In triangle PQR, if the length of all the three sides are distinct int

In triangle PQR, if the length of all the three sides are distinct integers, what is the area of triangle PQR?


I. \(PQ^2 = PR^2 + RQ^2\)
II. The perimeter of triangle PQR is equal to 12 units.

1) Since all three sides are distinct integer, so we can infer this is a right triangle.
2) 3 sides can only be 3, 4 and 5. So this is a right triangle. We can get the area from it. Sufficient.
B is the answer.

The video explanation of the problem can be seen here.

In triangle PQR, if the length of all the three sides are distinct integers, what is the area of triangle PQR?

I. \(PQ^2 = PR^2 + RQ^2\)
II. The perimeter of triangle PQR is equal to 12 units.

Given :- a,b,c (sides of the triangle) are distinct integers
To find :- Area of the triangle

i) We can just infer that the triangle is right-angled - Insufficient

ii) Perimeter = a+b+c = 12
Now we know, 3+4+5 = 12 => A right angled triangle is definitely one of the cases.

To try out others, we need to make sure that the "Sum of any 2 sides of the triangle is greater the third side and Difference of any 2 sides is less than the third side"

Let's try with the shortest side

Let's start with a=1 - Whenever we have a side as 1 it will be the shortest side and the difference of other 2 sides can not be less than 1 since the other 2 sides can not be equal (when the diff would be 0; else difference would \(geq{1}\)). For eg.
a = 1; b=2; c=9 => c-b = 7 > 1
a = 1; b=3; c=8 => c-b = 5 > 1

When a = 2; b + c = 10 but b and c have to be consecutive so that the difference is 1 and not \(geq{2}\) i.e. the third side
a = 2; b=3; c=7 => c-b = 4 > 2 (b and c are not consecutive and we can see why this can not form a triangle)
in fact b and c can not be consecutive and add upto 10

When a=3; b+c = 9 - When the shortest side is 3, we can not have any sum other than 3,4,5 values that will add upto 12 and have distinct integers at the same time.
a=3; b=4; c=5
Sufficient

Answer - B

In triangle PQR, if the length of all the three sides are distinct integers, what is the area of triangle PQR?
ST I : PQ2=PR2+RQ2
If, (PQ,PR,RQ) = (5,3,4) (distinct ) area = 6
If, (PQ,PR,RQ) = (10,6,8) (distinct ) area = 24 insufficient
ST II: The perimeter of triangle PQR is equal to 12 units.
Let, the sides of the triangle be x, y and 12−(x+y)
Now, the sum of any two sides of a triangle is greater than its third side.
So,x+y>12−(x+y)
∴x+y>6
also, x+12−(x+y) > y
∴y<6
Also, the difference of the two sides should be less than the third.
So,x−y<12−(x+y)
∴y<6
The possible combinations of x and y and 12−(x+y)are:
(1, 6, 5 ) (rejected y not less than 6)
(2, 5, 5 ) (rejected not distinct )
(3, 4, 5 ) (accepted )
(3, 5, 4 ) (accepted )
(4, 4, 4 ) (rejected not distinct )
(4, 5, 3) (accepted )
(5, 5, 2) (rejected not distinct )

so only accepted combination = (3,4,5) which gives an unique area of triangle PQR. sufficient
correct answer is B

Given triangle PQR
All 3 sides of the triangle are distinct intrgers.
Area triangle PQR=?

Statement 1: PQ^2 =PR^2+RQ^2
PQR is a right triangle
We have a infinite numbers of right triangles with sides having integers value.
Insufficient
Statement 2: PQ+PR+RQ=12
We know that the sum of addition of 2 sides of a triangle must be greater than the third side.
The 3 sides have to be distinct integers as well.
3,4,5 is a possibility
2,5,5- not distinct
2,3,7- violates the triangle rule
Whatever combinations we try, we see that 3,4,5 is the only possibility.
We have the values of the 3 sides of a triangle.
Area can be calculated.
Sufficient
B

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