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Preeti Dabral 3 years, 1 month ago
Given:
(x-y)²-4(x-y)
To Find:
The value of the given expression.
Solution:
Expanding the given expression,
x² - 2xy + y² - 4x + 4y.
x² - 4x - xy + y² + 4y - xy.
x(x - 4 - y) + y(y + 4 - x).
x(x - y - 4) - y(x - y - 4).
(x - y)(x - y - 4).
Hence, the factors of the given expression is (x - y)(x - y - 4).
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Posted by Rita Vinod Sharma 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
| Histogram | Bar Graph |
| 1. It consists of rectangles touching each other from each other with equal space | It consists of rectangles normally separated |
| 2. The frequency is represented by the area of each rectangle | The frequency is represented by height. The width has no significance. |
| 3. It is two dimensional where both the width (base) and the length (height of the rectangle) are important | It is one dimensional in which only length(height) matters while width is arbitrary. |
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Preeti Dabral 3 years, 1 month ago
Let the other parallel side be x . We know that the area of trapezium is given by the formula :-
Posted by Soumya Soumya 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
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, 1 month ago
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Preeti Dabral 3 years, 1 month ago
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 Soumya Soumya 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
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}
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Bhargavi Yadav 3 years, 1 month ago
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