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Correct me if i am wrong.

The hypotenuse of the right angle triangle 28.5. The maximum length of the hypotenuse formed by the two sides (12 & 12) is always less than 24. How can this be the base of the first triangle?
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IMO C
A. Two longest sides are at 90 degree.Not Sufficient
B. The length of sides is 18, 12,12,22....we cannot find the area od quadrilaterial just by knowing length of sides...
A+B...two longest sides are at 90...therefor we can find the area of the triangle formed by these sodes with diagonal
as well as the lenght of diagonal. This diagonal forms a isosceles triangle with the other two sides(12,12)...and area of a isoscles triangle can be found if all of its sides are known. therefore Sufficient.
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mehtakaustubh,

I am confused about how to form the second triangle. The length of the diagonal of the right angled triangle is 28.5. How can we form the second triangle with the sides 12, 12 & 28.5 because i think it violates the basic rules of triangles. Here the third side is greater than the sum of the other two. Can you please clarify?
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TARGET730
mehtakaustubh,

I am confused about how to form the second triangle. The length of the diagonal of the right angled triangle is 28.5. How can we form the second triangle with the sides 12, 12 & 28.5 because i think it violates the basic rules of triangles. Here the third side is greater than the sum of the other two. Can you please clarify?

I agree with Target here. You cannot even form a closed profile with those measurements and a 90° angle between the two longest sides.
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TARGET730
Correct me if i am wrong.

The hypotenuse of the right angle triangle 28.5. The maximum length of the hypotenuse formed by the two sides (12 & 12) is always less than 24. How can this be the base of the first triangle?

Target730,

I think you're right. There is no way we can get a triangle with side lengths 12x12x28.8

I suppose, the correct answer is different from the provided one.
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desaichinmay22
What is the area of a plot of land which is in the shape of a quadrilateral?

(1) Only one angle, the one formed by the intersection of the two longest sides of the plot, measures 90 degree

(2) The sides of the quadrilateral measure 18m, 12m, 12m and 22m
(1)other angels /sides not known.....Not suff ....
(2)Angles not known

combined...
Please follow attached fig.
we can saperatly calculate area of both triangles=area of quadrilateral

Suff...
Ans C
Attachments

nw.jpg
nw.jpg [ 21.67 KiB | Viewed 4637 times ]

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TARGET730
IMO E

1 => NOT SUFFICIENT. The sides are not mentioned.
2 => NOT SUFFICIENT. We can have multiple quadrilaterals

1 & 2 => NOT SUFFICIENT. We cannot form a quadrilateral combining 1 & 2. See the attachment. The hypotenuse of the triangle formed by the sides 18 and 22 is square root of 808 = 28.5. The remaining two sides are 12 & 12, so we cannot create a quadrilateral from the remaining sides. is my logic correct?

Can't we calculate area by using the below formula?
sqrt (s(s-a)(s-b)(s-c)(s-d))

where s = (a+b+c+d)/2
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TARGET730
IMO E

1 => NOT SUFFICIENT. The sides are not mentioned.
2 => NOT SUFFICIENT. We can have multiple quadrilaterals

1 & 2 => NOT SUFFICIENT. We cannot form a quadrilateral combining 1 & 2. See the attachment. The hypotenuse of the triangle formed by the sides 18 and 22 is square root of 808 = 28.5. The remaining two sides are 12 & 12, so we cannot create a quadrilateral from the remaining sides. is my logic correct?

Can't we calculate area by using the below formula?
sqrt (s(s-a)(s-b)(s-c)(s-d))

where s = (a+b+c+d)/2

VeritasPrepKarishma

Can you please clarify my query?
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Answer is wrong. There are two quadilaterals possible under the two conditions given - one concave and one convex...
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