Question

Please solve this ques with explanation ​

Answer 1

Answer:

1/p² = 1/a² + 1/b²

Step-by-step explanation:

(i) Area of ΔABC with base as BC:

= (1/2) * BC * AC

= (1/2) * a * b

= (1/2)ab

(ii) Area of ΔABC with base as AB:

= (1/2) * AB * CD

= (1/2) * c * p

= (1/2)cp

From (i) & (ii), we get

⇒ (1/2)ab = (1/2)cp

⇒ ab = cp

⇒ c = (ab/p)

Now,

In right angled triangle ΔABC,

⇒ AB² = BC² + AC²

⇒ c² = a² + b²

⇒ (ab/p)² = a² + b²

⇒ a²b²/p² = a² + b²

⇒ 1/p² = (a² + b²)/a²b²

⇒ 1/p² = (1/b²) + (1/a²)

⇒ 1/p² = (1/a² + 1/b²)

Hope it helps!

Answer 2

Step-by-step explanation:

Area is 1/2 base * height . So for the same triangle area can be 1/2*b*a and can be 1/2*p*c. Hence cp = ab

2 . From above p2 = a2b2/c2

1/p2 = c2/ a2b2

but for right triangle c2 = a2 +b2

Hence 1/p2 = (a2+b2)/ a2b2

Hence 1/p2 = 1/a2+1/b2