Question

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

Answer 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.!!

Answer 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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