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Ask QuestionPosted by Shamshad Alam 3 years, 11 months ago
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Posted by Sushmit Dwivedi 3 years, 11 months ago
- 2 answers
Posted by Sameer मोहम्मद 3 years, 11 months ago
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Posted by Varni Doshi 3 years, 11 months ago
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Sahil Nara 3 years, 11 months ago
Posted by Arun J 3 years, 11 months ago
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Gaurav Seth 3 years, 11 months ago
Area of the field used for Wheat is 2.2 hector , for Potato is 1.7 hector and for onion is 1.7 hector.
Given:
Length of the triangular field for wheat = 240 m , 200 m , 360 m
Length of the triangular field for potato & onion = 240 m , 320 m , 400 m
To find: Area of the filed used for Wheat , Potato and Onion .
we use Heron's formula to find area of triangle.
s is semi perimeter and a , b, c are side,
For first field,
let, a = 240 , b = 200 , c = 360
s = 240+200+360/2 = 800/2 = 400 m
Area of field
1 hector = 10000 m²
So, 16000√2 m² = 22627.4 m² = 2.2 hector
For second field,
let, a = 240 , b = 320 , c = 400
s = 240+320+400/2 = 960/2 = 480 m
Area of field
Second field is divided in 2 equal areas.
So, Area of Potato = Area of Onion = 19200√3/2 = 9600√3 m² = 16627.69 m² = 1.7 hector
Therefore, Area of the field used for Wheat is 2.2 hector , for Potato is 1.7 hector and for onion is 1.7 hector.
Posted by Muskan Mishra 3 years, 11 months ago
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Yogita Ingle 3 years, 11 months ago
Green skills are those skills needed to adapt processes, services and products to climate change and the environmental regulations and requirements related to it. They include the knowledge, abilities, values and attitudes needed to live in, develop and support a sustainable and resource-efficient society. These skills are required in areas such as Renewable energy, climate change readiness, Wastewater treatment, Climate-resilient cities, Green construction, Solid waste management etc.
Posted by Shruti S 3 years, 11 months ago
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Posted by Roshni Maurya 3 years, 11 months ago
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Posted by Eshwar Yadav 3 years, 11 months ago
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Posted by Gopika Sunil 3 years, 11 months ago
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Eshwar Yadav 3 years, 11 months ago
Posted by Gk Vikreta 3 years, 11 months ago
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Posted by Shruti S 3 years, 11 months ago
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Yogita Ingle 3 years, 11 months ago
(x-2)² = x2 - 2 (x)(2) + (2)2
= x2 - 4x + 4
So, the coefficient of x² in the expansion of (x-2)² is 1.
Posted by Shruti S 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
x^a × y^a = (xy)^a
(25)⅓×(5)⅓
= (25×5) 1/3
= 125 1/3
= 53×1/3
= 51 = 5
Posted by Uday Dixit 3 years, 11 months ago
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Posted by Vinay ?? Ganiger 2 years ago
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Posted by Jai Ghosh 4 years ago
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Yogita Ingle 3 years, 11 months ago
As we know the equilateral triangle have all its sides equal. To find the area of equilateral triangle let us first find the semi perimeter of the equilateral triangle will be:
s = (a+a+a)/2
s=3a/2
where a is the length of the side.
Now, as per the heron’s formula, we know;
Area=√s(s−a)(s−b)(s−c)
Since, a = b = c
Therefore,
A = √[s(s-a)3]
which is the required formula.
Posted by Parmar Manali 4 years ago
- 3 answers
Yogita Ingle 4 years ago
( 1-2√5 ) + (3+7√5)
= 1-2√5 + 3+7√5
= 1+ 3 -2√5+7√5
= 4+ (-2+ 7)√5
= 4 + 5√5
Posted by Lucky Tailor 4 years ago
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Posted by Sumit Gupta 4 years ago
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Yogita Ingle 4 years ago
Pythagoras’ Theorem: In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
Given: Let XYZ be a triangle in which ∠YXZ = 90°.
YZ is the hypotenuse.
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To prove: XY2 + XZ2 = YZ2.
Construction: Draw XM ⊥YZ.
Therefore, ∠XMY = ∠XMZ = 90°.
Proof:
|
Statement |
Reason |
|
1. In ∆XYM and ∆XYZ, (i) ∠XMY = ∠YXZ = 90° (ii) ∠XYM = ∠XYZ |
1. (i) Given and by construction (ii) Common angle |
|
2. Therefore, ∆XYM ∼ ∆ZYX |
2. BY AA criterion of similarity |
|
3. Therefore, XYYZXYYZ = YMXYYMXY |
3. Corresponding sides of similar triangle are proportional |
|
4. Therefore, XY22 = YZ ∙ YM |
4. By cross multiplication in statement 3. |
|
5. In ∆XMZ and ∆XYZ, (i) ∠XMY = ∠YXZ = 90° (ii) ∠XZM = ∠XZY |
5. (i) Given and by construction (ii) Common angle |
|
6. Therefore, ∆XMZ ∼ ∆YXZ. |
6. BY AA criterion of similarity |
|
7. Therefore, XZ/YZ/ = MZ/XZ |
7. Corresponding sides of similar triangle are proportional |
|
8. Therefore, XZ2 = YZ ∙ MZ |
8. By cross multiplication in statement 7. |
|
9. Therefore, XY2 + XZ2 = YZ ∙ YM + YZ ∙ MZ ⟹ XY2 + XZ2 = YZ(YM+ MZ) ⟹ XY2 + XZ2 = YZ ∙ YZ ⟹ XY2 + XZ2= YZ2 |
9. By adding statements 4 and 8 |
Posted by Dilshad Khan 4 years ago
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Posted by Rajalakshmi Venkatraman 4 years ago
- 3 answers
Posted by Shreya Gupta 4 years ago
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Gaurav Seth 3 years, 11 months ago
10.10 theorem :
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment , the four points lie on a circle i.e (they are concyclic)
A , B , C , D are 4 points ( no 3 of them are collinear)
Lets draw a circle passing through A , B , C with Center O.
Assume D does not lie on circle
and circle intersect AD at E
=> A , B , C & E are concyclic
now ∠ACB = ∠AEB ( Angle formed by same chord in same arc segment)
∠ACB = ∠ADB given
in Δ DEB
∠AEB = ∠EDB + ∠EBD
=> ∠AEB = ∠ADB + ∠EBD
=> ∠ACB = ∠ACB + ∠EBD
=> ∠EBD = 0°
=> E & D Coincide
A , B , C & E are concyclic
E & D coincide
=> A , B , C & D are concyclic
Posted by Garima Agrawal 4 years ago
- 1 answers
Yogita Ingle 4 years ago
Rules to find the reflection of a point in the y-axis:
(i) Change the sign of abscissa i.e., x-coordinate.
(ii) Retain the ordinate i.e., y-coordinate.
So, mirror image of (-3,5) along y axis is (3,5)
Posted by Harsh Chaudhary 3 years, 11 months ago
- 3 answers
Yogita Ingle 4 years ago
The gold is $50000 per 10 gm
Rate of 1 gm gold = 50000 /10 = 5000
Hence the rate of 10000gm = 5000 × 10000 = $50000000
Posted by Harsh Srivastav Harsh Srivastav 4 years ago
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Posted by Neha Jangid ??? 4 years ago
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Posted by Sakti Swarupa Sahu 4 years ago
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Yogita Ingle 4 years ago
In Maths, integers are the numbers which can be positive, negative or zero. The examples of integers are, 1, 2, 5,8, -9, -12, etc. ...
Posted by Sumedha Kamble 4 years ago
- 1 answers
Gaurav Seth 4 years ago
Let O and O’ be the centres of the circles of radii 5 cm and 3 cm respectively and let PQ be their common chord.
OP = 5 cm, O’P = 3 cm, OO’ = 4 cm.
Let OL = x, so LO’ = 4 - x
Let PQ = y, so PL = y/2
In right triangle OLP,
OL2 = (OP2 – LP2) = 52 – (y/2)2
=> x2 = 25 – y2/4 …(i)
In right triangle O’LP, O’L2 = (32 – y2/4)
(4 – x)2 = 9 – y2/4 …(ii)
From (i) and (ii) we get
x2 – 16 + 8x – x2 = 25 – y2/4 – 9 + y2/4
=> 8x =32
=> x = 4
From equation (i)
16 = 25 – y2/4 => y2/4 = 9 => y2 = 36 => y = 6
Thus, the length of the common chord is 6 cm.
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