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.

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.

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).

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

= `"AD"^2+(("BC")/2)^2 - 2(("BC")/2) xx "MD"`

= `"AD"^2 + ("BC"/2)^2 - "BC" xx "MD"`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

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

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC)

= AB² + AC²

= `2("AM"^2 + "MD"^2) + (("BC")/2)^2 + (("BC")/2)^2 + 2"MD" ((-"BC")/2 + ("BC")/2) = "AB"^2 + "AC"^2`

`"2AD"^2 + ("BC"^2)/2 = "AB"^2 + "AC"^2`

Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

Draw perpendicular BD from the vertex B to the side AC. A – D

In right-angled ΔABC

seg BD ⊥ hypotenuse AC.

∴ By the similarity in right-angled triangles

ΔABC ~ ΔADB ~ ΔBDC

Now, ΔABC ~ ΔADB

∴ `"AB"/"AD" = "AC"/"AB"`   ...(c.s.s.t)

∴ AB² = AC × AD   ...(1)

Also, ΔABC ~ ΔBDC

∴ `"BC"/"DC" = "AC"/"BC"`   ...(c.s.s.t)

∴ BC² = AC × DC   ...(2)

From (1) and (2),

AB² + BC² = AC × AD + AC × DC

= AC × (AD + DC)

= AC × AC   ...(A – D – C)

∴ AB² + BC² = AC²

i.e., AC² = AB² + BC²

Given: In ΔPQR, ∠PQR = 90°.

To prove: PR² = PQ² + QR².

Construction:

Draw seg QS ⊥ side PR such that P–S–R.

Proof: In ΔPQR,

∠PQR = 90°   ...(Given)

Seg QS ⊥ hypotenuse PR   ...(Construction)

∴ ΔPSQ ∼ ΔQSR ∼ ΔPQR   ...(Similarity of right-angled triangles)   ...(1)

ΔPSQ ∼ ΔPQR   ...[From (1)]

∴ `"PS"/"PQ" = "PQ"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ PQ² = PS × PR   ...(2)

ΔQSR ∼ ΔPQR   ...[From (1)]

∴ `"SR"/"QR" = "QR"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ QR² = SR × PR   ...(3)

Adding (2) and (3), we get

PQ² + QR² = PS × PR + SR × PR

∴ PQ² + QR² = PR(PS + SR)

∴ PQ² + QR² = PR × PR   ...(P–S–R)

∴ PQ² + QR² = PR² OR PR² = PQ² + QR².

Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.

To prove: QR² = 4(PS² – PQ²)

in right-angled ΔPQS, by Pythagoras theorem,

PQ² + QS² = PS²

⇒ QS² = PS² – PQ²   .......(i)

Since S is the mid-point of side QR,

∴ QS = `(QR)/2`

Substituting the value of QS in equation (i),

`((QR)/2)^2 = PS^2 - PQ^2`

`(QR^2)/4 = PS^2 - PQ^2`

QR² = 4(PS² – PQ)²

Hence proved.

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