# The base of a triangle is increased by 66.6% and the altitude is decreased by 40%, then the change in the area of the triangle is ?

Jul 30, 2018

:. color(maroon)(" there is no change in area of the triangle due to the alteration."

#### Explanation:

Let the original base of the triangle be b and height be h.

$\text{Area of the triangle } = \frac{1}{2} b h$

After altering the triangle, "base " = b_1 = b + b * 66.6% = 1.666 b

"sCorresponding change in heifht " h_1 = h - h * 40% = 0.6 h

$\text{Area of changed triangle } = {A}_{1} = \left(\frac{1}{2}\right) 1.66 b \cdot 0.6 b = \frac{1}{2} b h$

:. color(maroon)(" there is no change in area of the triangle due to the alteration."

Jul 30, 2018

 % "decrease in Area"=0.04%.

#### Explanation:

Suppose that the base and altitude of the triangle are $b$ and $a$

units, resp.

Then, its area $A = \frac{1}{2} \cdot b \cdot a$ sq. unit.

The base $b$ is increased by 66.6%.

$\therefore \text{ the new base} = b + b \cdot \frac{66.6}{100} = 1.666 b$.

Similarly, the new altitude$= 0.6 a$.

$\therefore \text{ New Area A'} = \frac{1}{2} \left(1.666 b\right) \left(0.6 a\right) = \frac{1}{2} \left(0.9996\right) b a$ sq.unit.

$\therefore \text{Decrease in Area} = A - A ' = \frac{1}{2} b a - \frac{1}{2} \left(0.9996\right) b a$,

$= \frac{1}{2} \left(0.0004\right) b a$ sq.unit.

:. % "decrease in Area"={1/2(0.0004)ba}/(1/2*b*a)xx100,

=0.04%.

Jul 30, 2018

$\text{no change}$

#### Explanation:

"note that "66.6%~~2/3" and "40%=2/5

$\text{original area } = \frac{1}{2} b h$

$\text{new area with "b=5/3b" and } h = \frac{3}{5} h$

$\text{new area } = \frac{1}{2} \times \frac{5}{3} b \times \frac{3}{5} h = \frac{1}{2} b h$

$\text{the area remains unchanged}$