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9th Maths 13.2

Chapter 13 Surface Areas and Volumes NCERT Class 9th solution of Exercise 13.1 NCERT Class 9th solution of Exercise 13.3 NCERT Class 9th solution of Exercise 13.4 NCERT Class 9th solution of Exercise 13.5 NCERT Class 9th solution of Exercise 13.6 NCERT Class 9th solution of Exercise 13.7 NCERT Class 9th solution of Exercise 13.8 NCERT Class 9th Maths Projects Exercise 13.2 Assume π=227, unless stated otherwise. Q1. The curved surface area of a right circular cylinder of height 14cm is 88cm2. Find the diameter of the base of the cylinder. Sol. : Given: Height of Cylinder is h=14cm, Area=88cm2 To Find: Diameter of the base of Cylinder. Solve: Let Radius is r Curved Surface Area=2πrh 88=2×227×⋊ r=(88times7)/(2times22times14) r=1 Diameter of the base=2r=2times1=2cm Answer: Diameter of the base 2cm. Q2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the shee...

9th Maths 13.1

Chapter 13 Surface Areas and Volumes NCERT Class 9th solution of Exercise 13.2 NCERT Class 9th solution of Exercise 13.3 NCERT Class 9th solution of Exercise 13.4 NCERT Class 9th solution of Exercise 13.5 NCERT Class 9th solution of Exercise 13.6 NCERT Class 9th solution of Exercise 13.7 NCERT Class 9th solution of Exercise 13.8 NCERT Class 9th Maths Projects Exercise 13.1 Q1. A plastic box 1.5m long, 1.25m wide and 65cm deep is to be made. It is opened at the top. Ignoring the  thickness of the plastic sheet, determine: i) The area of the sheet required for making the box. ii) The cost of sheet for it, if a sheet measuring 1m^2 costs ₹20. Sol. : Given: L=1.5m, B=1.25m, H=65cm=0.65m To Find:  Area of the sheet. Solve: i) Surface Area of box open at the top SA =2(L+B)H+LB SA =2(1.5+1.25)0.65+1.5times1.25 SA =2times2.75times0.65+1.875 SA =3.575+1.875 SA =5.45m^2 Answer: Surface Area of the sheet required to making the box is 5.45m^2. ii) `Cost = Areat...

9th Maths 12.1

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NCERT Class 9th solution of Exercise 12.1 NCERT Class 9th Maths Projects Exercise 12.1 Q1. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side ' a '. Find the area of the signal board, using Heron's formula. If its perimeter is 180cm, what will be the area of the signal board? Sol. : Semi persmeter S=(a+a+a)/2=(3a)/2 By Heron's formula:  Area of triangle=sqrt(S(S-a)(S-b)(S-c)) Area of triangle=sqrt((3a)/2((3a)/2-a)((3a)/2-a)((3a)/2-a)) Area of triangle=sqrt((3a)/2(a/2)(a/2)(a/2)) Area of triangle=sqrt(3)/4a^2 square unit. The Perimeter of equilateral triangle=3a 180=3a a=180/3 a=60 Area of triangle=sqrt3/4a^2 Area of triangle=(60)^2/4sqrt3cm^2 Area of triangle=(60times60)/4sqrt3cm^2 Area of triangle=900sqrt3cm^2 Answer : Area of the signal board triangle=900sqrt3cm^2. Q2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122m, 22m and 120m[see...

9th Maths 11.1

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NCERT Class 9th solution of Exercise 11.1 NCERT Class 9th solution of Exercise 12.1 NCERT Class 9th Projects Exercise 11.1 Q1. Construct an angle of 90^circ at the initial point of a given ray and justify the construction. Sol. : Step of Construction: Draw ray BC, take centre B draw an arc with any radius it intersects BC at P. Take centre P draw arc with the same radius it intersects the previous arc at Q. Take centre Q draw arc with the same radius it intersects the previous arc at R. Take centres Q and R draw arcs with the same radius which intersects at A. Join AB this gives angleABC=90^circ. Justification:  triangleBQP is an equilateral triangle. [ By Construction] angleQBP=60^circ [ Angle of the equilateral triangle] angleABQ=1/2angleQBP [Ray BA is angle bisector] angleABQ=30^circ angleABC=angleABQ+angleQBP angleABC=30^circ+60^circ angleABC=90^circ. hence Justified. Q2. Construct an angle of 45^circ at the initial point of a given...

9th Maths 10.4

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NCERT Class 9th solution of Exercise 10.1 NCERT Class 9th solution of Exercise 10.2 NCERT Class 9th solution of Exercise 10.3 NCERT Class 9th Maths Projects Exercise 10.4 Q1. Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm. Find the length of the common chord. Sol. : Given: Let radius OP=5cm, O'P=3cm. The distance of OO' is 4cm. To Find: PQ Solve: In triangleOPO' (OP)^2=(OO')^2+(O'P)^2 [By Pythagoras Theorem] (5)^2=(4)^2+(3)^2 25=16+9 25=25  triangleOPO' is a right-angled triangle. this gives OO'⊥OP PQ is the diameter of the circle of radius 3cm as it passes through centre O'. PQ=2times3=6cm.  Answer: The length of the common chord is 6cm. Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. Sol. : Given: Chords AB=CD of a circle with centre O which inters...

9th Maths 10.3

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NCERT Class 9th solution of Exercise 10.1 NCERT Class 9th solution of Exercise 10.2 NCERT Class 9th solution of Exercise 10.4 NCERT Class 9th Maths Projects Exercise 10.3 Q1. Draw different pairs of circles. How many points does each pair have in common? what is the maximum number of common points? Sol. : Different pairs of circles Number of common points (a) 0  (zero) (b) 1  (one)  A, (c) 2  (two)  A,B Answer: The maximum number of common points is 2. Q2. Suppose you are given a circle. Give a construction to find its centre. Sol. : Step of Construction: 1) Take three points A, B and C on the cricumference of the given circle. 2) Join AB and BC. 3) Draw perpendicular bisector of AB and draw perpendicular bisector of BC, which intersect at O. 4) We get centre of circle O.  Q3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chords. Sol. : Given: Two circles with centr...

9th Maths 10.1

NCERT Class 9th solution of Exercise 10.1 NCERT Class 9th solution of Exercise 10.2 NCERT Class 9th solution of Exercise 10.3 NCERT Class 9th solution of Exercise 10.4 NCERT Class 9th solution of Exercise 9.1 NCERT Class 9th Maths Projects Exercise 10.1 Q1. Fill in the blanks: i) The centre of a circle lies in ____ of the circle. [exterior/ interior] ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle. [exterior/interior] iii) The longest chord of a circle is a ____ of the circle. iv) An arc is a ____ when its ends are the ends of a diameter. v) Segment of a circle is the region between an arc and ____ of the circle. vi) A circle divides the plane, on which it lies, in _____ parts. Answer:  i) Interior, ii) Exterior, iii) Diameter, iv) Semicircle, v) The chord,  vi) Three. Q2. Write True or Fales: Give reasons for your answers. i) Line segment joining the centre to any point on the circle is a radius of the cricle. ii) ...

9th Maths 10.2

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NCERT Class 9th solution of Exercise 10.1 NCERT Class 9th solution of Exercise 10.3 NCERT Class 9th solution of Exercise 10.4 NCERT Class 9th Maths Projects Exercise 10.2 Q1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. Sol. : Given: Two equal Chords AB and CD of a circle with centre at O. To Prove:  angleAOB=angleCOD Proof: In triangleAOB and triangleCOD OA=OC [Radii of a circle] OB=OD [Radii of a circle] AB=CD [Given] triangleAOBcongtriangleCOD  [By SSS rule] This gives angleAOB=angleCOD [By CPCT] Proved. Q2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. Sol. : Given: angleAOB=angleCOD To Prove: AB=CD Proof: In triangleAOB and triangleCOD OA=OC      [Radii of a circle] angleAOB=angleCOD [Given] OB=OD    [Radii of a circle] `triangleAOBcongtriangleC...

9th Maths 9.3

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NCERT Class 9th solution of Exercise 9.1 NCERT Class 9th solution of Exercise 9.2 NCERT Class 9th solution of Exercise 10.1 NCERT Class 9th Maths Projects Exercise 9.3 Q1. In Figur, E is any point on median AD of a triangleABC. Show that ar(ABC)=ar(ACE). Sol. : Given:  triangleABC, AD is a median. To Prove: ar(ABE)=ar(ACE). Proof: In triangleABCAD is median ar(ABE)=ar(ACE)__________(1) In triangleEBC, ED is median. ar(BDE)=ar(CDE)__________(2) By equation (1) - (2) ar(ABE)-ar(BDE)=ar(ACE)-ar(CDE) ar(ABE)=ar(ACE) Proved. Q2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED)=1/4(ar(ABC). Sol. : Given: triangleABCAD is median and E is the midpoint of AD. To Prove: ar(BED)=1/2ar(ABC) Proof: In triangleABCAD is the median ar(ADB)=ar(ACD) ar(ADB)=1/2(ABC)__________(1) In triangleADB E is the midpoint of AD, BE is the median of triangleADB ar(BED)=1/2ar(ADB)________(2) from equati...