By what percentage has the triangle's height decreased?

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The base of a triangle increases in length by $25$%. Its area stays the same. Evaluate by what percentage has its height decreased?

  • My answer is $20$% whereas my teacher's answer is $80$%. The question looks simple but it really made me perplexed about it. Any help would be appreciated. Thanks!
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area of triangle = Ar$\triangle$=${1\over2}*b*h$ ,

-where $b=$base lenght and $h=$height given , base increases by $25$% $ \implies b'=b+{b\over4}={5b\over4} $ ,$b'=$new base and $h'$=new height

-Ar$\triangle$=${1\over2}*b*h={1\over2}*{5b\over4}*h'\implies h'={4h\over5}$

  • $\implies$ decrease in height = $20$%
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The base of a triangle increases in length by 25%. Its area stays the same. Evaluate by what percentage has its height decreased?

Since the area is same, $$ \frac{1}{2} \ . \text{base} \ . \text{height} = \text{constant} \implies base * height = \text{constant}$$

Let the inital base and height be $ b, h$.

Let the final base and height be $ b_1, h_1$

$$ bh = b_1h_1 \implies \frac{b}{b_1} = \frac{h_1}{h} $$

Hence after taking reciprocal, we get: $$ \frac{h}{h_1} = \overbrace{\frac{1.25}{1}}^{25\% \ \text{increase}} \implies h = 1.25h_1 \implies 0.8h = h_1 $$

Hence percentage decrease in height = $$ \frac{h_1 - h}{h} * 100 $$

which amounts to $$ \frac{0.8h - h}{h} * 100 = -20\% $$

Or you have a decrease of 20%.

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For the two triangles, we have the following. $$A_1= \frac{1}{2}\cdot b_1\cdot h_1$$ $$A_2 = \frac{1}{2}\cdot b_2\cdot h_2$$ The two triangles have the same area, and the second triangle’s base is $25$% longer ($\frac{5}{4}$ times longer). $$A_1 = A_2$$ $$\frac{1}{2}\cdot b_1\cdot h_1 = \frac{1}{2}\cdot b_2\cdot h_2$$ $$b_1\cdot h_1 = b_2\cdot h_2$$ $$b_2 = \frac{5}{4}\cdot b_1$$ $$b_1\cdot h_1 = \frac{5}{4}\cdot b_1\cdot h_2$$ $$h_1 = \frac{5}{4}\cdot h_2$$ $$h_2 = \frac{h_1}{\frac{5}{4}} \implies \boxed{h_2 = \frac{4}{5}\cdot h_1}$$ The second triangle’s height is $80$%, or $\frac{4}{5}$, of the first triangle’s height, meaning it has decreased by $20$%, or $\frac{1}{5}$.