Find the Pythagorean triplet whose one member is 18.

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28 and 20

80 and 82

72 and 92

34 and 54

answer is B.

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Detailed Solution

Concept- Let's tackle this problem with the Pythagorean theorem. To determine the right response, we will examine each possibility.
The Pythagorean triplet of 18 is what the query now inquires about. So, in order to answer the prompt, we will use Pythagoras' theorem. The hypotenuse is the greatest side of a right-angled triangle, and according to Pythagoras's theorem, the sum of the squares of the base and perpendicular is equal to the square of the hypotenuse. However, the base and perpendicular are shorter than the hypotenuse. In other words, the Pythagoras theorem states that if a, b, and c are the lengths of the sides of a right-angled triangle, where c is the hypotenuse and a, b are the base and perpendicular, respectively, then

a^{2}

b^{2}

c^{2}

Checking the option (1) ,
Taking 28 as hypotenuse and 20 as base. Let the perpendicular be a. So, applying Pythagoras theorem,

\Rightarrow a^{2} + 20^{2} = 28^{2}

\Rightarrow a^{2} = 28^{2} - 20^{2} = (28 - 20) (28 + 20) as, a^{2} - b^{2} = (a - b) (a + b)

\Rightarrow a^{2} = (8) \times (48) = 384 \Rightarrow \Rightarrow a = \pm \sqrt 384

a^{2} - b^{2}

=(a−b) (a+b)
⇒a=19 as, side cannot be negative.
Using 80 as the base and 82 as the hypotenuse, we will check option (B) . Let a be the perpendicular. Applying Pythagoras' theorem, we may deduce that

a^{2}

80^{2}

82^{2}

a^{2}

82^{2}

80^{2}

=(82-80) (82+80)

a^{2}

=(2)

\times

(162) =324
a=

\sqrt

324=18 a=18
Since option (2) states that 18 is the other side and the question specifies that as the other side, option (B) is the right answer.
Similar to how we can tell that both alternatives (3) and (4) are incorrect by checking them. Therefore, option 1 is the correct one since 18 is the Pythagorean triplet of 80 and 82.
Hence, the correct option is 1) 80 and 82
ExamType: CBSE

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