If the height of a triangle is decreased by 4 | vedclass

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(C) The area of a triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Let the original base be $b$ and original height be $

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$16 \% 	ext{ decrease}$

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(C) The area of a triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Let the original base be $b$ and original height be $h$. The original area is $A_1 = \frac{1}{2}bh$.New base $b' = b + 0.40b = 1.4b$.New height $h' = h - 0.40h = 0.6h$.New area $A_2 = \frac{1}{2} 	imes (1.4b) 	imes (0.6h) = 0.84 	imes (\frac{1}{2}bh) = 0.84A_1$.The change in area is $A_2 - A_1 = 0.84A_1 - A_1 = -0.16A_1$.This represents a $16 \%$ decrease in the area.Alternatively,using the net percentage change formula: $	ext{Change} = x + y + \frac{xy}{100} = -40 + 40 + \frac{(-40)(40)}{100} = -16 \%$.

If the height of a triangle is decreased by $40 \%$ and its base is increased by $40 \%$,what will be the effect on its area?

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