Question

14 .Find the area of triangle bounded by a line 4x+3y=48 with coordinate axes. ​

Answer 1

Step-by-step explanation:

Given: Line:4x+3y=48

To find:Find the area of triangle bounded by a line 4x+3y=48 with coordinate axes. 

Solution:

To find the area,

Firstly draw the line in graph paper.

Put x=0

$4(0) + 3y = 48 \\ \\ 3y = 48 \\\\ y = \frac{48}{3} \\ \\ y = 16$

Point A(0,16)

Put y=0

$4x + 3(0) = 48 \\ \\ 4x = 48 \\ \\ x = \frac{48}{4} \\ \\ x = 12$

Point B(12,0)

Plot this line,Please see in attachment.

Now,To find the area of triangle ∆ABC,

as it is right triangle

$ar( \triangle \: ABC) = \frac{1}{2} \times base \times height \\ \\ ar( \triangle \: ABC) = \frac{1}{2} \times AC \times CB \\ \\ = \frac{1}{2} \times 16 \times 12 \\ \\ = 8 \times 12 \\ \\ ar( \triangle \: ABC) = 96 \: ( {unit)}^{2}$

Thus,area enclosed by given line and coordinate axes is 96 sq-unit.

Hope it helps you.

To learn more on brainly:

1)draw the graphs of 2x+y=6 and 2x-y+2=0. shade the region bounded by these lines and x axis. find the area of shaded region

https://brainly.in/question/5822114

2)Draw the graph of the following equations : x = 0 ,y= 0 ,x+y = 3 . Also find the area enclosed between the lines .

https://brainly.in/question/2765948

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

TO FIND THE AREA USING DEFINITE INTEGRAL

The area bounded by the curve y = f(x), x axis and by the lines x = a, x= b is

$= \displaystyle \int\limits_{a}^{b} y \, dx = \int\limits_{a}^{b} f(x) \, dx$

CALCULATION

The given equation of the line is

$4x + 3y = 48 \: \: .........(1)$

TO FIND THE POINT OF INTERSECTION OF THE GIVEN LINE AND X- AXIS

The equation of x axis is

$y = 0$

Putting y = 0 in Equation (1) we get

$4x = 48$

$\implies \: x = 12$

So the point of intersection is ( 12, 0 )

TO FIND THE POINT OF INTERSECTION OF THE GIVEN LINE AND Y - AXIS

The equation of y axis is

$x = 0$

Putting x = 0 in Equation (1) we get

$3y = 48$

$\implies \: y = 16$

So the point of intersection is ( 0 , 16 )

From Equation 1

$\displaystyle \: y = \frac{48 - 4x}{3} = 16 - \frac{4x}{3} \: \: \: ..... ...(2)$

The shaded region is bounded by given line ( Equation 1 ) and coordinate axes

RESULT

SO the required area is

$= \displaystyle \int\limits_{0}^{12} \bigg(16 - \frac{4x}{3} \bigg)\, dx \: \: \: \: \: \:$

$= \displaystyle \: \: \bigg[ 16x - \frac{ 2{x}^{2} }{3} \bigg] _0^{12} \: \: \: \: \: \sf{sq \: unit}$

$= \displaystyle \: \: \bigg[ (16 \times 12) - \frac{ 2 \times {(12)}^{2} }{3} \bigg] - 0 \: \: \: \: \sf{sq \: unit}$

$= (192 - 96) \: \sf{ \: sq \: unit}$

$= 96 \: \sf{ \: sq \: unit}$