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Ask QuestionPosted by Arun Kushwah 5 years, 3 months ago
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Posted by Pradeep Garg 5 years, 3 months ago
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Posted by Prithivi Raj 5 years, 3 months ago
- 4 answers
Posted by Sara Biju 5 years, 3 months ago
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Sara Biju 5 years, 3 months ago
Tannu Chauhan 5 years, 3 months ago
Posted by Prince Fozdar 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR.
Solution:
Consider the ΔPQR. PRS is the exterior angle and QPR and PQR are interior angles.
So, PRS = QPR+PQR (According to triangle property)
Or, PRS -PQR = QPR ———–(i)
Now, consider the ΔQRT,
TRS = TQR+QTR
Or, QTR = TRS-TQR
We know that QT and RT bisect PQR and PRS respectively.
So, PRS = 2 TRS and PQR = 2TQR
Now, QTR = ½ PRS – ½PQR
Or, QTR = ½ (PRS -PQR)
From (i) we know that PRS -PQR = QPR
So, QTR = ½ QPR (hence proved).
Posted by Prince Fozdar 5 years, 3 months ago
- 1 answers
Posted by Mahesh Gujjar 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR.
Solution:
Consider the ΔPQR. PRS is the exterior angle and QPR and PQR are interior angles.
So, PRS = QPR+PQR (According to triangle property)
Or, PRS -PQR = QPR ———–(i)
Now, consider the ΔQRT,
TRS = TQR+QTR
Or, QTR = TRS-TQR
We know that QT and RT bisect PQR and PRS respectively.
So, PRS = 2 TRS and PQR = 2TQR
Now, QTR = ½ PRS – ½PQR
Or, QTR = ½ (PRS -PQR)
From (i) we know that PRS -PQR = QPR
So, QTR = ½ QPR (hence proved).
Posted by Arsha S Kumar 5 years, 3 months ago
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Posted by Gurkaran Singh 5 years, 3 months ago
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Posted by Ankush Sharma 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above, taking the first interval as 0–5 (5 not included). What main features do you observe from this tabular representation ?
<hr />(i)
We observe the following main features from this tabular representation:
(i) The distances (in km) from their residence to their work place of the maximum number of engineers are in the third interval, i.e., 10–15.
(ii) The distances (in km) from their residence to their work place of the minimum number of engineers are in the intervals 20–25 and 25–30 each.
(iii) The frequencies of the intervals 20–25 and 25–30 are the same. (Each = 1)
Posted by Hridayansh Periwal 5 years, 3 months ago
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Posted by V. Suman V. Suman Raj 5 years, 3 months ago
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Posted by Anik Das 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
let AB and CD be two lines intersecting at O . They lead to two pairs of vertically opposite angles, namely,
(i) ∠ AOC and ∠ BOD (ii) ∠ AOD and ∠ BOC.
We need to prove that ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC.
Now, ray OA stands on line CD.
Therefore, ∠ AOC + ∠ AOD = 180° (Linear pair axiom) ………..(1)
We can write ∠ AOD + ∠ BOD = 180° (Linear pair axiom)……………(2)
From (1) and (2), we can write
∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD
This implies that ∠ AOC = ∠ BOD
Therefore it is proved lines intersect each other then the vertically opposite angles are equal.
Posted by Zoya Sayad 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
Multiplying both numerator and denominator with numerator
Posted by Shubham Raj 5 years, 3 months ago
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Shubham Raj 5 years, 3 months ago
Posted by Likhith Liki 5 years, 3 months ago
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Posted by Shivani Shah 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
Identity : (x +a) (x + b ) = x2 + (a+b)x + ab
(x+4)(x+10) = x2 + ( 4 + 10)x + 4(10)
= x2 + 14x + 40
Posted by Charmee Yadav 5 years, 3 months ago
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Posted by Taufeeq Lone 5 years, 3 months ago
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Sakshi Jagtap 5 years, 3 months ago
Posted by Taufeeq Lone 5 years, 3 months ago
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Posted by Akarshit Narula 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
1. Determine which of the following polynomials has (x + 1) a factor:
(i) x3+x2+x+1
Solution:
Let p(x) = x3+x2+x+1
The zero of x+1 is -1. [x+1 = 0 means x = -1]
p(−1) = (−1)3+(−1)2+(−1)+1
= −1+1−1+1
= 0
∴By factor theorem, x+1 is a factor of x3+x2+x+1
(ii) x4+x3+x2+x+1
Solution:
Let p(x)= x4+x3+x2+x+1
The zero of x+1 is -1. . [x+1= 0 means x = -1]
p(−1) = (−1)4+(−1)3+(−1)2+(−1)+1
= 1−1+1−1+1
= 1 ≠ 0
∴By factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1
For more click on the given link:
<a href="https://mycbseguide.com/blog/ncert-solutions-class-9-maths-exercise-2-4/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-class-9-maths-exercise-2-4/&ved=2ahUKEwjZsuKznvLsAhVFwzgGHeNlBs8QFjADegQIBRAC" rel="noopener" target="_blank">NCERT Solutions for Class 9 Maths Exercise 2.4 ...</a>
Posted by Shamshad Alam 5 years, 3 months ago
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Posted by Sushmit Dwivedi 5 years, 3 months ago
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Posted by Sameer मोहम्मद 5 years, 3 months ago
- 5 answers
Posted by Varni Doshi 5 years, 3 months ago
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Sahil Nara 5 years, 3 months ago
Posted by Arun J 5 years, 3 months ago
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Gaurav Seth 5 years, 3 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.
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