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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. $20.35$
| 518 views
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i am getting 183/12=15.25sq units...what is the ans?
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Yes it is 15.25 provided we take y axis as well for binding as well which is not mentioned in the question
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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
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what is the answer by d way @niharika
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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.
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15.25 sq units
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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

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

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

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

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

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

ans may be like this

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

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