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An isosceles right triangle has area 8cm square. The length of it's hypotenuse is?

Posted by Abhi Jain 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Meghna Thapar 3 years, 10 months ago

The hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle. The other two sides are called the opposite and adjacent sides. These sides are labeled in relation to an angle. In geometry, an isosceles triangle is a triangle that has two sides of equal length. Sometimes it is specified as having exactly two sides of equal length, and sometimes as having at least two sides of equal length, the latter version thus including the equilateral triangle as a special case.

ANSWER

Find two rational number between 3 and 4

Posted by Faiz Faiz 3 years, 10 months ago

CBSE > Class 09 > Mathematics

3 answers

X-O Harsh & Parveen 3 years, 10 months ago

A rational number between 3 and 4 is 1/2 (3 + 4) = 7/2. Then, 3 < 7/2 < 4 A rational number between 3 and 7/2 = 1/2 {3 + 7/2} = 1/2 (3/1 + 7/2) = 1/2 ((6 + 7)/2) = (1/2 × 13/2) = 13/4 A rational number between 7/2 and 4 = 1/2 {7/2 + 4} = 1/2 (7/2 + 4/1) = 1/2 ((7 + 8)/2) = {1/2 × 15/2} = 15/4 Therefore, 3 < 13/4 < 7/2 < 15/4 < 4 Hence, 13/4, 7/2 and 15/4 are the three rational numbers lying between 3 and 4.

Priya Rana 3 years, 10 months ago

A rational number between 3 and 4 is 1/2 (3 + 4) = 7/2. Then, 3 < 7/2 < 4 A rational number between 3 and 7/2 = 1/2 {3 + 7/2} = 1/2 (3/1 + 7/2) = 1/2 ((6 + 7)/2) = (1/2 × 13/2) = 13/4 A rational number between 7/2 and 4 = 1/2 {7/2 + 4} = 1/2 (7/2 + 4/1) = 1/2 ((7 + 8)/2) = {1/2 × 15/2} = 15/4 Therefore, 3 < 13/4 < 7/2 < 15/4 < 4 Hence, 13/4, 7/2 and 15/4 are the three rational numbers lying between 3 and 4.

Yogita Ingle 3 years, 10 months ago

A rational number between 3 and 4 =1/2(3+4)=7/2

Then 3<7/2<4

A rational number between 3 and 7/2

=1/23+7/2=1/2(3/1+72)

=1/2 (6+7)/2 =1/2+13/2=13/4

13/4, 7/2 are two rational number between 3 and 4

3<13/4<7/2<4.

ANSWER

2x-3x

Posted by Abhinav Sain 3 years, 10 months ago

CBSE > Class 09 > Mathematics

4 answers

Black Yadav 3 years, 10 months ago

2x - 3x = (-x)

Neeraj Rajput 3 years, 10 months ago

-x

Yukthi Raj 3 years, 10 months ago

2x-3x =-x

Ayush Thakur 3 years, 10 months ago

-x

ANSWER

Three angels of a quadrilateral abcd are equal is it a parallelogram why are why not

Posted by Murshitha Sheerin 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Aryan Singh 3 years, 10 months ago

Yes all angles of parallelogram can be equal to each other Eg:Rectangle, Square They both are also parallelogram

Samman Nirvan 3 years, 10 months ago

No, because parallelogram doesn't have all equal angles

ANSWER

E divides AB in the ratio 1:3 and also divides AC in the ratio 1:3, EF=2.8cm find BC

Posted by Dev Kumar 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 3 years, 10 months ago

AE/AB = AF / AC and

< BAC = < EAF

Triangle ABC similar to triangle AEF (SAS)

by BPT

EF is parallel to BC

AB: AE = BC/EF

Let l be the length of AB

the AE = l/4

1:(1/4)= BC:2.8

= BC/2.8

BC = 2.8 X 4 = 11.2m

ANSWER

ABCD is a rhombus with (angle ABC= 40•) . Find the measure of angle ACD .

Posted by Anuj Kumar 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Abhishek Chauvhan 3 years, 10 months ago

ABC is Currently

Yogita Ingle 3 years, 10 months ago

Given ABCD is a rhombus.
Diagonals bisect each other perpendicularly.
Hence ∠BOC = 90°
Given ∠OCB = 40°
AD||BC and BD is the transversal
∴ ∠ADB = ∠DBC (Alternate angles)
Hence in right angled DBOC,
∠BOC + ∠OCB + ∠OBC = 180°
⇒ 90° + 40° + ∠OBC = 180°
⇒130° + ∠OBC = 180°
⇒ ∠OBC = 180° - 130° = 50°
But ∠OBC = ∠DBC
Therefore, ∠ADB = 50°

ANSWER

Show that if the diagonals of quadrilateral bisect each other at right angles then it is a rhombus

Posted by Vimal Desai 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Preeti Singh 3 years, 10 months ago

It's is a right answer ?

Yogita Ingle 3 years, 10 months ago

Sol: We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.

AO = AO [Common]

OB = OD [Given that O in the mid-point of BD]

∠AOB = ∠AOD [Each = 90°]

ΔAOB ≌ ΔAOD [SAS criteria]

Their corresponding parts are equal.

AB = AD ...(1)

Similarly, AB = BC ...(2)

BC = CD ...(3)

CD = AD ...(4)

∴ From (1), (2), (3) and (4), we have AB = BC CD = DA

Thus, the quadrilateral ABCD is a rhombus.

ANSWER

Rational number between 3 and 5

Posted by Faiz Faiz 3 years, 10 months ago

CBSE > Class 09 > Mathematics

5 answers

Preeti Singh 3 years, 10 months ago

30/10: 50/10

Prince Kumar 3 years, 10 months ago

3.5

Shabd Kumar 3 years, 10 months ago

4

Nidhi Kaurav 3 years, 10 months ago

Infinite

Praveen _Raj 3 years, 10 months ago

3 and 5

ANSWER

i want 4.3,4.4 please i want to write

Posted by Kenil Baldaniya 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 3 years, 10 months ago

1. Draw the graph of each of the following linear equations in two variables:

(i) x+y = 4

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

x+y = 4

Substituting the values for x,

When x = 0,

x+y = 4

0+y = 4

y = 4

When x = 4,

x+y = 4

4+y = 4

y = 4–4

y = 0

x	y
0	4
4	0

The points to be plotted are (0, 4) and (4,0)

For for click on the given link:

EX 4.3:

<a href="https://mycbseguide.com/blog/ncert-solutions-for-class-9-maths-exercise-4-3/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-for-class-9-maths-exercise-4-3/&ved=2ahUKEwjYoO3KiqztAhXEyDgGHRSWDbcQFjAAegQIBRAC" rel="noopener" target="_blank">NCERT Solutions for Class 9 Maths Exercise 4.3 ...</a>

Ex 4.4 :

<a href="https://mycbseguide.com/blog/ncert-solutions-for-class-9-maths-exercise-4-4/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-for-class-9-maths-exercise-4-4/&ved=2ahUKEwjgrJHwiqztAhW-yDgGHcABA_cQFjAAegQIAxAC" rel="noopener" target="_blank">NCERT Solutions for Class 9 Maths Exercise 4.4 ...</a>

ANSWER

Please gave me the last year question paper set-1 marking scheme 2019

Posted by Aman Hindu 3 years, 10 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

3-2√2/3+2√2/3+2√2/3-2√2

Posted by Unknown Student 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Aryan Singh 3 years, 10 months ago

3-4√2

ANSWER

Calculate the area of quadrilateral abcd in which angle b= 90° BCD is an eq of side 20 cm and ad 250 m

Posted by Aarti Pirjapati 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Aarti Pirjapati 3 years, 10 months ago

Are here not answer this question

ANSWER

sqrt 11+4sqrt 7-sqrt 8-2sqrt 7=?

Posted by Vedant Vyas 3 years, 10 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

Find the value of k if x=2 ,y=1 is a solutions of the equation 2x+3y=k

Posted by Amrita Kumari 3 years, 10 months ago

CBSE > Class 09 > Mathematics

5 answers

Aditya Ranjan 3 years, 10 months ago

K=7

Preeti Singh 3 years, 10 months ago

K=7

Shadik Raza 3 years, 10 months ago

K=7

Shreshth Gupta 3 years, 10 months ago

K=7

Asmi . 3 years, 10 months ago

2x +3y=k On putting the value of x and y 2(2)+3(1)=k 4+3=k 7=k Or k=7 Hope it helps......?

ANSWER

Find two rational number between 8 and 5

Posted by Faiz Faiz 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Faiz Faiz 3 years, 10 months ago

Rational number between 3 and 5

Smriti Singh 3 years, 10 months ago

2 rational numbers betweeen 8 and 5 are 6 and 7

ANSWER

10000×1000

Posted by Avinash Thakur 3 years, 10 months ago

CBSE > Class 09 > Mathematics

5 answers

Aryan Singh 3 years, 10 months ago

1×10^7

Shreshth Gupta 3 years, 10 months ago

10000000

Sneha Pant 3 years, 10 months ago

10000*1000 10000000

Himanshi Seksaria 3 years, 10 months ago

10000000

Avinash Thakur 3 years, 10 months ago

10000000

ANSWER

EXERCISE 13.8 2nd problem

Posted by Bhaskar.H.L Nayak 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Pratyush Pradyun Subrata 3 years, 10 months ago

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

ANSWER

if angle adc = 23 degree find angle ABC

Posted by Vinayak Baghel 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Meghana Manohar 3 years, 10 months ago

Make sure that Ur question is proper

ANSWER

2/3×9/2×3/5

Posted by Sakshi Hiremath 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Himanshi Seksaria 3 years, 10 months ago

9/5

Yogita Ingle 3 years, 10 months ago

2/3×9/2×3/5

= 9/5

ANSWER

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

Posted by Gazala Anjum 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Gazala Anjum 3 years, 10 months ago

I don't know

ANSWER

Find the value of x3 +y3 15xy-125

Posted by Gazala Anjum 3 years, 10 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

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 3 years, 10 months ago

CBSE > Class 09 > Mathematics

3 answers

Yogita Ingle 3 years, 10 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.

Rahul Yadav 3 years, 10 months ago

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

Harsh Dixit 3 years, 10 months ago

9

ANSWER

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 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 3 years, 10 months ago

Solution of this question

Or

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

ANSWER

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

Posted by Varsha Pinku 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Ankit Kumar Sharma 3 years, 10 months ago

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

Yogita Ingle 3 years, 10 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

ANSWER

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

Posted by Prathmesh Chikhale 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Shabd Kumar 3 years, 10 months ago

40°

ANSWER

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

Posted by Pranav Mishra 3 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Pratyush Pradyun Subrata 3 years, 10 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.

ANSWER

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

Posted by N T 3 years, 10 months ago

CBSE > Class 09 > Mathematics

5 answers

Saransh Gupta 3 years, 10 months ago

4

Saswat Jyotiraditya Jena 3 years, 10 months ago

22 or 4

Sarthak Halder 3 years, 10 months ago

4

Akash Raj 3 years, 10 months ago

Laura

Ayush Raj 3 years, 10 months ago

4

ANSWER

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

Posted by Rakshit Kumar 3 years, 10 months ago

CBSE > Class 09 > Mathematics

5 answers

Shreshth Gupta 3 years, 10 months ago

5

Divya Khola 3 years, 10 months ago

Length of PQ = 4 cm

Prince Kumar Sah 3 years, 10 months ago

5

Sarthak Halder 3 years, 10 months ago

5

Prabhav Bairwa 3 years, 10 months ago

5

ANSWER

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 3 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Sarthak Halder 3 years, 10 months ago

0.5

Prabhav Bairwa 3 years, 10 months ago

0.5

ANSWER

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

Posted by Ngajum Yomgam 3 years, 10 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

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