prove that if x and y are both odd positive integers then x square + y square is even but not divisible by 4​

2 thoughts on “prove that if x and y are both odd positive integers then x square + y square is even but not divisible by 4​”

1. Given:–

   • x and y are odd positive integers.

   To Prove:–

   • x² + y² is even but not divisible by 4.

   step-by-step solution:–

   Let the two odd numbers be (2a+1) & (2b+1) because if we add 1 to any even no. it will be odd.

   x²+y²

   • → (2a + 1)² + (2b + 1)²
   • → (4a² + 4a + 1) + (4b²+ 4b + 1)
   • → 4(a² + b² + a + b)+2

   4 Is not a multiple of 2 it means clearly that 4 is not multiple of x²+y² , so x²+y² is even but not divisible by 4.

   Hence proved.!!

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2. Answer:

   Given:-

   ABC is an equilateral triangle.

   AB = BC = AC = 6cm.

   ∠A=∠B=∠C=60°.

   step-by-step explaination:-

   according to question,

   → PC=\frac{1}{3}BC

   3

   1

   BC

   therefore PC=2 cm.

   Now, using the cosine formula in ΔAPC, we have

   → cos∠C= \sf\frac{AC^{2}+PC^{2}-AP^{2}}{2(AC)(PC)}

   2(AC)(PC)

   AC

   2

   +PC

   2

   −AP

   2

   → cos60°=\sf\frac{6^{2}+2^{2}-AP^{2}}{2(6)(2)}

   2(6)(2)

   6

   2

   +2

   2

   −AP

   2

   → \sf\frac{1}{2}=\sf\frac{40-AP^{2}}{24}

   2

   1

   =

   24

   40−AP

   2

   → AP^{2}=40-12AP

   2

   =40−12

   → AP^{2}=28AP

   2

   =28

   → AP=2\sqrt{7}cmAP=2

   7

   cm

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