Question

If the height of a triangle is decreased by 40% and its base is increased by 40%, what will be the percentage change in its area?​

Answer 1

Answer:

Area decreased by 16%

Explanation:

Area of a triangle is given by,

$= \dfrac{1}{2} \times b \times h$

Where,

• ‘b’ denotes the base
• ‘h’ denotes the height

Then,

$= \dfrac{bh}{2}$

Now, height decreases by 40%

$= h - (40\% \: { \rm \: of} \: h)$

$= h - \dfrac{40h}{100}$

$= \dfrac{60h}{100}$

And also, the base increases by 40%

$=b + (40\%{ \rm \: of} \: b)$

$= b + \dfrac{40b}{100}$

$= \dfrac{140b}{100}$

∴ New area :-

$= \dfrac{1}{2} \times \dfrac{60h}{100} \times \dfrac{140b}{100}$

$= \dfrac{84bh}{200}$

Since, the new area is less than the original area. It's a decrease in area.

So, area decreased :- New area - Original area.

$= \dfrac{bh}{2} - \dfrac{84bh}{200}$

$= \dfrac{100bh - 84bh}{200}$

$= \dfrac{16bh}{200}$

Decrease% is given by,

$\rm = \dfrac{change \: in \: area}{original \: area} \times 100$

$= \dfrac{ \frac{16bh}{100} }{ \frac{bh}{2} } \times 100$

$= \dfrac{16 \cancel{bh}}{200} \times \dfrac{2}{ \cancel{bh}} \times 100$

$= 16\%$

${ \underline{ \sf{ \therefore{Area \: is \: decreased \: by \: \red{16\%}}}}}$