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pic is attached ...please solve with diagram ...please

pic is attached ...please solve with diagram ...please

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Grade:12

1 Answers

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

241-2005_pTIUp.png
Arrangement is something like this, but in place of y, we have x & vice-versa.
Area of rectangle:
xy…..........(1)
Area of triangle:
\frac{1}{2}bh…........(2)
Now, we have three other portion remaining in the triangle other than rectangle. We are going to compare the area.
Area of bottom two triangles:
\frac{1}{2}x.(b-y)….....(3)
Area of upper triangle:
\frac{1}{2}y.(h-x)..........(4)
Now, we have
(1) +(3)+(4) = (2)
xy + \frac{1}{2}y(h-x)+\frac{1}{2}x(b-y) = \frac{1}{2}bh
hy + bx = bh
y = \frac{b}{h}(h-x)
Perimeter of rectangle:
2(x+y)
2(x+b-\frac{b}{h}x)
2(b+(1-\frac{b}{h})x)
Area of rectangle:
xy
\frac{bx}{h}(h-x)

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