + BQ + CR) = AB + BC + CA
∴ AP + BQ + CR = \(\frac { 1 }{ 2 }\) (AB + BC + CA)
= \(\frac { 1 }{ 2 }\) Perimeter of Δ ABC.            Q.E.D.

Question 10.
In the figure of Q.9 if AB = AC then prove that BQ = CQ.
Solution:
Given:  A circle touches the sides AB, BC, CA of Δ ABC at P, Q and R respectively, and AB = AC

Question 11.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centres if 
(i) they touch each other externally,
(ii) they touch each other internally.
Solution:
Radius of bigger circle = 6.3 cm

and raduis of smaller circle = 3.6 cm.
(i) Two circles touch each other at P externally. 0 and O’ are the centres of the circles.
Join OP and OP’
OP = 3.6 cm, O’P = 6.3 cm.
Adding we get
OO’ = OP + O’P = 3.6 + 6.3 = 9.9 cm
(ii) If the circles touch each other internally at P.
OP = 3.6 cm and O’P = 6.3 cm.

∴ OO’ = O’P – OP
= 6.3 – 3.6 = 2.7 cm

Question 12.
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP,
(ii) OP is the ⊥ bisector of chord AB.
Solution:
Given:  A circle with centre 0. A point P out side the circle. From P, PA and PB are the tangents to the circle, OP and AB are joined.

To prove:
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of chord AB.
Proof : In ∆ AOP and ∆ BOP,
AP = BP (Tangents from P to the circle.)
OP = OP (Common)
OA = OB (Radii of the same circle)
∴ ∆ AOP s ∆ BOP (SSS postulate)
∴∠AOP = ∠BOP (C.P.C.T.)
Now in ∆ OAM and ∆ OBM,
OA = OB (Radii of the same circle)
OM = OM (Common)
∠AOM = ∠BOM (Proved ∠AOP = ∠BOP)
∴ ∆ OAM = ∆ OBM (S.A.S. Postulate)
∴ AM = MB (C.P.C.T.)
and ∠OMA = ∠OMB (C.P.C.T.)
But ∠OMA + ∠OMB = 180° (Linear pair)
∴ ∠OMA = ∠OMB = 90°
Hence OM or OP is the perpendicular bisector of AB. Q.E.D.

Question 13.
In the given figure, two circles touch each other externally at point P, AB is the direct common tangent of these circles. Prove that:
(i) tangent at point P bisects AB.
(ii) Angle APB = 90°

Solution:
Given : Two circles with centre O and O’ touch each other at P externally. AB is the direct common tangent touching the circles at A and B respectively.

AP, BP are joined. TPT’ is the common tangent to the circles.
To Prove : (i) TPT’ bisects AB (ii) ∠APB = 90°
Proof :
∵ TA and TP arc the tangents to the circle
∴ TA = TP …(i)
Similarly TP. = TB ….(ii)
From (i) and (ii)
TA = TB
∴ TPT’ is the bisector of AB.
Now in ∆ ATP
TA = TP
∴ ∠TAP = ∠TPA
Similarly in A BTP.
∠TBP = ∠TPB
Adding we get.
∠TAP + ∠TBP = ∠APB
But ∠TAP + ∠TBP + ∠APB = 180°
∴ ∠APB = ∠TAP + ∠TBP = 90°. Q.E.D.

Question 14.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ
Solution:
Given: A circle with centre O. two tangents PA and QA are drawn from a point A out side the circle OP, OQ. OA and PQ are joined.

Question 15.
Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.
Solution:
Given: A circle with centre O, AP and BQ are two parallel tangents. A third tangent PQ intersect them at P and Q. PO and QO are joined

Question 16.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.

Calculate the value of x, the radius of the inscribed circle.
Solution:
In ∆ ABC, ∠B = 90°
OL ⊥ AB, OM ⊥ BC and
ON ⊥ AC.

Question 17.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
(i) ∠ QOR
(ii) ∠ QPR given that ∠ A = 60°.
Solution:

Question 18.
In the following figure, PQ and PR are tangents to the circle, with centre O. If ∠ QPR = 60°, calculate:
(i) ∠ QOR
(ii) ∠ OQR
(iii) ∠ QSR.

Solution:

Question 19.
In the given figure, AB is the diameter of the circle, with centre O, and AT is the tangent. Calculate the numerical value of x.

Solution:

Question 20.
In quadrilateral ABCD; angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle. |1990]
Solution:
BQ and BR arc the tangenls from B to the circle.

∴ BR = BQ = 27 cm.
∴ RC = 38-27 = 11 cm.
Since CR and CS are the tangents from C to the circle
∴ CS = CR= 11 cm.
∴ DS = 25 – 11 = 14 cm.
DS and DP are the tangents to the circle
∴ DS = DP
∴ ∠ PDS = 90° (given)
and OP ⊥ AD, OS ⊥ DC
∴ Radius = DS = 14 cm

Question 21.
In the given, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given ∠ SPR = x° and ∠ QRP = y°;
Prove that
(i) ∠ ORS = y°
(ii) Write an expression connecting x and y. [1992]

Solution:
∠ QRP = ∠ OSR = y (Angles in the alternate segment)

But OS = OR (radii of the same circle)
∴ ∠ ORS = ∠ OSR = y°
∴ OQ = OR (radii of the same circle)
∴ ∠ OQR = ∠ ORQ = 90° – y° ….(i) (OR ⊥ PT)
But in ∆ PQR,
Ext. ∠ OQR = x° + y° ….(ii)
from (i) and (ii)
x° + y° = 90° – y°
⇒ x° + 2y° = 90°

Question 22.
PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50°; calculate :

Solution:

Question 23.
In the given figure. O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC 80°; find the angle BAC. [1996]

Solution:

Join OC
∴ PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quad APCO,
∴ ∠ APC + ∠ AOC = 180°
⇒ 80° + ∠ AOC = 180°
∴ ∠ AOC = 180° – 80° = 100°
∠ BOC = 360° – (∠ AOB + ∠ AOC)
= 360° – (140° + 100°)
= 360° – 240° = 120°
Now arc BC subtends ∠ BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠ BAC = \(\frac { 1 }{ 2 }\)∠BOC
= \(\frac { 1 }{ 2 }\) x 120° = 60°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A  are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20G

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 20G.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20E
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G

Question 1.
What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter 12 cm ?
Solution:
Diameter of cone = 12cm
∴ Radius (r₁) = \(\frac { 12 }{ 2 }\)= 6cm
Height (A) = 45 cm
∴  Volume = \(\frac { 1 }{ 3 }\)πR²h
= \(\frac { 1 }{ 3 }\) x π x 6 x 6 x 45 cm³
= 540 π cm²
∴ Volume of solid spheres = 540 π cm³
Diameter of one sphere = 6cm
∴  Radius = \(\frac { 6 }{ 2 }\) = 3 cm
∴ Volume of one spherical ball

Question 2.
A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. (Answer correct to the nearest integer).
Solution:
Radius of cylinder = 7 cm
and height 14 cm
By carving a largest sphere from it,
the radius of the sphere will be = 7 cm

Question 3.
A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a hemi­spherical shape on the top. Find the number of cones  required.
Solution:
Diameter of cylinder = 12 cm
∴ Radius (r) = \(\frac { 12 }{ 2 }\) = 6 cm
Height (h) = 15 cm
Volume of ice-cream in cylinder = πr²h = π x 6 x 6 x 15 = 540π cm³
∴  Volume of ice-cream = 540π cm³
Now diameter of cone = 6 cm

Question 4.
A solid is in the form of a cone standing on a hemi-sphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find, in terms of π , the volume of the solid.
Solution:
Radius of each cone and hemi-sphere (r) = 8 cm
Height of cone (h) = r = 8 cm

Question 5.
The diameter of sphere is 6 It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Solution:
Diameter of a sphere = 6 cm.
∴ Radius = \(\frac { 6 }{ 2 }\) = 3 cm.

Question 6.
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
Solution:
Let edge of the cube = a
∴ Volume of cube = a x a x a = a³
The sphere, which exactly fits in the cube, has radius = \(\frac { a }{ 2 }\)

Question 7.
An iron pole consisting of a cylindrical por­tion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1cm³ of iron has 8 gm of mass (approx).

Solution:
Radius of the base of poles (r) = \(\frac { 12 }{ 2 }\) = 6cm.

P.Q.
When a metal cube is completely submerged in water contained in a cylindrical vessel with diameter 30 cm, the level of water rises by 1 \(\frac { 41 }{ 99 }\) cm. Find:
(i) the length of the edge of the cube.
(ii) the total surface area of the cube.
Solution:
Diameter of cylinderical vessel = 30 cm

Question 8.
In the following diagram a rectangular plat­form with a semi-circular end on one side is 22 metres long from one end to the other end. If the length of the half circumference is 11 metres, find the cost of constructing the platform, 1.5 metres high at the rate of Rs. 4 per cubic metres.

Solution:
Length of platform = 22m
Circumference of semicircle = 11m

Question 9.
The cross-section of a tunnel is a square of side 7 m surmounted by a semi-circle as shown in the following figure. The tunnel is 80 m long. Calculate :
(i) Its volume
(ii) The surface area of the tunnel (excluding the floor) and
(iii) its floor area.

Solution:
Side of square = 7m
Radius of semicircle = \(\frac { 7 }{ 2 }\) m
Length of tunnel = 80 m.

Question 10.
A cylindrical water tank of diameter 2.8 m and height 4.2 m is being fed by a pipe of diam­eter 7 cm through which water flows at the rate of 4 ms^-1. Calculate, in minutes, the time it takes to fill the tank.
Solution:
Diameter of cylindrical tank = 2.8 m 2.8
∴ Radius (r) = \(\frac { 2.8 }{ 2 }\) = 1.4 m
and height (h) = 4.2 m
Volume of water filled in it = πr²h

Question 11.
Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area 25 cm². If this water is collected into a rectangular cistern of dimensions 7.5 m by 5 m by 4 m; calculate the rise in level in the cistern in 1 hour 15 minutes.
Solution:
Rate of flow of water = 9km / hr
∴ Water flow in 1 hr 15 minutes i.e. in \(\frac { 5 }{ 4 }\)  hr

Question 12.
The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter 10 cm, and the other dimensions are as shown. Calculate :
(a) the total surface area,
(b) the total volume of the solid and
(c) the density of the material if its total weight is 1.7 kg.

Solution:
Diameter = 10 cm
∴  Radius (r) =  \(\frac { 10 }{ 2 }\)   = 5 cm

Question 13.
A solid, consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of cone is 4 cm. Give your answer to the nearest cubic centimetre. [1998]
Solution:
Radius of cylinder = 3 cm
and height = 6 cm
Radius of hemisphere = 2 cm
and height of cone = 4 cm

Volume of water in the cylinder when it is full
= πr²h = π x (3)² x 6 cm³ = 54π cm³
Volume of water displaced = Volume of cone + volume of hemisphere

Question 14.
A metal container in the form of a cylinder is surmounted by a hemi-sphere of the same radius. The internal height of the cylinder is 7 m and the internal radius is 3.5 m. Calculate
(i) he total area of the internal surface, exclud­ing the base;
(ii) the internal volume of the container in m³ [1999]
Solution:
Radius of cylinder = 3.5 m
and height = 7 m.
(i) Total surface area of container excluding the base = curved surface area of cylinder + area of hemisphere = 2πrh + 2πr²

Question 15.
An exhibition tent is in the form of a cylin­der surmounted by a cone. The height of the tent above the ground is 85 m and height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for fold and for stitching. Give your answer to the nearest m².     [2001]
Solution:
Total height of the tent = 85 m.
Daimeter of the base = 168 m.
∴ Radius (r) = \(\frac { 168 }{ 2 }\) = 84 m
Height of cylindrical part = 50 m
Then height of conical part (h) = 85 – 50 = 35 m

P.Q.
The total surface area of a hollow cylinder, which is open from both sides is 3575 cm²; area of the base ring is 357.5 cm² and height is 14 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder = 3575 cm²
Area of base ring = 357.5 cm²
Height = 14 cm.
Let external radius = R
and internal radius = r
and let thickness of cylinder = (R – r) = d
∴  Total surface area = 2πRh + 2πrh + 2π (R²– r²)
= 2πh (R + r) + 2π (R + r) (R – r)
= 2π (R + r) [h + R – r]
= 2π (R + r) (h + d)
= 2π (R + r) (14 + d)
But 2π (R + r) (14 + d) = 3575                  ….(i)
and area of the base = π (R² – r²) = 357.5
⇒  π (R + r) ( R – r) = 357.5
⇒ π (R + r)d = 357.5                                 ….(ii)

Question 16.
A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube  is \(\frac { 5159 }{ 6 }\)cm³, and \(\frac { 4235 }{ 6 }\) cm³ of water is required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.
Solution:

Question 17.
A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone, which exactly coincides with hemisphere, is 7cm and its height is 8cm. The solid is placed in a cylindrical vessel of internal radius 7 cm and height 10cm. How much water, in cm³, will be required to fill the vessel completely.
Solution:
Diameter of hemisphere = 7cm
Diameter of the base of a cone = 7cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm = 3.5 cm
Height (h) = 8cm.

Question 18.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.   (2015)
Solution:
Radius of first sphere = 2 cm

Question 19.
A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.             (2016)
Solution:
Let the number of cones be n,
Let radius of the sphere be r_s, radius of a cone be r_c and h be the height of the cone.
Volume of sphere = n(Volume of a metallic cone)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

Question 1.
In the given circle with diameter AB, find the value of x. (2003)

Solution:

∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a ∆)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180°
⇒ x + 120° = 180°
∴ x = 180°- 120° = 60°

Question 2.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.

Solution:

Radius of the circle whose centre is O = 5 cm
OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.
Join OA and OC, then OA = OC=5cm.
∴ OP ⊥ AB
∴ P is the mid-point of AB.
Similarly, Q is the mid-point of CD.
In right ∆OAP,
OA² = OP² + AP² (Pythagorous theoram)
⇒ (5)² =OP² +(4)² ( ∵ AP = AB = \(\frac { 1 }{ 2 }\) x 8 = 4cm)
⇒ 25 = OP² + 16
⇒ OP³ = 25 – 16 = 9 = (3)²
∵ OP = 3 cm
Similarly, in right ∆ OCQ,
OC² = OQ² + CQ²
⇒ (5)² =OQ²+(3)² (∵ CQ = CD = \(\frac { 1 }{ 2 }\) x 6 = 3cm)
⇒ 25 = OQ² + 9
⇒ OQ³ = 25 – 9 = 16 = (4)²
∴ OQ = 4 cm
Hence, PQ = OP + OQ = 3-4 = 7 cm.

Question 3.
The given figure shows two circles with centres A and B; and radii 5cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Solution:

Join AP and produce AB to meet the bigger circle at C.
AB = AC – BC = 5 cm – 3 cm = 2 cm.
But, M is the mid-point of AB
∴ AM = \(\frac { 2 }{ 2 }\) = 1cm.
Now in right ∆APM,
AP2 = MP2 + AM2 (Pythagorous theorem)
⇒ (5)² = MP² -1- (1 )²
⇒ 25 = MP: + 1
⇒ MP: = 25 – 1 = 24

Question 4.
In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

Solution:

Given: In the figure ABC is a triangle in winch ∠A = 30°
To Prove: BC is the radius of circumcircle of ∆ABC whose O is the centre.
Const: Join OB and OC.
Proof: ∠BOC is at the centre and ∠BAC is at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC = 2 x 30° = 60°
Now in ∆OBC,
OB = OC (Radii of the same circle)
∴ ∠OBC = ∠OCB
But ∠OBC + ∠OCB + ∠BOC – 180°
∠OBC + ∠OBC + 60°- 180°
⇒ 2 ∠OBC = 180°- 60° = 120°
⇒ ∠OBC = \(\frac { { 120 }^{ circ } }{ 2 }\) = 60°
∴ ∆OBC is an equilateral triangle.
∴ BC = OB = OC
But OB and OC are the radii of the circumcircle
∴ BC is also the radius of the circumcircle.

Question 5.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:

Given: In ∆ABC, AB-AC and a circle with AB as diameter is drawn which intersects the side BC and D.
To Prove: D is the mid point of BC.
Const: Join AD.
Proof: ∠ 1 = 90° (Angle in a semi-circle)
But ∠ 1 + ∠ 2 – 180° (Linear pair)
∴ ∠ 2 = 90°
Now, in right ∆ ABD and ∆ ACD,
Hyp. AB – Hyp. AC (Given)
Side AD – AD (Common)
∴ ∆ABD = ∆ACD (RHS criterion of congruency)
∴ BD = DC (c.p.c.t.)
Hence D is he mid point of BC. Q.E.D.

Question 6.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Solution:
Join OE,
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∴ ∠EOC = 2 ∠EBC = 2 x 65° = 130°
Now, in ∆OEC, OE = OC (radii of the same circle)
∴ ∠OEC = ∠OCE
But ∠ OEC + ∠ OCE + ∠ EOC = 180°
⇒ ∠ OCE + ∠ OCE + ∠ EOC = 180°

Question 7.
Chords AB and CD of a circle intersect each other at point P, such that AP = CP. S.how that AB = CD.

Solution:
Given: Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP.

To Prove: AB = CD.
Proof: ∵ Two chords AB and CD intersect each other inside the circle at P.
∴ AP x PB = CP x PD ⇒ \(\frac { AP}{ CP }\) = \(\frac { AD }{ PB }\)
But AP = CP ….(i) (given)
∴ PD = PB or PB = PD …,(ii)
Adding (i) and (ii),
AP + PB = CP + PD
⇒ AB = CD Q.E.D.

Question 8.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Given: ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.
To Prove: PRQS is a cyclic quadrilateral.
Proof: In ∆APD,

∠1 +∠2 + ∠P= 180° ,…(i)
Similarly, in ∆BQC.
∠3 + ∠4 + ∠Q= 180° ….(ii)
Adding (i) and (ii), we get:
∠1 + ∠2 + ∠P + ∠3 + ∠4 ∠Q = 180° + 180° = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠Q = 360° ..(iii)
But ∠1+∠2+∠3+∠4 = \(\frac { 1 }{ 2 }\) (∠A+∠B+∠C+∠D)
= \(\frac { 1 }{ 2 }\) x 360°= 180°
∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)]
But these are the sum of opposite angles of quadrilateral PRQS
∴ Quad. PRQS is a cyclic quadrilateral. Q.E.D.

Question 9.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(I) ∠BDC
(ii) ∠BEC
(iii) ∠BAC (2014)

Solution:
∠DBC = 58°
BD is diameter
∴ ∠DCB=90° (Angle in semi circle)

(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°
(ii) ∠BEC =180°-32°
(opp. angle of cyclic quadrilateral)
= 148°
(iii) ∠BAC = ∠BDC = 32°
(Angles in same segment)

Question 10.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Solution:

Given: In ∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE, DE is joined.
To Prove: B,C,E,D. are concyclic.
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
Similarly in ∆ADE, AD = AE (given)
∴ ∠ADE = ∠AED
In ∆ABC,
∵ \(\frac { AP }{ AB }\) = \(\frac { AE }{ AC }\)
∴ DE || BC.
∴ ∠ADE = ∠B (Corresponding angles)
But ∠B = ∠C (Proved)
∴ Ext. ∠ADE = its interior opposite ∠C.
∴ BCED is a cyclic quadrilateral.
Hence B,C, E and D are concyclic.

Question 11.
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE’= 20°; determine ∠BCD. Give reason in support of your answer.

Solution:

In cyclic quad. ABCD,
AF || CB and DA is produced to E such that
∠ADC = 92° and ∠FAE – 20°
Now, we have to find the measure of ∠BCD In cyclic quad. ABCD,
∠B – ∠D = 180° ⇒ ∠B + 92° = 180″
⇒∠B = 180°-92° = 88°
∵ AF || CB.
∴∠FAB = ∠B = 88°
But ∠FAE – 20° (Given)
Ext. ∠BAE – ∠BAF + ∠FAE
= 88° + 20° = 108°
But Ext. ∠BAE – ∠BCD
∴ ∠BCD = 108°

Question 12.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC – 80°, calculate
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC.

Solution:

Join DB and DC, IB and IC.
∠BAC = 66°, ∠ABC = 80°. I is the incentre of the ∆ABC.
(i) ∵ ∠DBC and ∠DAC are in the same segment
∴ ∠DBC – ∠DAC.
But ∠DAC = \(\frac { 1 }{ 2 }\)∠BAC = \(\frac { 1 }{ 2 }\) x 66° = 33°
∴ ∠DBC = 33°.
(ii) ∵ I is the incentre of ∆ABC.
∴ IB bisect ∠ABC
∴ ∠ IBC = \(\frac { 1 }{ 2 }\) ∠ABC = \(\frac { 1 }{ 2 }\) x 80° = 40°.
(iii) ∴∠BAC = 66° ∠ABC = 80°
∴ In ∆ABC,
∠ACB = 180° – (∠ABC + ∠CAB)
= 180°-(80°+ 66°)= 180°- 156° = 34°
∵ IC bisects the ∠C
∴ ∠ ICB = \(\frac { 1 }{ 2 }\) ∠C = \(\frac { 1 }{ 2 }\) x 34° = 17°.
Now in ∆IBC,
∠ IBC + ∠ ICB + ∠ BIC = 180°
⇒ 40° + 17° + ∠BIC = 180°
⇒ ∠ BIC = 180° – (40° + 17°) = 180° – 57°
= 123°

Question 13.
In the given figure, AB = AD = DC= PB and ∠DBC = x°. Determine in terms of x :
(i) ∠ABD
(ii) ∠APB

Hence or otherwise prove that AP is parallel to DB.
Solution:
Given: In figure, AB = AD = DC = PB.
∠DBC = x. Join AC and BD.
To Find : the measure of ∠ABD and ∠APB.
Proof: ∠DAC=∠DBC= x(angles in the same segment)
But ∠DCA = ∠DAC (∵ AD = DC)
= x
But ∠ABD = ∠DAC (Angles in the same segment)
In ∆ABP, ext. ∠ABC = ∠BAP + ∠APB
But ∠ BAP = ∠APB (∵ AB = BP)
2 x x = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2x
⇒ ∠APB = x
∵ ∠APB = ∠DBC = x
But these are corresponding angles
∴ AP || DB. Q.E.D.

Question 14.
In the given figure, ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

Solution:
Given: In the figure, ABC, AEQ and CEP are straight lines.
To prove: ∠APE + ∠CQE = 180°.
Const: Join EB.
Proof: In cyclic quad. ABEP,
∠APE+ ∠ABE= 180° ….(i)
Similarly in cyclic quad. BCQE

∠CQE +∠CBE = 180° ….(ii)
Adding (i) and (ii),
∠APE +∠ABE + ∠CQE +∠CBE = 180° + 180° = 360°
⇒ ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°
But ∠ABE + ∠CBE = 180° (Linear pair)
∴ ∠APE + ∠CQE + 180° = 360°
⇒ ∠APE + ∠CQE = 360° – 180° = 180°
Hence ∠APE and ∠CQE are supplementary. Q.E.D.

Question 15.
In the given figure, AB is the diameter of the circle with centre O.

If ∠ADC = 32°, find angle BOC.
Solution:
Arc AC subtends ∠AOC at the centre and ∠ ADC at the remaining part of the circle

∴ ∠AOC = 2 ∠ADC
= 2 x 32° = 64°
∵ ∠AOC + ∠ BOC = 180° (Linear pair)
⇒ 64° + ∠ BOC = 180°
⇒ ∠ BOC=180° – 64° =116° .

Question 16.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.
If ∠ A : ∠ B = 2 : 1 ; find angles A and B.
Solution:
PQRS is a cyclic-quadrilateral in which ∠ PQR =135°
Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.

Question 17.
In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA. If AC produced and BD produced meet at point P ; show that ∠ APB = 60°.

Solution:
Given : In the figure, AB is the diameter of the circle with centre O.
CD is the chord with length equal to the radius OA.
AC and BD are produced to meet at P.
To prove : ∠ APB = 60°
Const : Join OC and OD
Proof : ∵ CD = OC = OD (Given)
∴ ∆OCD is an equilateral triangle
∴ ∠ OCD = ∠ ODC = ∠ COD = 60°
In ∆ AOC, OA = OC (Radii of the same circle)
∴ ∠ A = ∠ 1
Similarly, in ∆ BOD,
OB = OD
∴∠2= ∠B
Now in cyclic quadrilateral ACDB,
∠ A CD + ∠B = 180°

∠60°+ ∠ 1 + ∠B = 180°
= ∠ 1 + ∠B = 180° – 60°
⇒∠ 1 + ∠B = 120°
But ∠ 1 = ∠ A
∴ ∠ A + ∠B = 120° …(i)
Now, in ∆ APB,
∠ P + ∠ A + ∠ B = 180° (Sum of angles of a triangle)
⇒ ∠P+120°=180° [From (i)]
⇒ ∠P = 180°- 120°= 60°
Hence ∠ P = 60° or ∠ APB = 60° Hence proved.

Question 18.
In the following figure,

ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that :
(i) EF = FC
(ii) BF = DF
Solution:

Given : ABCD is a cyclic quadrilateral in which AD || BC.
Bisector of ∠ A meets BC at E and the given circle at F. DF and BF are joined.
To prove :
(i) EF = FC
(ii) BF = DF
Proof : ∵ ABCD is a cyclic -quadrilateral and AD || BC
∵ AF is the bisector of ∠ A
∴ ∠ BAF = ∠ DAF
∴ Arc BF = Arc DF (equal arcs subtends equal angles)
⇒ BF = DF(equal arcs have equal chords)
Hence proved

Question 19.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F. If ∠ DCF : ∠ F : ∠ E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Given : In a circle, ABCD is a cyclic quadrilateral AB and DC are produce to meet at E and BC and AD are produced to meet at F.

Question 20.
The following figure shows a circle with PR as its diameter.

If PQ = 7 cm, QR = 3 cm, RS = 6 cm. find the perimeter of the cyclic quadrilateral PQRS. (1992)
Solution:
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm,
QR = 3 RS = 6cm
∴ 3 RS = 6cm
and RS = \(\frac { 6 }{ 3 }\) = 2cm
Now in ∆ PQR,
∠ Q = 90° (Angle in a semi-circle)
∴ PR² = PQ² + QR² (Pythagoras theorem)
= (7)² + (6)² = 49 + 36 = 85
Again, in right ∆ PSQ, PR² = PS² + RS²
⇒ 85 = PS² + (2)²
⇒ 85 = PS² + 4
⇒ PS² = 85 – 4 = 81 = (9)²
∴ PS = 9cm
Now, perimeter of quad. PQRS = PQ + QR + RS + SP = (7 + 9 + 2 + 6) cm = 24cm

Question 21.
In the following figure, AB is the diameter of a circle with centre O.

If chord AC = chord AD, prove that :
(i)arc BC = arc DB
(ii) AB is bisector of ∠ CAD. Further, if the length of arc AC is twice the length of arc BC, find :
(a) ∠ BAC
(b) ∠ ABC
Solution:
Given : In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.
To prove : (i) arc BC = arc DB
(ii) AB is the bisector of ∠CAD
(iii) If arc AC = 2 arc BC, then find
(a) ∠BAC (b) ∠ABC

Construction: Join BC and BD.
Proof : In right angled A ABC and A ABD
Side AC = AD (Given)
Hyp. AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (R.H.S. axiom)
(i) ∴ BC = BD (C.P.C.T)
∴ Arc BC = Arc BD (equal chords have equal arcs)
(ii) ∠ BAC = ∠ BAD (C.P.C.T)
∴ AB is the bisector of ∠CAD.
(iii) If arc AC = 2 arc B
Then ∠ABC = 2 ∠BAC
But ∠ABC – 2 ∠BAC = 90°
∴ 2 ∠BAC + ∠BAC = 90°
⇒ 3 ∠BAC = 90° ⇒ ∠BAC = 30°
and ∠ABC = 2 ∠BAC = 2 * 30° = 60°

Question 22.
In cyclic-quadrilateral ABCD ; AD = BC,
∠ BAC = 30° and ∠ CBD = 70°, find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ ADC
Solution:
ABCD is a cyclic-quadrilateral and AD = BC
∠ BAC = 30°, ∠ CBD = 70°
∠ DAC = ∠ CBD , (Angles in the same segment)

But ∠ CBD = 70°
∴ ∠ DAC = 70°
⇒ ∠ BAD = ∠ BAC + ∠ DAC = 30° + 70° = 100°
But ∠ BAD + ∠ BCD = 180°
(Sum of opposite angles of a cyclic quad.)
⇒100°+ ∠ BCD=180° ⇒ ∠ BCD=180° – 100° = 80°
∴ ∠ BCD = 80°
∵ AD = BC (Given)
∴ ∠ ACD = ∠ BDC
(Equal chords subtends equal angles)
But ∠ ACB = ∠ ADB
(Angles in the same segment)
∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB
⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 80°)
∴ ∠ ADC = 80°
But in ∆ BCD,
∠ CBD + ∠ BCD + ∠ BDC = 180° (Angles of a triangle)
⇒ 70° + 80° + ∠ BDC = ∠ 180°
⇒ 150°+ ∠ BDC = 180°
∴ ∠ BDC = 180° – 150° = 30°
⇒ ∠ ACD = 30° (∵ ∠ ACD = ∠ BDC)
∴ ∠ BCA = ∠ BCD – ∠ ACD = 80° – 30° = 50°
∠ ADC + ∠ABC = 180°
(Sum of opp. angles of a cyclic quadrilateral)
⇒ 80°+ABC = 180°
⇒ ∠ ABC=180°-80° = 100°

Question 23.
In the given figure, if ∠ ACE = 43° and ∠CAF = 62°. Find the values of a, b and c.

Solution:
Now, ∠ ACE = 43° and ∠ CAF = 62° (given)
In ∆ AEC,

∠ ACE + ∠ CAE + ∠ AEC = 180°
∴ 43° + 62° + ∠ AEC = 180°
105° + ∠ AEC = 180°
⇒ ∠ AEC = 180°- 105° = 75°
Now, ∠ ABD + ∠AED=180°
(Opposite ∠ s of a cyclic quad, and ∠ AED = ∠ AEC)
⇒ 0 + 75°= 180° a = 180° – 75° = 105°
∠ EDF = ∠ BAE (Angles in the alternate segments)
∴ c = 62°
In ∆BAF, ∠a + 62° + ∠b = 180°
⇒ 105°+ 62°+ ∠b= 180°
⇒ 167° + ∠6 = 180°
⇒ ∠b= 180°-167°= 13°
Hence, a= 105°, 6=13° and c = 62°.

Question 24.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25° .

Find:
(i) ∠CAD
(ii)∠CBD
(iii) ∠ADC
Solution:

In the given figure,
ABCD is a cyclic quad, in which AB || DC
∴ ABCD is an isosceles trapezium AD = BC
(i) Join BD
and Ext. ∠BCE = ∠BAD
{ Ext. angle of a cyclic quad, is equal to interior opposite angle}
∴ ∠BAD = 80° (∵ ∠BCE = 80°)
But ∠BAC = 25°
∴ ∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°
(ii) ∠CBD = ∠CAD (Angles in the same segment)
= 55°
(iii) ∠ADC = ∠BCD (Angles of the isosceles trapezium)
= 180°- ∠BCE =180°- 80° = 100°

Question 25.
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AJD and BC produced meet at P, show that ∠APB = 60°.
Solution:

Given : In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle
To prove: ∠APB = 60°
Construction : Join OC and OD
Proof: ∵ chord CD = CO = DO (radii of the circle)
∴ ∆DOC is an equilateral triangle
∠DOC = ∠ODC = ∠OCD – 60°
Let ∠A = x and ∠B = y
∵ OA = OD = OC = OB (radii of the same circle)
∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC =y
∴∠AOD = 180° – 2x and ∠BOC = 180° – 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°
180°- 2x + 180° -2y + 60° = 180°
⇒ 2x + 2y = 240°
⇒ x + y = 120°
But ∠A + ∠B + ∠P = 180° (Angles of a triangle)
⇒ 120° + ∠P = 180°
⇒ ∠P = 180° – 120° = 60°
Hence ∠APB = 60°

Question 26.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.

Solution:
Given : In the figure,
CP is the bisector of ∠ACB
To prove : DP is the bisector of ∠ADB
Proof: ∵ CP is the bisector of
∴ ∠ACB ∠ACP = ∠BCP
But ∠ACP = ∠ADP {Angles in the same segment of the circle}
and ∠BCP = ∠BDP
But ∠ACP = ∠BCP
∴ ∠ADP = ∠BDP
∴ DP is the bisector of ∠ADB

Question 27.
In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ADB

Solution:
In the figure,
ABCD is a cyclic quadrilateral
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
∠CAD = ∠CBD = 70°
(Angles in the same segment)
Similarly ∠BAC = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
(i) Now ∠BCD + ∠BAD = 180° (opposite angles of cyclic quad.)
⇒ ∠BCD + 100°= 180°
⇒ ∠BCD = 180°-100° = 80°
(ii) ∵ AD = BC (given)
∴ ∠ABCD is an isosceles trapezium
and AB || DC
∴ ∠BAC = ∠DCA (alternate angles)
⇒ ∠DCA = 30°
∠ABD = ∠D AC = 30° (Angles in the same segment)
∴ ∠BCA = ∠BCD – ∠DAC = 80° – 30° = 50°
(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°
(iv) ∠ADB = ∠BCA = 50° (Angles in the same segment)

P.Q.
In the given below figure AB and CD are parallel chords and O is the centre.
If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Solution:
Given : AB = 24 cm, CD = 18 cm
⇒AM = 12 cm, CN = 9 cm
Also, OA = OC = 15 cm
Let MO = y cm, and ON = x cm
In right angled ∆AMO

(OA)² = (AM)² + (OM)²
(15)² = (12)² + (y²
⇒ (15)²-(12)²
⇒ y² = 225-144
⇒ y² = 81 = 9 cm
In right angled ∆CON
(OC)² = (ON)² + (CN)²
⇒(15 )²= x2 + (9)²
⇒ x² = 225-81
⇒x²= 144
⇒ x = 12 cm
Now, MN = MO + ON =y + x = 9 cm + 12 cm = 21 cm

Question 28.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:

(i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.
∴ ∠ODB = ∠CBD = 32° (Alternate angles)
In ∆OBD,
OD = OB (Radii of the same circle)
⇒ ∠ODB = ∠OBD = 32°
(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.
∴∠AOB = ∠OBC (Alternate angles)
∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
∴ ∠AOB = 64°
(iii) In AOAB,
OA = OB(Radii of the same circle)
∴ ∠OAB = ∠OBA = x (say)
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180° – 64°
⇒ 2x= 116°
⇒ x = 58°
∴ ∠OAB = 58°
That is ∠DAB = 58°
∴ ∠DAB = ∠BED = 58°
(Angles inscribed in the same arc are equal)

Question 29.
In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.

(i) Prove ∆TPS ~ ∆TRQ
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.(2016)
Solution:
(i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180°
(Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP …(i)
∠RQT + ∠RQP = 180°
(Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT …(ii)
From (i) and (ii),
180° – ∠RSP = 180° – ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In ∆TPS and ∆TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ ATPS ~ ATRQ (AA similarity criterion)
(ii) Since ∆TPS ~ ∆TRQ implies that corresponding sides are proportional that

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20F

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21F

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20E
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G

Question 1.
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find the volume of the ramaining solid.
Solution:
Height of the cylinder (h) = 10 cm
and radius of base (r) = 6 cm.
∴ Volume of cylinder = πr²h
Height of cone = 10 cm
and radius of base of cone = 6 cm

Question 2.
From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out Find the volume and total surface area of the remaining solid.
Solution:
Radius of solid cylinder (R) = 12 cm
and height (H) = 16 cm.

Question 3.
A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at Rs. 15 per metre if the width is 1.5 m.
Solution:
Radius of the cylindrical part of tent (r) = \(\frac { 105 }{ 2 }\)m

Question 4.
A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:
(i) total surface area .of the tent,
(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.
Solution:
Total height = 13 m
Diameter of base of the tent = 24 m
∴ Radius (r) = \(\frac { 24 }{ 2 }\) = 12 m
Height of cylindrical part h₁ = 8 m
and height of conical part (h₂) = 13 – 8 = 5 m

Question 5.
A cylindrical boiler, 2 m high, is 3.5 m in diameter. It has a hemi-spherical lid. Find the volume of its interior, including the part covered by the lid.
Solution:
Diameter of cylinderical boiler = 3.5 m

Question 6.
A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylindrical part is 4 \(\frac { 2 }{ 3 }\) m and the diameter of hemisphere is 3.5 m. Calculate the capacity and the internal surface area of the vessel.
Solution:

Question 7.
A wooden toy is in the shape of a cone mounted on a cylinder as shown alongside.

If the height of the cone is 24 cm, the total height of the toy is 60 cm and the radius of the base of the cone = twice the radius of the base of the cylinder = 10 cm ; find the total surface area of the toy.  [Take π = 3.14]
Solution:
Height of the conical part (h₁)= 24 cm
total height of the toy = 60 cm
∴ Height of cylinderical part (h) = 60-24 = 36 cm
Radius of the cone (r) = twice the radius of the cylinder = 10 cm
∴ Radius of cylinder (r₁) = 5 cm

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to sub­merge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:
Diameter of cylinderical container = 42cm
∴  Radius (r) = \(\frac { 42 }{ 2 }\) = 21 cm.
Dimension of a rectangular solid = 22 cm x 14cm x 10.5 cm
∴ Volume of solid
= 22 x 14 x 10.5 cm³        ….(i)
Let the height of water = h
∴ Volume of water in the container

Question 9.
Spherical marbles of diameter 1.4 cm are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.
Solution:
Diameter of spherical marble = 1.4 cm.
∴ Radius = \(\frac { 1.4 }{ 2 }\) = 0.7 cm.
Volume of one ball = \(\frac { 4 }{ 3 }\) πr³

Question 10.
The cross-section of a railway tunnel is a rectangle 6 m broad and 8 m high surmounted by a semi-circle as shown in the figure. The tun­nel is 35 m long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of Rs. 2.25 per m².

Solution:
Breadth of’tunnel = 6 m
and height = 8m
Length of tunnel = 35 m
Radius of semicircle = \(\frac { 6 }{ 2 }\) = 3 m.
Circumference of semicircle = πr = \(\frac { 22 }{ 7 }\) x 3 = \(\frac { 66 }{ 7 }\) m
∴ Internal surface area of the tunnel

Question 11.
The horizontal cross-section of a water tank is in the shape of a rectangle with semi-circle at one end, as shown in the following figure. The water is 2.4 metres deep in the tank. Calculate the volume of water in the tank in gallons.


(Given : 1 gallon = 4.5 litres)
Solution:
Length = 21m
breadth = 7 m
Depth of water = 2.4 m
∴ Radius of semicircle = \(\frac { 7 }{ 2 }\) m.
Area of the cross section = Area of rectangle + area of semicircle

Question 12.
The given figure shows the cross-section of a water channel consisting of a rectangle and a semi-circle. Assuming that the channel is always full, find the volume of water discharged through it in one minute if water is flowing at the rate of 20 cm per second. Give your answer in cubic metres correct to one place of decimal.

Solution:
Rate of water flow = 20 cm/sec.
Period = 1 min. = 60 sec.
Radius of semi-circular part (r) = \(\frac { 21 }{ 2 }\)  cm
Height of channel (h) = 7 cm
Length of channel = 20 x 60 = 1200 cm

Question 13.
An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is 3 \(\frac { 1 }{ 2 }\) cm and height 8 cm. Find the volume of water required to fill the vessel.
If this cone is replaced by another cone, whose height is 1 \(\frac { 3 }{ 4 }\) cm and the radius of whose base is 2 cm, find the drop in the water level.  [1993]
Solution:
Diameter of cylinder = 7 cm
∴ Radius (R) = \(\frac { 7 }{ 2 }\) cm
Height (h) = 8 cm

Question 14.
A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can. [1997]
Solution:
Radius of the cylindrical can = 3.5 cm
∴  Radius of the sphere which fits in it

Question 15.
A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level’of water, when it is inverted upside down, common diameter is 7 cm and height of the cylinder is 20  cm.
Solution:
(i) Diameter of the cylinder = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E

Question 1.
Show that:
(i) tan 10° tan 15° tan 75° tan 80° = 1
(ii) sin 42° sec 48°+cos 42° cosec 48°= 2

Solution:
(i) tan 10° tan 15° tan 75° tan 80°= 1
L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan (90° – 80°) tan (90° – 75°) tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80°
= tan 80° cot 80° x tan 75° cot 75°
= 1 x 1 = 1= R.H.S. (∵ tan A cot A = 1)
(ii) sin 42° sec 48°+ cos 42° cosec 48°= 2
L.H.S. = sin 42° sec 48°+ cos 42° cosec 48°
= sin 42° sec (90° – 42°) + cos 42° cosec (90° – 42°)
= sin 42° cosec 42°+ cos 42° sec 42°
=1 + 1=2 R.H.S. (∵ sin A cosec A=1, cos A sec A=1)

Question 2.
Express each of the following in terms of angles between 0°and 45°.
(i) sin 59° + tan 63°
(ii) cosec 68° + cot 72°
(iii) cos 74° + sec 67°
Solution:
(i) sin 59° + tan 63°
= sin (90° – 31°) + tan (90° – 27°)
= cos 31°+ cot 27°
(ii) cosec 68° + cot 72°
= cosec (90° – 22°) + cot (90° – 18°)
= sec 22°+ tan 18°
(iii) cos 74°+ sec 67°
= cos (90° – 16°) + sec (90° – 23°)
= sin 16°+ cosec 23°

Question 3.
Show that:

Solution:

Question 4.
For triangle ABC, Show that:

Solution:

Question 5.
Evaluate:

Solution:

Question 6.

Solution:

Question 7.
Find (in each case, given below) the value of x, if:
(i) sin x = sin 60° cos 30° – cos 60° sin 30°
(ii) sin x = sin 60° cos 30° + cos 60° sin 30°
(iii) cos x = cos 60° cos 30° – sin 60° sin 30°

(v) sin 2x = 2 sin 45° cos 45° 
(vi) sin 3x = 2 sin 30° cos 30°
(vii) cos (2x – 6°) = cos² 30° – cos² 60°
Solution:

Question 8.
In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
(i) sin (90° – 3A). cosec 42° = 1
(ii) cos (90° – A). sec 77° = 1
Solution:

Question 9.
Prove that:

Solution:

Question 10.
Evaluate:

Solution:

Question 11.
Without using trigonometric tables, evaluate sin² 34° + sin² 56° + 2 tan 18° tan 72° – cot² 30°.
Solution:

Question 12.
Without using trigonometrical tables, evaluate: cosec² 57° – tan² 33° + cos 44° cosec 46° – \( \sqrt{2} \) cos45°- tan² 60°
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E

Question 1.
Prove that: 

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Question 2.
If xcosA + ysinA = m and xsinA-ycosA = n, then prove that: x²+y² = m² + n²
Solution:
x cos A + y sin A = m    …(i)
x sin A – y cos A = n     ….(ii)
squaring (i) and (ii)
x² cos² A + y² sin² A + 2 xy cosA sinA = m²
x² sin² A + y² cos² A – 2 xy cos A sin A = n²
Adding we get,
x² (sin² A + cos² A) + y² (sin² A + cos² A) = m²+n²
∴ x²+y² = m² + n²(∵ sin²A + cos²A= 1)
Hence proved.

Question 3.
If m = a sec A +b tan A and n=atanA + bsecA, then prove that: m²-n² = a²-b²
Solution:
m = asec A + btan A         ……(i)
n = a tan A + b sec A       …..(ii)
squaring (i) and (ii)
m² = a² sec² A + b² tan² A + 2ab sec A tan A
n² = a² tan² A + b² sec² A + 2 ab tan A sec A
Subtracting, we get
m² – n² = a² (sec² A – tan² A) + b² (tan² A – sec² A)
= a²x 1 +b²(-1) = a²-b² ( ∵ sec²A-tan²A= 1)  .
Henceproved

Question 4.
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that: x² + y² + z² = r.
Solution:
x = r sin A cos B      ….(i)
y = r sin A sin B      ….(ii)
z = r cosA               …….(iii)
Squaring, (i), (ii) & (iii)
x²=r² sin² A cos² B,
y² = r²sin²Asin²B,
z² = r²cos²A
Adding, we get,
x²+y² + z²=r² (sin²A cos²E + sin² A sin² B+cos²A)
= r[sin²A (cos² B + sin²B) + cos²A]
= r [sin² A x 1 + cos² A]
= r² [sin² A + cos² A] = r² x 1  = r²        ( ∵ sin² A + cos² A = 1)
Hence proved.

Question 5.
If sin A + cos A = m and sec A + cosec A=n, show that n (m²-1) = 2m
Solution:

Question 6.
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that x² + y² + z² = r²
Solution:
x = r cosAcosB              ….(i)
y = r cosAsinB             ….(ii)
z = r sinA                 ….(iii)
Squaring (i), (ii), (iii)
x² = r² cos² A cos² B, y² = r² cos² A sin²B
z² = r²sin²A
Adding, we get
x² + y² + z² = r² (cos² A cos²B + cos² A sin² B + sin² A)
= r² [cos² A (cos² B + sin²B) + sin² A]
= r²[cos²Ax 1+sin²A]
= r² (1) = r^{2    `}Hence proved.

Question 7.

Solution:

P.Q.

Solution:

P.Q.
Evaluate:

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E

Prove the following Identities :
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
sin⁴A – cos⁴ A = 2 sin²A-1 
Solution:
L.H.S. = sin⁴ A – cos⁴A = (sin²A)²-(cos²A)²
= (sin²A + cos²A) (sin²A – cos²A)     [(a² – b² = (a + b) (a – b)]
= 1 (sin² A – cos²A) [∵ sin²A + cos²A = 1]
= sin² A – (1- sin²A) (∵ cos²A = 1 – sin²A)
= sin² A – 1 + sin² A
= 2 sin²A-1 = R.H.S.

Question 6.
(1 – tan A)² + (1 + tanA)² = 2sec²A
Solution:
LHS = (1 -tanA)² + (1 +tanA)²
= 1 + tan² A- 2 tan A + 1 + tan² A + 2 tanA
= 2 + 2 tan² A = 2 (1+tan²A)
= 2 sec²A (∵ l+tan²A=sec²A)
= R.H.S.

Question 7.
Cosec⁴ A – cosec² A = cot⁴ A + cot² A
Solution:
L.H.S. = cosec⁴ A -cosec² A
= (cosec²A)² – cosec²A
= (1 + cot²A)² – (1 + cot²A)
= 1 + cot⁴ A + 2 cot²A – 1- cot²A
= cot⁴ A + cot² A = R.H.S.

Question 8.
sec A (1-sin A) (sec A + tan A) = 1
Solution:

Question 9.
cosec A (1 + cos A) (cosec A – cot A) = 1
Solution:

Question 10.
sec² A + cosec²A = sec² A cosec² A
Solution:

Question 11.

Solution:

Question 12.
tan²A – sin²A = tan² A. sin² A
Solution:

Question 13.
cot² A – cos² A = cos² A. cot² A
Solution:

Question 14.
(cosecA + sinA) (cosec A – sinA) = cot² A + cos²A
Solution:
L.H.S. = (cosec A + sin A) (cosec A – sin A)
= (cosec²A – sin² A) [∵ (a + b) (a – b) = a²– b²]
= 1 + cot² A – sin² A = cot² A + 1 – sin²A
= cot² A + cos² A (∵ 1-sin²A = cos² A)
= R.H.S.

Question 15.
(sec A – cosA) (sec A + cosA) = sin² A + tan²A
Solution:
L.H.S. = (sec A-cos A) (sec A + cos A)
= sec² A – cos² A
= 1 + tan²A-cos² A
= 1-cos² A + tan² A
= sin² A + tan² A  (∵ 1- cos²A=sin²A)
= R.H.S.

Question 16.
(cos A + sin A)² + (cos A – sin A)² = 2
Solution:
LHS = (cos A + sin A)² + (cos A – sin A)²
= cos² A + sin² A + 2 cos A sin A + cos² A + sin² A – 2 cos A sin A
= 2 sin² A + 2 cos² A
= 2 (sin²A+cos²A)
= 2 x 1=2 = R.H.S. (∵ sin²A + cos² A = 1)

Question 17.
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.
(sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Solution:
L.H.S. = (sin A + cosecA)² + (cosA+ secA)²
= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
= sin² A+cosec² A+2 sin A x \(\frac { 1 }{ sinA }\) + cos² A+sec²A + 2cosA x \(\frac { 1 }{ cosA }\)
= sin²A + cos² A + cosec² A + sec²A+ 2 + 2   (∵ sin² A + cos²A= 1)
= 1 +cosec²A + sec²A + 4
= (1 + cot² A) + (1 + tan² A) + 5 [∵ cosec²A = 1 + cot² A and sec² A = 1 + tan²A]
= 1 + cot² A + 1 + tan² A + 5
= 7 + tan²A + cot²A = R.H.S.

Question 22.
sec²A. cosec²A = tan²A + cot²A + 2
Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.

Solution:

Question 36.

Solution:

Question 37.

Solution:

Question 38.
(1 +cot A-cosec A) (1 + tan A + sec A) = 2
Solution:

Question 39.

Solution:

Question 40.

Solution:

Question 41.

Solution:

Question 42.

Solution:

Question 43.

Solution:

Question 44.

Solution:

Question 45.

Solution:

Question 46.

Solution:

Question 47.

Solution:

Question 48.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Question 1.
Find AD.

Solution:

Question 2.
In the following diagram.
AB is a floor-board. PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.

Solution:

Question 3.
Calculate BC.

Solution:

Question 4.
Calculate AB .

Solution:

Question 5.
The radius of a circle is given as 15cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, Calculate :
(i) the length of AB;
(ii) the distance of AB from the centre C.
Solution:
Chord AD substends an angle of 131° at the centre. Join CA, CB and draw CD ⊥ AB which bisects AB at D.
(In ∆CAB)
∵ CA = CB (radii of the same circle)
∴ ∠CAB = ∠CBA
But ∠CAB + ∠CBA = 180°- 131° = 49°

Question 6.
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \(\frac { 5 }{ 12 }\). On walking 192 metres towards the tower; the tangent of the angle is found to be \(\frac { 3 }{ 4 }\). Find the height of the tower.
Solution:

Question 7.
A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is a and that of the top of flagstaff is β. Prove that the height of the tower is :

Solution:

Question 8.
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m.
The man’s eye is 2 m above the ground. He observes the angle of elevation of C. The top of the pole, as x°, where tan x° = \(\frac { 2 }{ 5 }\) . calculate:

(i) the distance AB in m;
(ii) the angle of elevation of the top of the pole
when he is standing 15 m from the pole. Give your answer to the nearest degree. [1999]
Solution:

Question 9.
The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same striaght line with it are complementary.Prove that height of the tower is \( \sqrt{ab} \) metre.
Solution:

Question 10.
From a window A. 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x = \(\frac { 5 }{ 2 }\) and the angle of depression of the foot D of the tower is y°, where tany° = \(\frac { 1 }{ 4 }\).(See the figure given below). Calculate the height CD of the tower in metres. [2000]

Solution:
Let CD be the height of the tower and height of window A from the ground = 10m
In right ∆AEC,

Question 11.
A vertical tower is 20 m high, A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? [2001]
Solution:

Question 12.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate :
(i) the width of the river and
(ii) the height of the tree.
Solution:

Let TR be the tree of height x m and y be the width AR of the river,
then ∠B = 30° and A = ∠60° , AB = 50 m.
Now in right ∆ATR,

Question 13.
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is 60° and the angle of elevation of the top of the pole as seen from the foot of the tower is 30°. Find
(i) the height of the tower.
(ii) the horizontal distance between the pole and the tower.
Solution:
Let PQ is the pole and TS is the tower. PQ = 20 m.
Let TS = h and QS = x Angles of elevation from Q to T A T is 60° and from S to P is 30°.
In the ∆PQS

Question 14.
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°.
Find : (i) the height of the tower, if the height of the pole is 20m;
(ii) the height of the pole, if the height of the tower is 75 m.
Solution:
Let PQ is the pole and TW is the tower
Angle of elevation from T to P is 60° and angle of depression from P to W is 30°
∴ ∠PWQ = 30° = ∠RPW ( ∠ Altanate angles)
(i) In first case when height of pole OQ = 20m, Then in right ∆ PQW

Question 15.
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.
Solution:
Let AQ is the sea-level
P is a point 36 m above sea-level
∴ PQ = 36
Let B be the bird and R is its reflection in the water and angle of elevation of the bird B at P is 30° and angle of depression of the reflection of the bird at R is 60°

Question 16.

Solution:

Question 17.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of the light house in horizontal line with its base , are 30° and 40° respectively . Find the distance between the two ships.Give your answer correct to the nearest meter. [2012]
Solution:

Question 18.
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :

(i) the horizontal distance between AB and CD.
(it) the height of the lamp post.
Solution:

Question 19.
An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:

Question 20.
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24° respectively. Find the height of the two towers. Give your answer correct to 3 significant figures. (2015)

Solution:
AB and CD are two towers which are 120 m apart
i.e. BD= 120m
Angles of elevation of the top and angle of depression of bottom of the first tower observed from the top of second tower is 30° and 24°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C are helpful to complete your math homework.

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