CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m (Assume π =22/7) Solution: (i) Diameter = 28 cm Radius, r = 28/2 cm = 14cm Volume of the solid spherical ball = (4/3) πr3 Volume of the ball = (4/3)×(22/7)×143 = 34496/3 Hence, volume of the ball is 34496/3 cm3 (ii) Diameter = 0.21 m Radius of the ball =0.21/2 m= 0.105 m Volume of the ball = (4/3 )πr3 Volume of the ball = (4/3)× (22/7)×0.1053 m3 Hence, volume of the ball = 0.004851 m3

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if angle adc = 23 degree find angle ABC

Posted by Vinayak Baghel 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Meghana Manohar 5 years, 2 months ago

Make sure that Ur question is proper

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2/3×9/2×3/5

Posted by Sakshi Hiremath 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Himanshi Seksaria 5 years, 2 months ago

9/5

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Yogita Ingle 5 years, 2 months ago

2/3×9/2×3/5

= 9/5

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Find the value of X³ +y³ +15xy -125 if x+y =5

Posted by Gazala Anjum 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gazala Anjum 5 years, 2 months ago

I don't know

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Find the value of x3 +y3 15xy-125

Posted by Gazala Anjum 5 years, 2 months ago

CBSE > Class 09 > Mathematics

0 answers

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Two sides of triangle are 13cm and 14 cm and its semi perimeter is 18cm. Find the third side of the triangle

Posted by Tejasv Choubey 5 years, 2 months ago

CBSE > Class 09 > Mathematics

3 answers

Yogita Ingle 5 years, 2 months ago

Let a =13 cm, b =14 cm, and third side =c cm

Semiperimeter is half of perimeter and is given by,

s=(a+b+c)/3

⇒18= (13+14+c)/2

⇒c=36−27

⇒c=9 cm

∴ Third side of the triangle is 9 cm.

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Rahul Yadav 5 years, 2 months ago

##i#m#.p#b name n engl

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Harsh Dixit 5 years, 2 months ago

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Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.

Posted by Aiman Junaid 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 2 months ago

Solution of this question

radius will be 4.2 /2 =2.1
since amount of water displaced = volume of body submerged = 4/3 ×pi× 2.1^3=38.772 cm cube

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can we construct a traingle with side 2.6 3.4 and 6.1 cm ?

Posted by Varsha Pinku 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Ankit Kumar Sharma 5 years, 2 months ago

Yes,because sum of two angles is greater than the third angle so it is a triangle.

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Yogita Ingle 5 years, 2 months ago

2.6 + 3.4 = 6.0

No,because in a triangle sum of any two sides should be greater than third side.

2.6 + 3.4 = 6.0 < 6.1

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ABCD is a rhombus such that angle ACB = 40° , then angle ADB is ----- (a) (b) (c)

Posted by Prathmesh Chikhale 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Shabd Kumar 5 years, 2 months ago

40°

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Ex. 8.1 question NO 6 maths class 9 plz. Tell

Posted by Pranav Mishra 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Pratyush Pradyun Subrata 5 years, 2 months ago

GRAMMAR CBSE NOTES CLS 6 7 8 9 10 11 12 MCQ QUESTIONS NCERT BOOKS STUDY MATERIAL NUMBERS CALCULATOR Learn CBSE NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12 NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1. NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Ex 8.1 Class 9 Maths Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution: Let the angles of the quadrilateral be 3x, 5x, 9x and 13x. ∴ 3x + 5x + 9x + 13x = 360° [Angle sum property of a quadrilateral] ⇒ 30x = 360° ⇒ x = 360∘30 = 12° ∴ 3x = 3 x 12° = 36° 5x = 5 x 12° = 60° 9x = 9 x 12° = 108° 13a = 13 x 12° = 156° ⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°. Ex 8.1 Class 9 Maths Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle. Solution: Let ABCD is a parallelogram such that AC = BD. In ∆ABC and ∆DCB, AC = DB [Given] AB = DC [Opposite sides of a parallelogram] BC = CB [Common] ∴ ∆ABC ≅ ∆DCB [By SSS congruency] ⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1) Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram] ∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles] From (1) and (2), we have ∠ABC = ∠DCB = 90° i.e., ABCD is a parallelogram having an angle equal to 90°. ∴ ABCD is a rectangle. Ex 8.1 Class 9 Maths Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Solution: Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O. ∴ In ∆AOB and ∆AOD, we have AO = AO [Common] OB = OD [O is the mid-point of BD] ∠AOB = ∠AOD [Each 90] ∴ ∆AQB ≅ ∆AOD [By,SAS congruency ∴ AB = AD [By C.P.C.T.] ……..(1) Similarly, AB = BC .. .(2) BC = CD …..(3) CD = DA ……(4) ∴ From (1), (2), (3) and (4), we have AB = BC = CD = DA Thus, the quadrilateral ABCD is a rhombus. Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus. Ex 8.1 Class 9 Maths Question 4. Show that the diagonals of a square are equal and bisect each other at right angles. Solution: Let ABCD be a square such that its diagonals AC and BD intersect at O. (i) To prove that the diagonals are equal, we need to prove AC = BD. In ∆ABC and ∆BAD, we have AB = BA [Common] BC = AD [Sides of a square ABCD] ∠ABC = ∠BAD [Each angle is 90°] ∴ ∆ABC ≅ ∆BAD [By SAS congruency] AC = BD [By C.P.C.T.] …(1) (ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram] ∴ ∠1 = ∠3 [Alternate interior angles are equal] Similarly, ∠2 = ∠4 Now, in ∆OAD and ∆OCB, we have AD = CB [Sides of a square ABCD] ∠1 = ∠3 [Proved] ∠2 = ∠4 [Proved] ∴ ∆OAD ≅ ∆OCB [By ASA congruency] ⇒ OA = OC and OD = OB [By C.P.C.T.] i.e., the diagonals AC and BD bisect each other at O. …….(2) (iii) In ∆OBA and ∆ODA, we have OB = OD [Proved] BA = DA [Sides of a square ABCD] OA = OA [Common] ∴ ∆OBA ≅ ∆ODA [By SSS congruency] ⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3) ∵ ∠AOB and ∠AOD form a linear pair. ∴∠AOB + ∠AOD = 180° ∴∠AOB = ∠AOD = 90° [By(3)] ⇒ AC ⊥ BD …(4) From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles. Ex 8.1 Class 9 Maths Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Solution: Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles. Now, in ∆AOD and ∆AOB, We have ∠AOD = ∠AOB [Each 90°] AO = AO [Common] OD = OB [ ∵ O is the midpoint of BD] ∴ ∆AOD ≅ ∆AOB [By SAS congruency] ⇒ AD = AB [By C.P.C.T.] …(1) Similarly, we have AB = BC … (2) BC = CD …(3) CD = DA …(4) From (1), (2), (3) and (4), we have AB = BC = CD = DA ∴ Quadrilateral ABCD have all sides equal. In ∆AOD and ∆COB, we have AO = CO [Given] OD = OB [Given] ∠AOD = ∠COB [Vertically opposite angles] So, ∆AOD ≅ ∆COB [By SAS congruency] ∴∠1 = ∠2 [By C.P.C.T.] But, they form a pair of alternate interior angles. ∴ AD || BC Similarly, AB || DC ∴ ABCD is a parallelogram. ∴ Parallelogram having all its sides equal is a rhombus. ∴ ABCD is a rhombus. Now, in ∆ABC and ∆BAD, we have AC = BD [Given] BC = AD [Proved] AB = BA [Common] ∴ ∆ABC ≅ ∆BAD [By SSS congruency] ∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5) Since, AD || BC and AB is a transversal. ∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles] ⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)] So, rhombus ABCD is having one angle equal to 90°. Thus, ABCD is a square. Ex 8.1 Class 9 Maths Question 6. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. Solution: We have a parallelogram ABCD in which diagonal AC bisects ∠A ⇒ ∠DAC = ∠BAC (i) Since, ABCD is a parallelogram. ∴ AB || DC and AC is a transversal. ∴ ∠1 = ∠3 …(1) [ ∵ Alternate interior angles are equal] Also, BC || AD and AC is a transversal. ∴ ∠2 = ∠4 …(2) [ v Alternate interior angles are equal] Also, ∠1 = ∠2 …(3) [ ∵ AC bisects ∠A] From (1), (2) and (3), we have ∠3 = ∠4 ⇒ AC bisects ∠C. (ii) In ∆ABC, we have ∠1 = ∠4 [From (2) and (3)] ⇒ BC = AB …(4) [ ∵ Sides opposite to equal angles of a ∆ are equal] Similarly, AD = DC ……..(5) But, ABCD is a parallelogram. [Given] ∴ AB = DC …(6) From (4), (5) and (6), we have AB = BC = CD = DA Thus, ABCD is a rhombus.

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2+2= ???? Tell if you are a intelligent

Posted by N T 5 years, 2 months ago

CBSE > Class 09 > Mathematics

5 answers

Saransh Gupta 5 years, 2 months ago

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Saswat Jyotiraditya Jena 5 years, 2 months ago

22 or 4

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Sarthak Halder 5 years, 2 months ago

3Thank You

Akash Raj 5 years, 2 months ago

Laura

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Ayush Raj 5 years, 2 months ago

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In triangle PQR, if angle R= angle P and QR=4cm and PR=5 cm. find the length of PQ.

Posted by Rakshit Kumar 5 years, 2 months ago

CBSE > Class 09 > Mathematics

5 answers

Shreshth Gupta 5 years, 2 months ago

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Divya Khola 5 years, 2 months ago

Length of PQ = 4 cm

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Prince Kumar Sah 5 years, 2 months ago

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Sarthak Halder 5 years, 2 months ago

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Prabhav Bairwa 5 years, 2 months ago

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A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius us 1m . Then find the volume of the iron used to make the tank

Posted by Ritika Swami 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Sarthak Halder 5 years, 2 months ago

0.5

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Prabhav Bairwa 5 years, 2 months ago

0.5

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Find the value of,k if x_1 is a factor of p(x) in each of the following cases

Posted by Ngajum Yomgam 5 years, 2 months ago

CBSE > Class 09 > Mathematics

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Gkggjfjhfgh gkcjggj kee Kshatriyas hdhdjdgffn

Posted by Usha Upadhyay 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 2 months ago

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Find the total surface area and lateral surface area of a cuboid whose length, breath and height are 12cm,8cm and 5cm respectively

Posted by Nazneen Khan 5 years, 2 months ago

CBSE > Class 09 > Mathematics

3 answers

Divya Khola 5 years, 2 months ago

T.S.A.= 392 sq.cm L.S.A.= 200 sq.cm

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Ankit Dubey 5 years, 2 months ago

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Shivani Yadav 5 years, 2 months ago

T.S.A of cuboid= 2(lb×bh×hl) =2(12×8+8×5+5×12) =2(196) =396cmsq. L.S.A of cuboid=2(l+b)×h =2(12+8)×5 =2(20)×5 =200cmsq.

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The curved surface area of a right circular cylinder is 4400 cm ^2 if the circumference of its base is 110 cm then it's height is

Posted by K Sriprada 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Shivani Yadav 5 years, 2 months ago

Given, Circumference of base(2πr)=110cm C.S.A.of cylinder=2πrh 4400 =(2πr)×h 4400=110×h h=4400÷110 h=40cm

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ABCD is a trapezium with AB||DC. A line parallel to AC intersects AB at X and BC at Y Prove that ar (ADX) = ar (ACY)

Posted by Řøýåľ Jāâţ 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 2 months ago

It can be observed that ΔADX and ΔACX lie on the same base AX and are between the same parallels AB and DC.

⊥ Area (ΔADX) = Area (ΔACX) ... (1)

ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XY.

⊥ Area (ΔACY) = Area (ACX) ... (2)

From equations (1) and (2), we obtain

Area (ΔADX) = Area (ΔACY)

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A rational number between 17 and 27

Posted by Anesh Fernandez 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Tanay Lathi 5 years, 2 months ago

18, 18.5, 18.75, 19, 19.25, 19.30, 20, 21, 22, 23, 24, 25, 26, 27.... are rational numbers between 17 and 27 . Pick anyone from these and apply your brain such ladoo questions asked in Class 6.

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Prashant Mishra 5 years, 2 months ago

41/2

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The decimal expansion of an irrational number is

Posted by Anesh Fernandez 5 years, 2 months ago

CBSE > Class 09 > Mathematics