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Find the area bounded by the lines $3x + 2y=14, 2x - 3y = 5$ in the first quadrant.

  1. $14.95$
  2. $15.25$
  3. $15.70$
  4. $20.35$
in Numerical Ability by Boss (30.8k points) | 518 views
+1
i am getting 183/12=15.25sq units...what is the ans?
0
Yes it is 15.25 provided we take y axis as well for binding as well which is not mentioned in the question
+1
but it z not mentioned in the question if we will take x-axis for binding the area would b different

so if we will take the given two line and 1st quadrant the answer would be 1/2*(14/3)*(7)= 49/3
0
what is the answer by d way @niharika
0
It was actually asked in virtualgate test series and the answer given was 15.25 sq units .That is why I was mentioning that axis should also be mentioned.
0
15.25 sq units
0
solve for x and y, observe that two given lines are perpendicular. so it will form something like rectangle in first quadrant.

now coordinates of rectangle are (0,0) (x,0) (0,y) (x,y)

find area = length * breadth

3 Answers

+2 votes
Best answer
  • $3x+2y=14\rightarrow(1)$
  • $2x−3y=5\rightarrow(2)$

To get the intersection point, we solve the equations and get $x=4,y=1$

We can draw the required area like this 

 

From the above diagram,

Area $=\triangle ABC +\Box OBCE-\triangle CDE$

Area $=\left(\dfrac{1}{2}\times 4\times 6\right)+\left(1\times 4\right)-\left(\dfrac{1}{2}\times\dfrac{3}{2}\times 1\right)$

Area $=12+4-\dfrac{3}{4}=16-0.75=15.25$

So, the correct answer is $(B).$

by Veteran (58.9k points)
edited by
+1

 

@ why u didn't consider triangle DEC, as it is also satisfying question condition?

0

@Shubhm because $DEC$ is not bounded by two lines and $y$-axis.

0
sorry! I made mistake on asking question.

My question is about triangle DCF, say F is point where line AC meets the x-axis. we can find area of that DCF by heron's formula.
+1

My question is about triangle DCF, say F is point where line AC meets the x-axis. we can find area of that DCF by heron's formula.

We can only apply the Heron's formula when all the sides are given in the triangle.

Here, $\triangle DCF = \dfrac{1}{2}(\text{ Base} \times \text{Height}) = \dfrac{1}{2}\bigg(\dfrac{13}{6} \times 1\bigg) = \dfrac{13}{12} $ 

Required area $ = \triangle AOF - \triangle DCF = \dfrac{1}{2}\times \dfrac{14}{3}\times 7 - \dfrac{13}{12} = \dfrac{49}{3}-\dfrac{13}{12} $

$= \dfrac{196-13}{12} = \dfrac{183}{12}=15.25\:\text{Unit}^{2}.$

+6 votes

ans may be like this

by Boss (49.3k points)
0
Hello can the picture be more clear please?
+3 votes
Bt how it will come 15.25 square units can someone expkain plz
by Loyal (5.7k points)
Answer:

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