Question

hello, please help me to solve this problem.​

Answer 1

Given Question :-

Find the area of triangle whose sides are along the lines x = 2, y = 0 and 4x + 5y = 20.

$\large\underline{\sf{Solution-}}$

Given that, three sides of a triangle are along the line

$\rm \: 4x + 5y = 20 - - - (1)$

$\rm \: x = 2 - - - - (2)$

$\rm \: y = 0 - - - - (3)$

[ See the attachment ]

So, required area of triangle ABC is given by

$\rm \: = \: \displaystyle\int_2^5\rm y_{(line \: 1)} \: dx$

$\rm \: = \: \displaystyle\int_2^5\rm \dfrac{20 - 4x}{5} \: dx$

$\rm \: = \: \frac{1}{5} \displaystyle\int_2^5\rm (20 - 4x) \: dx$

$\rm \: = \: \frac{1}{5}\bigg[20x -4 \times \dfrac{ {x}^{2} }{2} \bigg] _2^5$

$\rm \: = \: \frac{1}{5}\bigg[20x - {2x}^{2} \bigg] _2^5$

$\rm \: = \: \frac{1}{5}\bigg[20(5 - 2) - 2({5}^{2} - {2}^{2}) \bigg]$

$\rm \: = \: \frac{1}{5}\bigg[20 \times 3 - 2(25 - 4) \bigg]$

$\rm \: = \: \frac{1}{5}\bigg[60 - 2(21) \bigg]$

$\rm \: = \: \frac{1}{5}\bigg[60 - 42 \bigg]$

$\rm \: = \: \frac{1}{5} \times 18$

$\rm \: = \: 3.6 \: square \: units$

Hence,

$\bf\implies \:Required\:Area = \: 3.6 \: square \: units$

$\rule{190pt}{2pt}$

Alternative Method :-

$\rm \: Required\:Area \: of \: \triangle \: ABC$

$\rm \: = \: \dfrac{1}{2} \times AB \times BC$

$\rm \: = \: \dfrac{1}{2} \times 2.4 \times 3$

$\rm \: = \: 1.2 \times 3$

$\rm \: = \: 3.6 \: square \: units$

Hence,

$\bf\implies \:Required\:Area = \: 3.6 \: square \: units$