Question

Prove that “In a right-angled triangle, the square of the hypotenuse side is

equal to the sum of squares of the other two sides”

Using the result of this theorem find the length of the altitude of an

equilateral triangle of side 2a.​

Answer 1

Pythagoras theorem statement:-

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

_____________________

Proof:-

Let, ABC is a right angled triangle,right-angled at B

Therefore,

• AC = hypotenuse
• AB = adjacent side
• BC = perpendicular

_____________________

To Prove :

AC²= AB² + BC²

_____________________

Construction: Draw a perpendicular BD meeting AC at D

_____________________

From the diagram,

→ΔADB ~ ΔABC

As the corresponding sides of similar triangles are proportional,

⇒AD/AB = AB/AC

⇒AB²= AD × AC -------(1)

Also,

→ΔBDC ~ΔABC

As the corresponding sides of similar triangles are proportional,

⇒CD/BC = BC/AC

⇒ BC²= CD × AC -------(2)

Adding the equations (1) and (2),

⇒AB² + BC²= AD × AC+ CD × AC

⇒AB²+ BC² = AC (AD + CD)

We know,

AD + CD = AC

Therefore,

${ab}^{2} + {bc}^{2} = {ac}^{2}$

Hence proved !

---------------------------

Altitude of an equilateral triangle whose side is 2a:-

Given:

• ABC is an equilateral triangle
• Side = 2a

To find:

Altitude

Solution:

Let ABC be a equilateral triangle .

Therefore,AB=BC=AC=2a

AD is perpendicular on BC.

We know,

Altitude of an equilateral triangle bisects opposite side

Hence,

→BD=CD=a

In ΔADB,

We have:

• AB = hypotenuse
• AD = perpendicular
• BC = base

By Pythagoras theorem,

$⇝ {ab}^{2} = {ad}^{2} + {bd}^{2} \\ ⇝ {2a}^{2} = {ad}^{2} + {bd}^{2} \\ ⇝ {4a}^{2} = {ad}^{2} + {a}^{2} \\ ⇝ {4a}^{2} - {a}^{2} = {ad}^{2} \\ ⇝ {3a}^{2} = {ad}^{2} \\ ⇝ ad = \sqrt{3a}$

Hence,

Length of altitude is √3a.

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Answer 2

Step-by-step explanation:

Pythagoras theorem statement:-

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

_____________________

Proof:-

Let, ABC is a right angled triangle,right-angled at B

Therefore,

AC = hypotenuse

AB = adjacent side

BC = perpendicular

_____________________

To Prove :

AC²= AB² + BC²

_____________________

Construction: Draw a perpendicular BD meeting AC at D

_____________________

From the diagram,

→ΔADB ~ ΔABC

As the corresponding sides of similar triangles are proportional,

⇒AD/AB = AB/AC

⇒AB²= AD × AC -------(1)

Also,

→ΔBDC ~ΔABC

As the corresponding sides of similar triangles are proportional,

⇒CD/BC = BC/AC

⇒ BC²= CD × AC -------(2)

Adding the equations (1) and (2),

⇒AB² + BC²= AD × AC+ CD × AC

⇒AB²+ BC² = AC (AD + CD)

We know,

AD + CD = AC

Therefore,

{ab}^{2} + {bc}^{2} = {ac}^{2}ab

2

+bc

2

=ac

2

Hence proved !

---------------------------

Altitude of an equilateral triangle whose side is 2a:-

Given:

ABC is an equilateral triangle

Side = 2a

To find:

Altitude

Solution:

Let ABC be a equilateral triangle .

Therefore,AB=BC=AC=2a

AD is perpendicular on BC.

We know,

Altitude of an equilateral triangle bisects opposite side

Hence,

→BD=CD=a

In ΔADB,

We have:

AB = hypotenuse

AD = perpendicular

BC = base

By Pythagoras theorem,

\begin{gathered} ⇝ {ab}^{2} = {ad}^{2} + {bd}^{2} \\ ⇝ {2a}^{2} = {ad}^{2} + {bd}^{2} \\ ⇝ {4a}^{2} = {ad}^{2} + {a}^{2} \\ ⇝ {4a}^{2} - {a}^{2} = {ad}^{2} \\ ⇝ {3a}^{2} = {ad}^{2} \\ ⇝ ad = \sqrt{3a} \end{gathered}

⇝ab

2

=ad

2

+bd

2

⇝2a

2

=ad

2

+bd

2

⇝4a

2

=ad

2

+a

2

⇝4a

2

−a

2

=ad

2

⇝3a

2

=ad

2

⇝ad=

3a

Hence,

Length of altitude is √3a.

______________