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 `pi=22/7`, unless stated otherwise. Q1. The curved surface area of a right circular cylinder of height `14cm` is `88cm^2`. Find the diameter of the base of the cylinder. Sol. : Given: Height of Cylinder is `h=14cm,` ` Area=88cm^2` To Find: Diameter of the base of Cylinder. Solve: Let Radius is `r` Curved Surface Area`=2pirh` `88=2times22/7timesrtimes14` `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

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 = AreatimesRate`

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

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 ray

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

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 centre `O` and `O'`. `AB` is a

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) A circle

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] `triangleAOBcongtriangleCOD` [By `SAS` rule] This gives `AB

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 `triangleABC`,  `AD` 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: `triangleABC`,  `AD` is median and `E` is the midpoint of `AD`. To Prove: `ar(BED)=1/2ar(ABC)` Proof: In `triangleABC`,  `AD` 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 equation (1) and (2) `ar(