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Ask QuestionPosted by Harsh Maurya 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 11 months ago
In △ADF and △ECF , we have
∠ADF = ∠ECF [alt.int.∠s]
AD = EC [∵ AD = BC and BC = EC]
∠DFA = ∠CFE [vert. opp. ∠s]
∴ By AAS congruence rule ,
△ADF ≅ △ECF
⇒ DF = CF [c.p.c.t.]
⇒ ar(△ADF) = ar(△ECF)
Now, DF = CF
⇒ BF is a median in △BDC
⇒ ar(△BDC) = 2 ar(△DFB)
= 2 × 3 = 6 cm2 [∵ar(△DFB) = 3 cm2]
Thus, ar(||gm ABCD) = 2 ar(△BDC)
= 2 × 6 = 12 cm2
Posted by Mamta Patidar 3 years, 11 months ago
- 0 answers
Posted by Mamta Patidar 3 years, 11 months ago
- 3 answers
Posted by Jahanvi Thakur 3 years, 11 months ago
- 3 answers
Chaitanya Saini 3 years, 11 months ago
Yogita Ingle 3 years, 11 months ago
(x + a) ( x + b) = x2 + (a+b)x + ab
(x+4) (x+10) = x2 + ( 4 + 10)x + 4(10)
= x2 + 14x + 40
Yaduvanshi## Vansh$$ 3 years, 11 months ago
Posted by Aiswarya Raanee .K 3 years, 11 months ago
- 4 answers
Khushi Gupta 3 years, 11 months ago
Posted by Aman Rangi 3 years, 11 months ago
- 2 answers
Posted by Shubham O 3 years, 11 months ago
- 3 answers
Khushi Gupta 3 years, 11 months ago
Ayushi ... 3 years, 11 months ago
Posted by Amaan Ahmad Salmani 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 11 months ago
Given,
area of curved surface of a right circular cylinder is =6600 cm^2...(1)
circumference of its base is =55 cm...(2)
volume of a right circular cylinder is= 1100.......(3)
curved surface of a right circular cylinder is=2πrh. ... from (1)
=6600 cm^2...(4)
where, r=radius of the base of the cylinder,
h=height of the cylinder
circumference of the base of the cylinder=2πr=55 cm..... from (2)....(5)
where, r=radius of the base of the cylinder
2π×r=2×3.14159265×r
=6.2831853×r=55
r=55÷6.28=8.75 cm {it's no need to find radius of the base of the cylinder for finding out height of the cylinder} ...(6)
from (4),
2πrh=6600
55×h=6600... from (5)
.•. h=6600÷55=125 cm
Therefore the height of the cylinder is=125 cm
Posted by Harsh Jha 3 years, 11 months ago
- 2 answers
Vinod Saini 3 years, 11 months ago
Posted by Suyog Sonar 3 years, 11 months ago
- 3 answers
Posted by Mahisha G.T. 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 11 months ago
Here , class width = 20
class mark = 70
Half of the class width =20 /2 =10
Upper limit of first class interval = 70 + 10 = 80
Lower limit of first class interval = 70 – 10 = 60
Thus, class interval becomes 60 – 80
So, frequency distribution table becomes :
(a) Number of oranges weights more than 180 g = 1 + 1 = 2
(b) Number of oranges weights less than 100 g = 3 + 10 = 13
Posted by Devansh Singh 3 years, 11 months ago
- 1 answers
Posted by Vagisha Choudhary 3 years, 11 months ago
- 0 answers
Posted by Krishna Parchani 3 years, 11 months ago
- 0 answers
Posted by Akshat Panwar 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
Given : LCM =225 and HCF =15
Let x=LCM/HCF = 225/15 =15.
So, x can be written as a product of 3,5 and also the product of 1,15 which are relatively prime.
Thus, there exist two such pairs where LCM =225 and HCF =15.
Posted by Chanda Kahar 3 years, 11 months ago
- 1 answers
Yogita Ingle 3 years, 11 months ago
Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC
Area of ∆ABC ::•• •●
heron's formula
{where, s is semi perimeter}
area of ∆ABC
Area of∆ADC ::•• •●
Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC
So the Area of Quadrilateral ABCD = 15.2cm .sq
Posted by Ritika Arya? 3 years, 11 months ago
- 3 answers
Posted by Vipresh Yadav 3 years, 11 months ago
- 0 answers
Posted by Parth Mahawer 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 11 months ago
➧ Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'.
➧ Hence OA = OB = 5 cm
➾ O'A = O'B
➾ 3 cm
➧ OO' is the perpendicular bisector of chord AB.
➧ Therefore, AC = BC
➧ Given, OO' = 4 cm
➧ Let OC = x
➧ Hence O'C = 4 − x
➧ In right angled ΔOAC, by Pythagoras theorem OA²
➾ OC² + AC²
➾ 5² = x² + AC²
➾ AC² = 25 − x²à
➧ (1) In right angled ΔO'AC, by Pythagoras theorem O'A²
➾ AC² + O'C²
➾ 3² = AC² + (4 – x)²
➾ 9 = AC² + 16 + x² − 8x
➾ AC² = 8x − x² − 7 à
➧ (2) From (1) and (2),
➧ we get 25 − x² = 8x − x² − 7 8x = 32
➧ Therefore, x = 4
➧ Hence, the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.
➧ AC² = 25 − x²
➾ 25 − 4²
➾ 25 − 16 = 9
➧ Therefore, AC = 3 m Length of the common chord,
➾ AB = 2AC
➾ 6 cm
Posted by Rani ? 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
3x+5/2x+1=1/3
3(3x + 5) = 1 (2x + 1)
3(3x) + 3(5) = 2x + 1
9x + 15 = 2x + 1
9x - 2x = 1 - 15
7x = -14
x = -14/7 = -2
Posted by Sahil Nara 3 years, 11 months ago
- 1 answers
Mamta ... 3 years, 11 months ago
Posted by Kumkum Prajapati 3 years, 11 months ago
- 3 answers
Posted by Sahil Singh 3 years, 11 months ago
- 1 answers
Posted by Mukul Kumar 😎 3 years, 11 months ago
- 3 answers
Posted by Mukul Kumar 😎 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
The area of an isosceles triangle is given by the following formula:
| Area = ½ × base × Height |
Posted by Mukul Kumar 😎 3 years, 11 months ago
- 5 answers
Posted by Rishab Bagul 3 years, 11 months ago
- 3 answers
Manshi Dangi 3 years, 11 months ago
Mukul Kumar 😎 3 years, 11 months ago
Posted by Unknown . 3 years, 9 months ago
- 3 answers
Yogita Ingle 3 years, 11 months ago
We have integers 1,2, 3,…1000
We have integers 1,2, 3,…1000, n(S) = 1000
Number of integers which are multiple of 2 = 500
Let the number of integers which are multiple of 9 be n.
nth term = 999
=> 9 + (n -1)9 = 999
=> 9 + 9n – 9 = 999
=> n = 111
From 1 to 1000, the number of multiples of 9 is 111.
The multiple of 2 and 9 both are 18, 36,…, 990.
Let m be the number of terms in above series.
.’. mth term = 990
=> 18 + (m- 1)18 = 990
=> 18+18m-18 = 990
=> m = 55
Number of multiples of 2 or 9 = 500 +111-55 = 556 = n(E)
Required probability = n(E)/n(S) = 556/1000 = 0.556
Posted by Jeevitha R 3 years, 11 months ago
- 2 answers
Pradhuman Sharma 3 years, 11 months ago
Sakshi Jagtap 3 years, 11 months ago
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Varun Panshotra 3 years, 11 months ago
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