Question

If the height of a triangle is decreased by 40% and it's base is increased by 40%, what will be the effect on its area ?​

Answer 1

Step-by-step explanation:

hence area decrease by 16%

Answer 2

Step-by-step explanation:

$let \: the \: original\: height\: \& \:base \: of \: the \: triangle$ $be \: h \: \& \: b$.

$Then, \: ar. \: of \: original \: ∆ = \frac{1}{2}×b×h$

$\underline{\bold{After \: change}}$

$new \: height \: of \: the \: ∆ = & h - 40 \% \: of \: h & \\ &= h -\frac{40h}{100} \\ &= h- \frac{4h}{10} & \\ &= \frac{6h}{10} &$

$new \: base \: of \: the \: ∆ & = b+ 40 \% of b \\ & = b+\frac{40b}{100} \\ & = b+\frac{4b}{10} \\ & = \frac{14b}{10}$

$So, \: ar. \: of \: new \: ∆ & = \frac{1}{2}×\frac{14b}{10}×\frac{6h}{10} \\ & = \frac{1}{2}×\frac{84×b×h}{100} \\ & = \frac{84}{100}×\big(\frac{1}{2}×b×h\big)$

$\% \: increment \: in \: ar. \: of \: ∆$$& = \frac{ar. \: of \: original \: ∆ - ar. \: of \: new \: ∆}{ar. \: of \: original \: ∆ }×100 \\ & = \frac{(\frac{1}{2}×b×h) - \frac{84}{100}×(\frac{1}{2}×b×h)}{\frac{1}{2}×b×h}×100 \\ & =\frac{(1-\frac{84}{100})×(\frac{1}{2}×b×h)}{\frac{1}{2}×b×h}×100 \\ & = 1-\frac{84}{100}×100 \\ & =\frac{100-84}{100}×100 \\ & = 16\%$