Question

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?

• A. Decreased by $16\%$
• B. Increased by $16\%$
• C. Decreased by $20\%$
• D. Increased by $20\%$

Answer 1

The correct option is A Decreased by $16\%$

Let height be $x$ and base be ${}^{\prime }{y}^{\prime }$

$\therefore initialarea=\frac{1}{2}\times x\times y$

$=\frac{xy}{2}$

$Newheight=x(1-\frac{40}{100})$

$=\frac{3x}{5}$

$Newbreadth=y(1+\frac{40}{100})$

$=\frac{7y}{5}$

$Newarea{A}_2=\frac{1}{2}\times \frac{3x}{5}\times \frac{7y}{5}$

$=\frac{21xy}{50}$

$changeinarea$

$A_1=\frac{xy}{2}{A}_2=\frac{21xy}{50}$

taking LCM and bringing $A_1{A}_2$ to denominator we get:

$A_1=\frac{xy\times 25}{2\times 25}$

$A_2=\frac{xy\times 21}{2\times 25}$

=> Area decreases since $A_2<A_1$

$\therefore decreaseinarea=A_1-A_2$

$=\frac{25xy}{50}-\frac{21xy}{50}$

$=\frac{4xy}{50}$

$changeinarea=\frac{decrease}{initial}\times 100\%$

$=\frac{\frac{4xy}{50}}{\frac{xy}{2}}\times 100\%$

$=\frac{4}{50}\times 2\times 100\%$

$=16\%$

$Hence,areadecreasesby16\%$