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The area of a triangle is given by, A=\frac{1}{2}bh,
The change in area (if there is any) is given by, (1+\frac{\delta A}{A})=(1+\frac{\delta b}{b})(1+\frac{\delta h}{h}).
Here, \frac{\delta b}{b}=+0.4 and \frac{\delta h}{h}=-0.4.
Hence, \frac{\delta A}{A}=-0.16.
Therefore, Area decreases by 16%.
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Let initial base=b,
area, A₁ = 0.5 bh

after height is decreased by 40% and base is increased by 40%,
base = 1.4b
height = 0.6h
Area, A₂ = 0.5×1.4b×0.6h = 0.42 bh

A₂ - A₁ = 0.42 bh - 0.5 bh = -0.08 bh
-ve sign indicates that area decreased.

% decrease = (0.08bh/0.5bh)×100 = 16%
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