Question : If the hypotenuse of a right-angled triangle is 29 cm and the sum of the other two sides is 41 cm, then the difference between the other two sides is:
Option 1: 5 cm
Option 2: 1 cm
Option 3: 2 cm
Option 4: 10 cm
Correct Answer: 1 cm
Solution :
Let $a$ and $b$ be the two sides other than the hypotenuse.
Use the Pythagoras theorem: Hypotenuse
2
= Base
2
+ Perpendicular
2
According to the question,
$(29)^{2} = a^{2} + b^{2}$
⇒ 841 = $a^{2} + b^{2}$
Also, $a + b$ = 41
⇒ $a = 41 – b$
Now,
⇒ $(41−b)^{2} + b^{2} = 841$
⇒ $1681 − 82b + 2b^{2}$ = 841
⇒ $2b^{2} − 82b + 840$ = 0
⇒ $b^{2} − 41b + 420$ = 0
⇒ $(b − 20)(b − 21)$ = 0
If $b$ = 20, then
⇒ $a$ = 41– 20 = 21
If $b$ = 21, then
⇒ $a$ = 41– 21 = 20
The difference between the two sides is (|a – b|) = |20 – 21| = 1
Hence, the correct answer is 1 cm.
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