Pythagoras Theorem

For three numbers a, b and c if a² + b² = c². then (a, b, c) is known as a Phythagorean Triplet. Also, order here doesn't matter and we take c > a, b.

1^st case: 3² + 4² = 5². Thus this a triplet.

2^nd case: 4² + 9² ≠ 12²

3^rd case: 5² + 12² = 13² . Thus this is a triplet.

4^th case: 24^{2 +} 70² = 74². Thus this is a triplet.

5^th case: 10² + 24² ≠ 27²

6^th case: 11² + 60² = 61². Thus this is a triplet.^.

In ∆MNP, ∠ MNP = 90⁰,

MN² + NP² = MP²

⇒ MN² + NP² = (MQ + QP)²

⇒ MN² + NP² = (13)²

⇒ MN² + NP² = 169 … (1)

In ∆MQN, ∠MQN = 90⁰,

QN² + MQ² = MN²

⇒ QN² + 9² = MN²

⇒ QN² + 81 = MN² …(2)

In ∆PQN, ∠PQN = 90⁰,

QN² + PQ² = PN²

⇒ QN² + 4² = PN²

⇒ QN² + 16 = PN² … (3)

Now (2) + (3)

⇒ QN² + 81 + QN² + 16 = MN² + PN²

⇒ 2QN² + 97 = 169 [from(1)]

⇒ 2QN² = 72

⇒ QN² = 36

Thus NQ = 6.

In ∆PMQ, ∠PMQ = 90⁰

So PM² + QM² = PQ<sup>2

⇒</sup> 10² + 8² = PQ²

PQ² = 164 …(1)

In ∆PQR, ∠RPQ = 90⁰

So PQ² + PR² = QR²

⇒ 164 + PR² = QR²

⇒ PR² = QR² – 164 …(2)

In ∆PMR, ∠PMR = 90⁰

So PM² + MR² = PR²

⇒ 100 + (QR – QM)² = QR² – 164

⇒ 100 + QR² – 2.QR.QM + QM² = QR² – 164

⇒ 100 – 2.QR.8 + 64 = – 164

⇒ 16QR = 2×164

⇒ QR = 20.5

Thus QR = 20.5

In ∆PSR, ∠PSR = 90⁰

So PS² + SR² = RP²

⇒ 6² + (RP cos(30°))² = RP²

⇒ 6² + RP² = RP²

⇒ 6² =

⇒ RP² = 4×36

Thus RP = 12.

PS = RP cos(30⁰)

⇒ PS = 12×

PS = 6√3.

In ∆ABC, ∠ABC = 90⁰

So AB² + BC² = AC²

⇒ 2AB² = 5

⇒ AB² =

⇒ AB = √ = X(Say)

AB = BC = √

∠BAC = 45⁰ Since AB = BC

NOW √ = X√5

X =

Similarly, X2√2 = √

X =

Similarly, X√8 = √

X =

In a square of side say a cm, any diagonal divide the square into two right triangles of equal dimensions.

Thus a² + a² = 10²

⇒ 2a² = 100

⇒ a² = 50

a = 5 cm

perimeter = 4a

= 4×5

₌ 20

_{Perimeter of square =} 20_cm

_{In ∆DGF,∠DGF = 90}⁰

_FD² = DG² + GF²

_⇒FD² = 64 + 144

_⇒FD² = 208

_{FD =}

_{In ∆DEF,∠DFE = 90}⁰

_ED² = DF² + EF²

⇒ (EG + 8)² = 208 + EF² … (1)

_{In ∆EGF,∠FGE = 90}⁰

_EF² = EG² + GF²

_⇒ (EG + 8)² – 208 = _EG² + 144

_⇒ EG² + 2.EG.8 + 64 – 208 = _EG² + 144 (As we know (a+b)²= a²+b²+2ab

_{EG = 18}

_{From (1)}

⇒ (EG + 8)² = 208 + EF²

_{EF = 6√13.}

The diagonal = √[length² + breadth²]

= √(35² + 12²)
= √(1225 + 144)

Thus the diagonal is 37 cm.

In ∆PRQ, ∠PRQ = 90⁰

PQ² = PR² + QR² – – – 1

In ∆PRM, ∠PRM = 90⁰

PM² = PR² + MR²

⇒ PM² = PR² + ² [ M is midpoint]

⇒ 4(PM² – PR²) = QR² – – – 2

1 And 2 implies

PQ² = PR² + 4(PM² – PR²)

⇒ PQ² = 4PM² – 3PR²

PROVED.

Let us consider a distance x m on the street from one building and a distance y m from the other one.

Now according to question,

In the 1^st case,

5.8² = 4² + x²

⇒ x² = 17.64

⇒ x = 4.2

Similarly for the second building,

5.8² = 4.2² + y²

⇒ y² = 16

⇒ y = 4

Total width = x + y

= 4 + 4.2

= 8.2

Thus the total width is 8.2m.

Given PS = 13, PQ = 11,PR = 17

By Apollonius's Theorem,

PS² =

⇒ 169 =

⇒ = 36

⇒ QR² = 144

QR = 12

The figure is given below:

According to Pythagoras theorem,

Median2 =

⇒ Median2 =

⇒ Median2 = 40

Median = 2√10

Thus the median is 2√10

Perpendicular² + Base² = Hypotenuse²

⇒ PT² + TR² = PR² [1]

Similarly, In ΔPTS

PT² + TS² = PS²

⇒ PT² = PS² – ST² [2]

Using [2] in [1], we have

⇒ PS² – ST² + (ST + SR)² = PR²

⇒ PS² – ST² + ST² + SR² + (2 × ST × SR) = PR²

Now, Since PS is a median we can write

In ΔPQT, By Pythagoras we have

PQ² = PT² + QT²

⇒ PQ² = PS² – ST² + (QS – ST)² [From 2]

⇒ PQ² = PS² – ST² + QS² + ST² – (2 × QS × ST)

Given AB² + AC² = 290 cm²,AM = 8 cm, BM = MC

According to formula,

AM² =

⇒ 64 =

⇒ 64 –

⇒ BC² = 324

BC = 18.

Thus BC = 18 cm.

From figure,

In ∆PAT, ∠PAT = 90⁰

TP² = AT² + PA² …1

In ∆AST, ∠SAT = 90⁰

TS² = AT² + SA² …2

In ∆QBT, ∠QBT = 90⁰

TQ² = BT² + QB² …3

In ∆BTR, ∠RBT = 90⁰

TR² = BT² + BR² …4

TS² + TQ² = AT² + SA² + BT² + QB² [Adding 2 and 3]

⇒ TS² + TQ^{2 =} AT² + PA² + BT² + BR² [SA = BR, QB = AP]

⇒ TS² + TQ² = TP² + TR² [From 1 and 4]

PROVED.

A Pythagorean triplet consists of three positive integers (l, b, h) such that

l² + b² = h²

and (3, 4, 5) is a Pythagorean triplet as,

5² = 3² + 4²

Given,

Sum of the squares of the sides making right angle = 169

⇒ (base)² + (perpendicular)² = 169

But we know, By Pythagoras's theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ (Hypotenuse)² = 169

⇒ Hypotenuse = 13 units.

A Pythagorean triplet consists of three positive integers (l, b, h) such that

l² + b² = h²

and 15/08/17 is a Pythagorean triplet as,

15² + 8² = 17²

i.e. 225 + 64 = 289

As, the sides of right-angled triangles satisfies the Pythagoras theorem, i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

We know that,

Diagonal of a square = √2 a

Where 'a' is the side of the triangle.

⇒ √2 a = 10√2

⇒ a = 10 cm

Also, we know

Perimeter of square = 4a

Where 'a' is the side of the triangle

∴ Perimeter of given square = 4(10) = 40 cm

Let ABC be a right-angled triangle, at B, and BP be the altitude on hypotenuse that divides it in two parts such that,

AP = 4 cm

PC = 9 cm

As, ABC, ABP and CBP are right-angled triangles, therefore they all satisfy Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

∴ In ΔABC

AB² + BC² = AC²

⇒ AB² + BC² = (AP + CP)²

⇒ AB² + BC² = (4 + 9)² = 13²

⇒ AB² + BC² = 169 [1]

∴ In ΔABP

AP² + BP² = AB²

AP² + 4² = AB² [2]

∴ In ΔCBP

CP² + BP² = BC²

⇒ 9² + BP² = BC² [3]

Adding [2] and [3], we get

AP² + 4² + 9² + BP² = AB² + BC²

⇒ 2AP² + 16 + 81 = 169 [From 1]

⇒ 2AP² = 72

⇒ AP² = 36

⇒ AP = 6 cm

Hence, length of Altitude is 6 cm.

By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

Given,

Base = 18 cm

Perpendicular = Height = 24 cm

⇒ Hypotenuse² = 24² + 18²

⇒ Hypotenuse² = 576 + 324

⇒ Hypotenuse² = 900

⇒ Hypotenuse = 30 cm

As,

(6√3)² + 6² = 12²

⇒ AB² + BC² = AC²

i.e. sides of the triangle ABC satisfy the Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

∴ ABC is a right-angled triangle with hypotenuse as AC

Now,

By converse of 30°-60°-90° triangle theorem i.e.

In a right-angled triangle, if one side is half of the hypotenuse then the angle

opposite to that side is 30°.

∠A = 30°

Let ABC be an equilateral triangle,

Let AP be a perpendicular on side BC from A.

To find : Height of triangle = AP

As, ABC is an equilateral triangle we have

AB = BC = CA = 2a

Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side

Now, In ΔABP, By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ AB² = BP² + AP²

⇒ (2a)² = a² + AP²

⇒ AP² = 4a² - a²

⇒ AP² = 3a²

⇒ AP = a√3

Yes,

Because

7² + 24² = 25² [i.e. 49 + 576 = 625]

As, sides satisfy the Pythagoras theorem, i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

They do form a right-angled triangle.

Let ABCD be a rectangle, with

AB = CD = 60 cm

BC = DA = 11 cm

And AC be a diagonal.

As, ∠A = 90°

ADC is a right-angled triangle, By Pythagoras Theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

AC² = (CD)² + (DA)²

⇒ AC² = 60² + 11²

⇒ AC² = 3600 + 121

⇒ AC² = 3721

⇒ AC = 61 cm

In a right-angled triangle

By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

Given,

Other sides are 9 cm and 12 cm

⇒ Hypotenuse² = 9² + 12²

⇒ Hypotenuse² = 81 + 144

⇒ Hypotenuse² = 225

⇒ Hypotenuse = 15 cm

In a right-angled triangle

By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

As, the triangle is isosceles

Base = Perpendicular = x

[Hypotenuse can't be equal to any of the sides, because hypotenuse is the greatest side in a right-angled triangle and it must be greater than other two sides]

⇒ (Hypotenuse)² = x² + x²

⇒ (Hypotenuse)² = 2x²

⇒ Hypotenuse = x√2

As,

(√5)² + (√3)² = (√8)²

⇒ QR² + PR² = PQ²

i.e. sides of the triangle ABC satisfy the Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

∴ PQR is a right-angled triangle at R [As hypotenuse is PQ].

As, ∠S = 90°, and ∠T = 30° and RT = 12 cm is given.

Clearly, RTS is a 30°-60°-90° triangle.

We know, Property of 30°-60°-90° triangle i.e.

If acute angles of a right angled-triangle are 30° and 60°, then the side opposite

30° angle is half of the hypotenuse and the side opposite to 60° angle is times of hypotenuse.

And

Given,

Length of rectangle, l = 16 cm

Breadth of rectangle = b

Area of rectangle = length × breadth

⇒ 192 = 16b

⇒ b = 12 cm

Also, we know that

Length of diagonal = √(l² + b²)

Where, l = length and b = breadth

⇒ length of diagonal = √(16² + 12²)

= √(256 + 144) = 20 cm

Let ABC be an equilateral triangle,

Let AP be a perpendicular on side BC from A.

To find : Height of triangle = AP

As, ABC is an equilateral triangle we have

AB = BC = CA = 'a'

Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side

Now, In ΔABP, By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ AB² = BP² + AP²

Given,

Height = √3

⇒ a = 2 cm

Also, Perimeter of equilateral triangle = 3a

Where 'a' depicts side of equilateral triangle.

∴ Perimeter = 3(2) = 6 cm

We know, By Apollonius theorem

In ΔABC,

if P is the midpoint of side BC, then AB² + AC² = 2AP² + 2BP ²

Given that, AP is median i.e. P is the mid-point of BC

And BC = 18 cm

and AB² + AC² = 260

⇒ 260 = 2AP² + 2(9)²

⇒ 260 = 2AP² + 162

⇒ 98 = 2AP²

⇒ AP² = 49

⇒ AP = 7 units

ABC be an equilateral triangle,

Point P is on base BC, such that

Let us construct AM perpendicular on side BC from A.

As, ABC is an equilateral triangle we have

AB = BC = CA = 6 cm

Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side

Now, In ΔACM, By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ CA² = CM² + AM²

⇒ (6)² = (3)² + AM²

⇒ 36 = 9 + AM²

⇒ AM² = 27 [1]

As,

We have,

CM - PC = PM

⇒ PM = 1 cm

Now, In right angled triangle AMP, By Pythagoras theorem

(AP)² = (AM)² + (PM)²

⇒ (AP)² = 27 + 1²

⇒ AP² = 28

⇒ AP = 2√7 cm

In ΔPQS and ΔPSR, By Pythagoras theorem

i.e. (Hypotenuse)² = (base)² + (Perpendicular)²

PQ² = QS² + PS² [1]

PR² = SR² + PS² [2]

Subtracting [2] from [1],

PQ² - PR² = QS² - SR²

⇒ a² - a² = QS² - SR²

⇒ QS² = SR²

⇒ QS = SR

Also,

MS = MQ + QS

And

SN = SR + RN

In ΔPSM and ΔPSN, By Pythagoras theorem

PM²= PS² + MS²

[3]

PN²= PS² + SN²

[4]

From [3] and [4]

PM² = PN²

⇒ PM = PN

Hence Proved.

Let ABCD be a parallelogram, with AB = CD ; AB || CD and BC = AD ; BC || AD.

Construct AE ⊥ CD and extend CD to F such that, BF ⊥ CF.

In ΔAED and ΔBCF

AE = BF [Distance between two parallel lines i.e. AB and CD]

AD = BC [opposite sides of a parallelogram are equal]

∠AED = ∠BFC [Both 90°]

ΔAED ≅ ΔBCF [By Right Angle - Hypotenuse - Side Criteria]

⇒ DE = CF [Corresponding sides of congruent triangles are equal] [1]

In ΔBFD, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

BD² = DF² + BF²

⇒ BD² = (CD + CF)² + BF² [2]

In ΔAEC, By Pythagoras theorem

AC² = AE² + CE²

⇒ AC² = AE² + (CD - AE)²

⇒ AC² = BF² + (CD - CF)² [As, AE = BF and CF = AE] [2]

In ΔBCF, By Pythagoras theorem,

BC² = BF² + CF²

BF² = BC² - CF² [3]

Adding [2] and [3]

BD² + AC² = 2BF² + (CD + CF)² + (CD - CF)²

⇒ BD² + AC²= 2BC² - 2CF² + CD² + CF² + 2CD.CF + CD² + CF² - 2CD.CF

⇒ BD² + AC² = 2BC² + 2CD²

⇒ BD² + AC² = BC² + BC² + CD² + CD²

⇒ BD² + AC² = AB² + BC² + CD² + AD² [since BC = AD and AB = CD]

Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Let their speed be 'x' km/h

We know, distance = speed × time

In two hours,

Distance travelled by both = '2x' km

Let their starting point be 'O', and Pranali and Prasad reach the point A in the East and point B in the north direction respectively.

Clearly, AOB is a right-angled triangle, So By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

(AB)² = (OA)² + (OB)²

As, AB = distance between them = 15√2 km

And OA = OB = distance travelled by each = 2x

⇒ (15√2)² = (2x)² + (2x)²

⇒ 450 = 8x²

⇒ x² = 56.25

⇒ x = 7.5 km/h

We know, By Apollonius theorem

In ΔABC, if L is the midpoint of side AC, then AB² + BC² = 2BL² + 2AL ²

Given that, BL is median i.e. L is the mid-point of CA

⇒ AB² + BC² = 2BL² + 2AL²

[1]

Also, if M is the midpoint of side AB, then AC² + BC² =2CM² + 2BM²

Given that, CM is median i.e. M is the mid-point of BA

⇒ AC² + BC² = 2CM² + 2BM²

[2]

Also, In ΔABC, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ BC² = AC² + AB² [3]

Adding [1] and [2]

⇒ AB² + AC² + 4BC² = 4(BL² + CM²)

⇒ BC² + 4BC² = 4(BL²+ CM²) [From 3]

⇒ 5BC² = 4(BL²+ CM²)

Hence Proved.

Let ABCD be a parallelogram, with AB = CD ; AB || CD and BC = AD ; BC || AD.

Construct AE ⊥ CD and extend CD to F such that, BF ⊥ CF.

Given: sum of squares of adjacent side = 130

⇒ CD² + BC² = 130 and

Length of one diagonal = 14 cm [let it be AC]

To Find: length of the other diagonal, BD

In ΔAED and ΔBCF

AE = BF [Distance between two parallel lines i.e. AB and CD]

AD = BC [opposite sides of a parallelogram are equal]

∠AED = ∠BFC [Both 90°]

ΔAED ≅ ΔBCF [By Right Angle - Hypotenuse - Side Criteria]

⇒ DE = CF [Corresponding sides of congruent triangles are equal] [1]

In ΔBFD, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

BD² = DF² + BF²

⇒ BD² = (CD + CF)² + BF² [2]

In ΔAEC, By Pythagoras theorem

AC² = AE² + CE²

⇒ AC² = AE² + (CD - AE)²

⇒ AC² = BF² + (CD - CF)² [As, AE = BF and CF = AE] [2]

In ΔBCF, By Pythagoras theorem,

BC² = BF² + CF²

BF² = BC² - CF² [3]

Adding [2] and [3]

BD² + AC² = 2BF² + (CD + CF)² + (CD - CF)²

⇒ BD² + AC²= 2BC² - 2CF² + CD² + CF² + 2CD.CF + CD² + CF² - 2CD.CF

⇒ BD² + AC² = 2BC² + 2CD²

⇒ BD² + 14² = 2(130)

⇒ BD² + 196 = 260 [Using given data]

⇒ BD² = 64

⇒ BD = 8 cm

Hence, length of other diagonal is 8 cm.

Given,

DB = 3CD

Also,

BC = CD + DB = CD + 3CD

⇒ BC = 4CD [1]

As, AD ⊥ BC, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

In Δ ACD

AC² = AD² + CD² [2]

In ΔABD

AB² = AD² + DB²

⇒ AB² = AD² + (3CD)²

⇒ AB² = AD² + 9CD² [3]

Subtracting [2] from [3]

⇒ AB² - AC² = 9CD² - CD²

⇒ AB² = AC² + 8CD²

⇒ 2AB² = 2AC² + 16CD²

⇒ 2AB² = 2AC² + (4CD)²

⇒ 2AB² = 2AC² + BC² [From 1]

Hence Proved.

Let ABC be an isosceles triangle, In which AB = AC = 13 cm

And BC = 10 cm

Let AM be median on BC such that

Let P be centroid on median BC

To Find : AP [Distance between vertex opposite the base and centroid]

We know, By Apollonius theorem

In ΔABC, if M is the midpoint of side BC, then AB² + AC² = 2AM² + 2BM ²

Putting values, we get

(13)² + (13)² = 2AM² + 2(5)²

⇒ 169 + 169 = 2AM² + 50

⇒ 2AM² = 288

⇒ AM² = 144

⇒ AM = 12 cm

Let P be the centroid

As, Centroid divides median in a ratio 2 : 1

⇒ AP : PM = 2 : 1

⇒ AP = 2PM

Now, AM = AP + PM

Construct DE ⊥ AB and CF ⊥ AB

In ΔADB, as BD ⊥ AD, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

(AB)² = (AD)² + (BD)²

⇒ 25² = 15² + BD²

⇒ BD² = 625 - 225 = 400

⇒ BD = 20 cm

Similarly,

AC = 20 cm

Now, In ΔAED and ΔABD

∠AED = ∠ADB [Both 90°]

∠DAE = ∠DAE [Common]

ΔAED ~ ΔABD [By Angle-Angle Criteria]

[Property of similar triangles]

As AD = 15 cm, BD = 20 cm and AB = 25 cm

⇒ DE = 12 cm

Also,

⇒ AE = 9 cm

Similarly, BF = 9 cm

Now,

DC = EF [By construction]

DC = AB - DE - AE

DC = 25 - 9 - 9 = 7 cm

Also, we know

Area of trapezium

As, PQR is an equilateral triangle,

Point S is on base QR, such that

PT is perpendicular on side QR from P.

As, PQR is an equilateral triangle we have

PQ = QR = PR [1]

Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side

Now, In ΔPTQ, By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ PQ² = PT² + QT²

[2]

As,

We have,

QT - QS = ST

Now, In right angled triangle PST, By Pythagoras theorem

(PS)² = (ST)² + (PT)²

[From 2]

⇒ 36 PS² = 28 PQ²

⇒ 9 PS² = 7 PQ²

Hence Proved.

We know, By Apollonius theorem

In ΔPQR, if M is the midpoint of side QR, then PQ² + PR² = 2PM² + 2QM ²

Given that, PM is median i.e. M is the mid-point of QR

And PQ = 40, PR = 42, PM = 29

Putting values,

⇒ (40)² + (42)² = 2(29)² + 2(QM)²

⇒ 1600 + 1764 = 1682 + 2QM²

⇒ QM² = 1682

⇒ QM = 29

⇒ QR = 2(29) = 58

We know, By Apollonius theorem

In ΔABC, if M is the midpoint of side BC, then AB² + AC² = 2AM² + 2BM ²

Given that,

AB = 22, AC = 34, BC = 24

AP is median i.e. P is the mid-point of BC

Putting values in equation

⇒ 22² + 34² = 2AM² + 2(12)²

⇒ 484 + 1156 = 2AM² + 288

⇒ 1352 = 2AM²

⇒ AM² = 676

⇒ AM = 26