Question

State and prove Pythagoras theorem.
​

Answer 1

Answer:

$\huge\underline{\overline{\mid {\bold {\green {☆ansWer☆}}\mid}}}$

${h}^{2} = {p}^{2} + {b}^{2}$

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

Given: ABC is a triangle in which ∠ABC=90

∘

Construction: Draw BD⊥AC.

Proof:

In △ADB and △ABC

∠A=∠A [Common angle]

∠ADB=∠ABC [Each 90

∘

]

△ADB∼△ABC [A−A Criteria]

So,

AB

AD

=

AC

AB

Now, AB

2

=AD×AC ..........(1)

Similarly,

BC

2

=CD×AC ..........(2)

Adding equations (1) and (2) we get,

AB

2

+BC

2

=AD×AC+CD×AC

=AC(AD+CD)

=AC×AC

∴AB

2

+BC

2

=AC

2

[henceproved]

Answer 2

________ Answer _______

☆☆☆☆☆☆Important ☆☆☆☆☆☆

Theroem :- 1

= 》 If a perpendicular is drawn

from the vertex of the right

angle to the hypotenuse then

triangles on the both side of the

perpendicular are similar to the

whole triangle and to each other.

☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆

Pythagoras Theorem

=》In a right triangle ,the square

of hypotenuse is equal to the sum

of the squares of the other

two sides.

《《___________________________》》

Proof : We are given a right triangle ABC right angled at B.

We need to prove that AC²= AB² + BC²

Let us draw BD ⊥ AC (see Fig.)

Now, ∆ ADB ~ ∆ ABC. (Theorem 1)

So,

AD/AB =AB/AC (Sides are proportional)

or, AD . AC = AB2. ........(1)

Also, ∆ BDC ~ ∆ ABC (Theorem 6.7)

So,

CD/BC =BC/AC

or, CD ×AC = BC² .........(2)

Adding (1) and (2),

=> AD . AC + CD . AC = AB² + BC²

=> AC (AD + CD) = AB² + BC²

=> AC . AC = AB² + BC²

=> AC² = AB² + BC²

Hence, proved

__________××××__________

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