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Susai Raj 7 years, 6 months ago
ax + ay = a^2 + ab ...(1)
ax + by =a^2 + b^2 ...(2)
(1) - (2) gives
y(a+b) = b(a+b)
So y =b (a+b)/(a+b)
ie. y = b
Substitute y = b in the equation x+y=a+b
x+b = a+b
x = a+b-b. ie x= a
So x=a and y = b
Posted by Dhwani Khushlani 5 years, 8 months ago
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Susai Raj 7 years, 6 months ago
Here a=3, b = 2√5 and c=-5
x= {-b+or-√(b^2-4ac)}/2a
={-2√5+or-√(20-4×3×(-5)}/6
={-2√5 +or-√80}/6
={-2√5 +or- 4√5}/6
x=(-2√5+4√5)/6
or (-2√5-4√5)/6
x =2√5/6 or x= -6√5/6
x=√5/3 or x = -√5
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Sia ? 6 years, 6 months ago
Let the four shaded regions be denoted by I,II,III and IV as shown in the figure.
Given, side of square =10 cm
Diameter of each semi-circle =10 cm
Therefore, Radius of each semi-circle = {tex}\frac{10}{2}=5\;cm{/tex}
Now, area of I region + area of III region= Area of ABCD - Area of two semi-circles each of radius 5 cm
{tex}=(10\times 10-2\times \frac{1}{2}\pi \times 5^2){/tex}
{tex}=(100-3.14\times 25){/tex}
{tex}=100-78.5{/tex}
{tex}=21.5\;cm^2{/tex}
Similarly, area of II region + area of IV region = 21.5 cm2
Therefore, Area of the shaded region = Area of ABCD - Area of (I + II + III + IV) region
{tex}=(100-2\times 21.5){/tex}
{tex}=100-43{/tex}
{tex}=57\;cm^2{/tex}
The area of shaded design in the given figure is 57 cm2.
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