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T R E N D _____F F 3 years, 2 months ago (13854947)
Posted by Shivam Singh 3 years, 2 months ago (13531220)
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Posted by Soumya Soumya 3 years, 2 months ago (13857107)
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Preeti Dabral 3 years, 2 months ago (2983787)
Let the other parallel side be x . We know that the area of trapezium is given by the formula :-
Posted by Shivam Singh 3 years, 2 months ago (13531220)
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Preeti Dabral 3 years, 2 months ago (2983787)
Cloud computing is on-demand access, via the internet, to computing resources—applications, servers (physical servers and virtual servers), data storage, development tools, networking capabilities, and more—hosted at a remote data center managed by a cloud services provider (or CSP).
Posted by Radhika Rawat 3 years, 2 months ago (9712867)
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Posted by Soumya Soumya 3 years, 2 months ago (13857107)
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Preeti Dabral 3 years, 2 months ago (2983787)
The parallel sides of trapezium are 25 cm and 13 cm and the distance between them is 8 cm.
{tex}\therefore{/tex}Area of trapezium = {tex}\frac{1}{2}{/tex} (sum of parallel sides) {tex}\times{/tex} (perpendicular distance between them)
{tex}=\frac{1}{2}(25+13) \times 8{/tex}
= {tex}38 \times 4{/tex}
= 152 cm2
Posted by Soumya Soumya 3 years, 2 months ago (13857107)
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Preeti Dabral 3 years, 2 months ago (2983787)
Given : Length of the rectangular lawn = 25 m
Breadth of the rectangular lawn = 16 m
If lawn is surrounded externally by a path which is 3 m wide. , then length of the path l= 25 m+2(3) m = 31 m
Width of the path b= 16 m +2(3m) = 22 cm
Area of path = l x b - (Area of lawn)
= 31 x 22 - (25 x 16) [area of rectangle = length x breadth]
=682 -400 = 282 sq. m
Rate of levelling = 3.50 per m square
Then, Cost of leveling = (Rate of levelling ) x (Area of path)
= 3.50 x (282)= Rs 987
Hence, the cost of leveling is Rs 987.
Posted by Chand Vishal 3 years, 2 months ago (13857227)
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Chahat Gee 3 years, 1 month ago (14034523)
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Posted by Riya Yadav 3 years, 2 months ago (13857222)
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Gurman Singh 3 years, 2 months ago (13834107)
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Preeti Dabral 3 years, 2 months ago (2983787)
Given, {tex}3x -{/tex} {tex}\frac{y+7}{11}{/tex} {tex}+ 2 = 10{/tex}
{tex}\Rightarrow \frac{33 x-(y+7)+22}{11}=10{/tex}
{tex}\Rightarrow{/tex} {tex}33x - y - 7 + 22 = 110{/tex}
{tex}\Rightarrow{/tex} {tex}33x - y = 95{/tex} ..(i)
Also, {tex}2 y+\frac{x+11}{7}{/tex} = 10
{tex}\Rightarrow \quad \frac{14 y+x+11}{7}{/tex} = 10
{tex}\Rightarrow{/tex} {tex}14y + x + 11 = 70{/tex}
{tex}14y + x = 59{/tex}
{tex}14y = 59 - x{/tex}
{tex}\Rightarrow{/tex} y = {tex}\frac{59-x}{14}{/tex}
eq. (i) becomes
33x - {tex}\left(\frac{59-x}{14}\right){/tex} = 95
{tex}\Rightarrow \quad \frac{462 x-59+x}{14}{/tex} = 95
{tex}\Rightarrow{/tex} 463x - 59 = 1330
{tex}\Rightarrow{/tex} x = 3
When x = 3 eq. (i) becomes
33(3) - y = 95
{tex}\Rightarrow{/tex} y =4
Posted by Soumya Soumya 3 years, 2 months ago (13857107)
- 2 answers
Preeti Dabral 3 years, 2 months ago (2983787)
Radius of the larger semicircle 14 cm
{tex}\therefore {/tex} Area of the larger semicircle
{tex}= \frac { 1 } { 2 } \pi ( 14 ) ^ { 2 } = \frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 196{/tex}
{tex}= 308 \mathrm { cm } ^ { 2 }{/tex}
Radius of each smaller semicircle = 7 cm
{tex}\therefore {/tex} Area of two smaller semicircles
{tex}= 2 \left[ \frac { 1 } { 2 } \times \pi ( 7 ) ^ { 2 } \right]{/tex}
{tex}= \frac { 22 } { 7 } \times ( 7 ) ^ { 2 } = 154 \mathrm { cm } ^ { 2 }{/tex}
{tex}\therefore {/tex} The area of the shaded region
{tex}= 308 \mathrm { cm } ^ { 2 } + 154 \mathrm { cm } ^ { 2 } = 462 \mathrm { cm } ^ { 2 }{/tex}
Posted by Sukhmanjeet Singh 3 years, 2 months ago (13245109)
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Preeti Dabral 3 years, 2 months ago (2983787)
Grooming is a term associated with neat and clean appearance.
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Jay Srivastav 1 year, 9 months ago (10536962)
4 Thank You