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

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

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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Hello Student,
Please find answer to your question below

241-723_pTIUp.png

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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