Question

prove tha :- h2=p2+b2

Answer 1

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides of the triangle.

Now,
In a right triangle, side opposite to the 90° is always the hypotenuse. The square of the side opposite of the 90° in a right triangle is equal to the sum of squares of the other two sides.

Here,
∆ ABC is a right triangle, right angled at B. AC is the hypotenuse of the right ∆ ABC.

So, the square of AC = the sum of the squares of the other two sides AB and BC.

that is, AC² = AB² + BC²
or, hypotenuse (h)² = perpendicular (p) ² + base (b)² [Pythagoras Theorem]

Hence, proved.

Answer 2

Answer:

h2=p2 +b2

Step-by-step explanation:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides of the triangle.

Now,

In a right triangle, side opposite to the 90° is always the hypotenuse. The square of the side opposite of the 90° in a right triangle is equal to the sum of squares of the other two sides.

Here,

∆ ABC is a right triangle, right angled at B. AC is the hypotenuse of the right ∆ ABC.

So, the square of AC = the sum of the squares of the other two sides AB and BC.

that is, AC² = AB² + BC²

or, hypotenuse (h)² = perpendicular (p) ² + base (b)² [Pythagoras Theorem]

Hence, proved.