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Differential Calculus> please solve.........reply soon sir.........

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please solve.........reply soon sir.............

milind , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

In place of y, we have x & vice – versa.
Area of triangle ABC
$\frac{1}{2}bh$ …..........(1)
Area of rectangle:
$xy$ …...........(2)
Area of triangle CDE
$\frac{1}{2}y(h-x)$ …............(3)
Area of bottom two triangle:
$\frac{1}{2}x(b-y)$ ….............(4)
Now we are going to compare the area’s
$(1)=(2)+(3)+(4)$
$\frac{1}{2}bh = xy + \frac{1}{2}y(h-x) + \frac{1}{2}x(b-y)$
$bh = hy + bx$
$y = \frac{b}{h}(h-x)$
Perimeter of rectangle:
$2(x+y)$
$2(x+b-\frac{bx}{h})$
$2(b+x(1-\frac{b}{h}))$
Area of rectangle:
$xy$
$\frac{bx}{h}(h-x)$

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