Revision: Pythagoras Theorem Geometry Maths 2 SSC (English Medium) 10th Standard Maharashtra State Board

Three natural numbers (a,b,c) form a Pythagorean triplet if:

c² = a² + b²
(where c is the largest number)

Examples:
(3,4,5), (5,12,13), (8,15,17)

If a > b, then:

(a² + b²,  a² − b²,  2ab) is a Pythagorean triplet.

Given: m > n, the sides of the triangle are m² + n², m² – n² and 2 mn.

To prove: The triangle has a right angle.

Proof: (m² + n²)² = (m² – n²)² + (2 mn)²

(m²)² + (n²)² + 2 m²n² = (m²)² + (n²)² – 2 m²n² + 4 m²n²

m⁴ + n⁴ + 2 m²n² = m⁴ + n⁴ + 2 m²n²

The fact that both sides are equal.

So, the triangle supplied is a right-angled triangle according to the converse of the Pythagoras theorem.

Two Pythagorean triplets are, therefore (6, 8, 10) and (8, 15, 17).

ΔABC is a right-angled triangle and ∠ABC = 90°.

So, by the Pythagoras' theorem,

AB² + BC² = AC²

Substituting 6 cm for AB and 8 cm for BC in L.H.S.

AB² + BC² = 6² + 8²

= 36 + 64

= 100

Substituting 10 cm for AC in R.H.S.

AC² = 10²

= 100

Since, L.H.S. = R.H.S.

Hence, the Pythagoras theorem is verified.

We have, ΔABC is an equilateral Δ and AD ⊥ BC

In ΔADB and ΔADC

∠ADB = ∠ADC [Each 90°]

AB = AC [Given]

AD = AD [Common]

Then, ΔADB ≅ ΔADC [By RHS condition]

∴ BD = CD = BC/2              .......(i) [Corresponding parts of similar Δ are proportional]

In, ΔABD, by Pythagoras theorem

AB² = AD² + BD²

⇒ BC² = AD² + BD²                  [AB = BC given]

⇒ [2BD]² = AD² + BD²             [From (i)]

⇒ 4BD² − BD² = AD²

⇒ 3BD² = AD²

We have, D as the mid-point of BC

(i) AC² = AE² + EC²

b² = AE² + (ED + DC)²           [By pythagoras theorem]

b² = AD² + DC² + 2DC × ED

`b^2=p^2+(a/2)^2+2(a/2)xxx`      [BC = 2CD given]

`rArrb^2=p^2+a^2/4+ax`            .........(i)

(ii) In ΔAEB, by pythagoras theorem

AB² = AE² + BE²

⇒ c² = AD² − ED² + (BD − ED)²          [By pythagoras theorem]

⇒ c² = p² − ED² + BD² + ED² − 2BD × ED

`rArrc^2=p^2+(a/2)^2-2(a/2)xx x`

`rArrc^2=p^2+a^2/4-ax`                    .........(ii)

(iii) Add equations (i) and (ii)

`b^2+c^2=p^2+a^2/4+ax + p^2+a^2/4-ax`

`b^2+c^2=2p^2+(2a^2)/4`

`b^2+c^2=2p^2+a^2/2`

We have,

∠C = 90° and D is the mid-point of BC

In ΔACB, by Pythagoras theorem

AB² = AC² + BC²

⇒ AB² = AC² + (2CD)²                  [D is the mid-point of BC]

AB² = AC² + 4CD²

⇒ AB² = AC² + 4(AD² − AC²)       [In ΔACD, by Pythagoras theorem]

⇒ AB² = AC² + 4AD² − 4AC²

⇒ AB² = 4AD² − 3AC²

We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB =BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC  

∴` DC=1/2 a ` 

Now, In right triangle ADC
By using Pythagoras theorem, we have 

`AC^2=CD^2+DA^2` 

⇒` a^2-(1/2 a)^2+DA^2` 

⇒ `DA^2=a^2-1/4 a^2` 

⇒` DA^2=3/4 a^2` 

⇒`DA=sqrt3/2 a` 

𝑁𝑜𝑤,𝑎𝑟𝑒𝑎 (Δ𝐴𝐵𝐶)=`1/2xxBCxxAD` 

=` 1/2xxaxxsqrt3/2 a` 

=`sqrt3/4 a^2`

We have, ∠B = 90° and AD² = AB² + BC² + CD²

∴ AD² = AB² + BC² + CD²   ...[Given]

But AB² + BC² = AC²   ...[By pythagoras theorem]

Then, AD² = AC² + CD²

By converse of pythagoras theorem

∠ACD = 90°

Statement: 
In a right-angled triangle, if the altitude is drawn to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other.

△ADB ~ △ АBС
△ BDC ~ △ ABC
△ ADB ~ △ BDC

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

Hypotenuse² = (base)² + (perpendicular)²

Converse of Pythagoras Theorem:

In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right-angled triangle

Statement:
In a right-angled triangle, the altitude drawn to the hypotenuse is the geometric mean of the two segments of the hypotenuse.

(Altitude)² = (segment₁​) × (segment₂​)

Statement:
In a triangle, the sum of the squares of any two sides is equal to twice the square of the median to the third side plus twice the square of half the third side.

AB² + AC² = 2AM² + 2BM²

(where M is the midpoint of BC)

(I)Property of 30°-60°-90° triangle

• Side opposite 30° = \[\frac{1}{2}\] ​× hypotenuse
• Side opposite 60° = \[\frac{\sqrt{3}}{2}\] × hypotenuse.

(II) Property of 45°-45°-90°

• Each perpendicular side = \[\frac{1}{\sqrt{2}}\] × hypotenuse