Similarity And The Theorem Pythagoras

m∠B = 90

AD = 4DC (Given) …Equation (i)

∴ BD² = AD × DC

⇒ BD² = 4DC² (From Equation (i))

Taking Square Root both sides we get

∴ BD = 2DC

Hypotenuse (AB) = 13

Perpendicular (AC) = 12

Base (BC) = 5

∴ AC² + BC² = 5² + 12²

⇒ AC² + BC² = 25 + 144

⇒ AC² + BC² = 169 = AB²

Hence Pythagoras theorem is valid

m∠C = 90

Let AD = x, BD = 13 – x

Applying Pythagoras theorem

∴ Altitude CD² = 12² – x² = 5² – (13 – x) ²

⇒ 144 – x² = 25 – 169 – x² + 26x

⇒ 26x = 288

⇒ x = 11.07 units

So Altitude CD = √21.43 = 4.63 units

m∠Q = 90

RM = 12

MP = 8

∴ RP = (RM + MP) = 20

Let QM be x

Applying Pythagoras theorem in ΔQMP, we get

QP² = x² + 8² = x² + 64

Applying Pythagoras theorem in ΔQMR, we get

QR² = x² + 12² = x² + 144

Applying Pythagoras theorem in ΔPQR, we get

RP² = (x² + 64) + (x² + 144)

⇒ 400 = 208 + 2x²

⇒ x² = 96

⇒ x = 4√6

⇒ QM = 4√6

PQ² = x² + 64

⇒ PQ² = 160

⇒ PQ = 4√10

QR² = x² + 144

⇒ QR² = 240

⇒ QR = 4√15

mB = 90

AM — MC = 7

Applying Pythagoras theorem in ΔABM and ΔCBM

AB² = BM² + AM² …Equation (i)

BC² = BM² + MC² …Equation (ii)

Subtracting Equation (ii) from Equation (i) we get

AB² – BC² = AM² – MC²

⇒ (AM – MC)(AM + MC) = AB² – BC²

Putting the values we get :

7 × (AM + MC) = 175

⇒ AM + MC = 25

⇒ AC = 25

m∠A = 90

AB = √5

BD = 2

Applying Pythagoras theorem in ΔADB we get

AD² = AB² – BD²

⇒ AD² = 5 – 4 = 1

⇒ AD = 1

Let AC = x

Applying Pythagoras theorem in ΔADC we get

DC² = AC² – AD²

⇒ DC² = x² – 1

⇒ DC = √(x² – 1)

Applying Pythagoras theorem in ΔABC we get

AB² + AC² = BC²

⇒ 5 + x² = (2 + √(x² – 1))²

⇒ 5 + x² = 4 + x² – 1 + 4√(x² – 1)

Hypotenuse BC = BD + DC

⇒ Hypotenuse BC

Applying Pythagoras theorem in ΔABM

AB² = AM² + BM²

⇒ AB² = 64 + 64 = 128

Let CM = x

Applying Pythagoras theorem in ΔCBM

⇒ BC² = BM² + CM²

⇒ BC² = x² + 64

Applying Pythagoras theorem in ΔABC

AC² = AB² + BC²

(x + 8)² = x² + 64 + 128

⇒ x² + 64 + 16x = x² + 192

⇒ 16x = 128

⇒ x = 8 units

AC = AM + CM = 16 units

Let AM = x, CM = 34 – x

BM² = AM × MC

⇒ x² – 34x = – 225

⇒ x² – 34x + 225 = 0

⇒ x² – 25x – 9x + 225 = 0

⇒ x(x – 25) – 9(x – 25) = 0

⇒ (x – 9)(x – 25) = 0

⇒ x = 9, 25

AB² = 15² + 9²

⇒ AB² = 306

AB = √306 units

BM = 2

MC = 6

Applying Pythagoras theorem :

BM² = CM × AM

⇒ 120 = 6 × AM

⇒ AM = 20

AC = AM + MC = 26 units

AB =

AM = 2.5

Applying Pythagoras theorem in ΔABM

AB² = AM² + BM²

⇒ BM² = 10 – 6.25 = 3.75

Let CM = x

AC = x + 2.5

Applying Pythagoras theorem in ΔCBM

⇒ BC² = BM² + CM²

⇒ BC² = x² + 3.75 …Equation (i)

Applying Pythagoras theorem in ΔABC

AC² = AB² + BC²

(x + 2.5)² = BC² + 10

⇒ x² + 6.25 + 5x = BC² + 10

⇒ BC² = x² – 3.75 + 5x …Equation (ii)

Equating Equation (i) and Equation (ii)

x² + 3.75 = x² – 3.75 + 5x

⇒ 5x = 7.5

⇒ x = 1.5

So CM = 1.5

mQ = 90

Applying Pythagoras theorem

PR = √(x² + y²)

PQ² = PD × PR

QR² = RD × PR

QD² = PD × RD

⇒ QD²

mQ = 90

PQ = 4QR

PQ² = PR × PM …Equation(i)

QR² = PR × RM …Equation(ii)

Dividing Equation (ii) by Equation(i) we get

Hence PM = 16RM

PQ + QR = 7

PR + QS = 10

Since Diagonals of a rectangle are of equal length, so

PR = QS = 5

Let PQ be x, QR = 7 – x

Applying Pythagoras theorem :

x² + (7 – x)² = 5²

⇒ 2x² – 14x + 24 = 0

⇒ x² – 7x + 12 = 0

⇒ x² – 4x – 3x + 12 = 0

⇒ x(x – 4) – 3(x – 4) = 0

⇒ (x – 3)(x – 4) = 0

x = 3, 4

∴ The two sides are 3 and 4

Area of the Rectangle = (3 × 4) = 12 sq. units

Using Pythagoras theorem :

AO² + OB² = AB² …Equation(i)

DO² + OC² = CD² …Equation(ii)

AO² + OD² = AD² …Equation(iii)

BO² + OC² = BC² …Equation(iv)

Adding Equation(i) and Equation(ii)

AB² + CD² = AO² + OB² + DO² + OC² …Equation(v)

Adding Equation(iii) and Equation(iv)

AD² + BC² = AO² + OB² + DO² + OC² …Equation(vi)

The RHS of equation (v) and (vi) are similar

So we can say that

AB² + CD² = AD² + BC²

mQ = 90

Applying Pythagoras theorem

PM² = QM² + PQ²

RN² = QN² + QR²

Adding the above two equations we get

PM² + RN² = QM² + PQ² + QN² + QR²

⇒ PM² + RN² = (QM² + QN²) + (PQ² + QR²)

⇒ PM² + RN² = MN² + PR²

Let a² + b² = p, 2ab = q, a² — b² = r

Since a, b ϵ R ^{+ ,} hence

p is the largest side of the triangle

p² = (a² + b²)² = a⁴ + b⁴ + 2a²b²

q² + r² = 4a²b² + (a² – b²)²

⇒ q² + r² = a⁴ + b⁴ + 2a²b²

So p² = q² + r²

Hence the Pythagoras theorem is established

Hence the angle opposite to the side p = a² + b² is a right angle

mB = 90

is a median

∴ AE = EC

⇒ AC = 2AE

Applying Pythagoras Theorem :

AB² + BC² = AC² …Equation (i)

Applying Appoloneous theorem

AB² + BC² = 2(AE² + BE²)

⇒ AC² = 2(AE² + BE²)

Now since AC = 2AE

⇒ (2AE)² = 2(AE² + BE²)

⇒ 2AE² = 2BE²

⇒ AE = BE …Equation (ii)

∴ AB² + BC² + AC² = 2AC² (From Equation (i))

⇒ AB² + BC² + AC² = 2 × (2AE)² = 8AE²

From Equation (ii) we can say that

⇒ AB² + BC² + AC² = 8BE²

mA = 90

Let AB = AC = x

BC = a

Applying Pythagoras Theorem :

∴ x² + x² = 2a²

⇒ 2x² = 2a²

⇒ x² = a²

⇒ x = a

∴ Area of the triangle

⇒ Area of the triangle

AB + BC = 23

AC + BD = 34

Since Diagonals of a rectangle are of equal length, so

AC = BD = 17

Let AB be x, BC = 23 – x

Applying Pythagoras theorem :

x² + (23 – x)² = 17²

⇒ 2x² – 46x + 240 = 0

⇒ x² – 23x + 120 = 0

⇒ x² – 15x – 8x + 120 = 0

⇒ x(x – 15) – 8(x – 15) = 0

⇒ (x – 8)(x – 15) = 0

x = 8, 15

∴ The two sides are 8 and 15

Area of the Rectangle = (8 × 15) = 120 sq. units

Let m∠A = x

Sum of interior angles = mA = mB + mC

⇒ mA + mB + mC = 180 (Sum of interior angle of triangle is 180)

⇒ 2x = 180

⇒ x = 90

So ∠A is right angled

Applying Pythagoras theorem

AC² = BC² – AB²

⇒ AC² = 25² – 7²

⇒ AC² = 576

⇒ AC = 24

Perimeter of ΔABC = (25 + 24 + 7) = 56 units.

Length of staircase = 6.5 metres

Height of Wall = 6 metres

Let the distance of base be d

Applying Pythagoras theorem

d² = 6.5² – 6²

⇒ d² = 6.25

⇒ d = 2.5 metres

Distance of base = 2.5 m

D is the mid – point of BC

AB = 7

AC = 5

AD = 5

Let BD = x

Applying Appoloneus theorem :

AB² + AC² = 2(AD² + BD²)

⇒ 7² + 5² = 2(5² + x²)

⇒ 74 = 50 + 2x²

⇒ 2x² = 24

⇒ x² = 12

⇒ x = 2√3 units

BC = 2 × BD = 4√3 units

BD : DC = 1 : 2

Let BD = x, DC = 2x –

⇒ AB = BC = AC = 3x

Let M be the mid point of BC

BM

∴ DM

⇒ DM

Applying Pythagoras theorem :

AB² = BM² + AM²

⇒ AM²

⇒ AM²

Applying Pythagoras theorem :

AD² = AM² + DM²

⇒ AD²

⇒ AD² = 7x²

⇒ AD = √7x

So

AB : AD = 3x : √7x

⇒ AB : AD = 3 : √7

So 3AD = AB

AB = 17

BC = 15

AC = 8

Let the median meet the side AB at point D

AD = 9.5

∴ CD is the median on the largest side

Applying Appoloneus theorem :

AC² + BC² = 2(AD² + CD²)

⇒ 8² + 15² = 2(9.5² + CD²)

⇒ 144.5 = 90.25 + CD²

⇒ CD² = 54.25

⇒ CD = 7.36 units

AB² + AC² = 148

AD = 7

Applying Apploneus theorem :

AB² + AC² = 2(AD² + BD²)

⇒ 148 = 2(49 + BD²)

⇒ BD² = 74 – 49 = 25

⇒ BD = 5 units

BC = 2 × BD = 10 units

AC = 25

CD = 7

Applying Pythagoras theorem :

AD² = AC² – CD²

⇒ AD² = 25² – 7² = 576

⇒ AD = 24 units

Perimeter of rectangle = 2(AD + CD) = 62 units

XZ = 14

⇒ XO = 7

YW = 48

⇒ YO = 24

Since Diagonals of a Rhombus bisect at Right angles,

So we apply Pythagoras theorem in ΔXOY

XY² = XO² + YO²

⇒ XY² = 49 + 576 = 625

⇒ XY = 25 units

Given: Ratio of angles of triangles = mQ : mR : mP = 1 : 2 : 1

Let mQ = x, mR = 2 x, mP = x

mQ + mR + mP = 180°

⇒ 4x = 180

⇒ x = 45

⇒ mR = 90
⇒ mQ = 45°
⇒ mP = 45°

It is an isosceles right angled triangle at R.

Let PR = QR = d

Applying Pythagoras theorem

PR² + QR² = PQ²

2d² = PQ²

⇒ 2d² = 24

⇒ d² = 12

⇒ d = 2√3

So PR = 2√3 units

and QR = 2√3 units

From the given info,

As AB² + BC² + AC² = 600

and BC = 10,

⇒ AB² + AC² = 500 ...... (1)

As AD is median,

⇒ BD = 5 ...... (2)

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB² + AC² = 2(AD² + BD²)

Substituting values for (1) and (2) in above equation,

⇒ 500 = 2 (AD² + 5²)

⇒ 250 = AD² + 25

⇒ AD² = 225

⇒ AD² = 15²

⇒ AD = 15

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB² = BD² + AD² ...... (1)

And, AC² = AD² + DC² ...... (2)

It is given that AD² = BD.DC …… (3)

From equation (1) and (3), we get,

AB² = BD² + BD.DC

⇒ AB² = BD(BD + DC) = BD. BC

Now, from equation (2) and (3),

AC² = BD.DC + DC²

⇒ AC² = DC (BD + DC)

⇒ AC² = DC(BC)

Also, BC = BD + DC,

Substituting the value of BD and DC in above equation, we get,

⇒ BC² = AB² + AC²

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB² = BD² + AD² ...... (1)

And, AC² = AD² + DC² ...... (2)

It is given that AB² = BD.BC

From equation (1) and (2), we get,

AB² – BD² = AC² – DC²

⇒ AB² – BD² = AC² – (BC – BD)²

⇒ BC² + BD² – 2BC.BD = AC² + BD² – AB²

⇒ BC² – 2BC (BD) = AC² – AB²

Substituting the value of BD above, we get,

⇒ BC² – 2(AB)² = AC² – AB²

⇒ BC² = AB² + AC²

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB² = BD² + AD² ...... (1)

And, AC² = AD² + DC² ...... (2)

It is given that AC² = CD.BC

From equation (1) and (2), we get,

AB² – BD² = AC² – CD²

⇒ AB² – (BC – CD)² = AC² – CD²

⇒ AB² – BC² – CD² + 2.BC.CD = AC² – CD²

⇒ AB² – BC² + 2.BC.CD = AC²

Substituting the value of CD above, we get,

⇒ AB² – BC² + 2AC² = AC²

⇒ AB² + AC² = BC²

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB² + AC² = 2(AD² + BD²)

As it is given that AD = BD,

⇒ AB² + AC² = 4.BD²

Since AD is the median to line BC, we have,

Substituting the value of BD in above equation, we get,

AB² + AC² = BC²

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

Let the length of upper part of pole be x,

The length of AB becomes 25 – x.

Applying Pythagoras theorem in ΔABC,

(AC’)² = AB² + (BC’)²

⇒ x² = (25 – x)² + 15²

⇒ x² = 625 + x² – 50x + 225

⇒ 50x = 850

⇒ x = 17 m

∴ The length of the upper part of the pole is 17 meters.

From the given information, we construct the figure as:

As AM is perpendicular BC, ΔACM and ΔABM are right triangle.

By using Pythagoras Theorem, we get,

AC² = CM² + AM² ...... (1)

Also, AB² = BM² + AM² ...... (2)

Eliminating AM² from equations (1) and (2), we get,

AC² – CM² = AB² – BM²

⇒ AC² – AB² = CM² – BM²

By using the identity a² – b² = (a + b)(a – b) on above equation,

⇒ AC² – AB² = (CM + BM)(CM – BM)

⇒ AC² – AB² = BC(CM – (BC – CM))

⇒ AB² – AC² = BC(2CM – BC)

As D is mid point of BC,

⇒ AB² – AC² = BC(2CM – 2BD)

⇒ AB² – AC² = 2BC(CM – BD)

⇒ AB² – AC² = 2BC(CM – CD)

⇒ AB² – AC² = 2BC(–DM)

⇒ AB² — AC² = 2BC.DM

Hence, proved.

ΔABC is a right triangle, right angle at B and BD is perpendicular from B vertex to hypotenuse AC.

So, as we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

⇒ BD² = AD.CD

⇒ BD² = (AC – CD).CD

As, AC = 5CD,

⇒ BD² = (5CD – CD).CD

⇒ BD² = 4CD²

⇒ BD² = (2CD)²

⇒ BD = 2CD

Hence, proved.

In ΔABC, we know that sum of all interior angles in triangle is equal to 180°

⇒ ∠P + ∠Q + ∠R = 180

Also, given that, ∠P + ∠Q = ∠R,

⇒ 2∠R = 180

⇒ ∠R = 90°

Using Pythagoras Theorem,

PQ² = PR² + QR²

⇒ PQ² = 7² + (24)²

⇒ PQ² = 49 + 576

⇒ PQ² = 625

⇒ PQ = 25

∴ Option (b) is correct.

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of each side other than the hypotenuse is the geometric mean of length of hypotenuse and segment of hypotenuse adjacent to the side.

⇒ AB² = BD.BC

⇒ 20 = 4. BC

⇒ BC = 5

Also, CD = BC – BD

CD = 5 – 4

CD = 1

∴ Option (d) is correct.

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB² + AC² = 2(AD² + BD²)

⇒ 50 = 2(3² + BD²)

⇒ 50 = 2 (9 + BD²)

⇒ 25 = 9 + BD²

⇒ BD² = 16

⇒ BD = 4

⇒ BC = 2BD = 8

∴ Option (c) is correct.

In ΔABC, applying Pythagoras Theorem, we get,

AC² = AB² + BC²

As AB = BC,

⇒ AC² = 2AB²

⇒ AB:AC = 1:√2

∴ Option (c) is correct.

In Δ ABC, using the theorem of Apolloneous, we know that,

For BM to be the median, we have,

AB² + AC² = 2(BM² + CM²)

Also, as Δ ABC is right angled at B,

⇒ AB² + AC² = AC² = 10² = 100

⇒ 100 = 2 (BM² + 5²)

⇒ 50 = BM² + 25

⇒ BM = 5

∴ Option (a) is correct.

Let us assume,

⇒ AB = BC = x and AC = √2x

As we see that, AC² = AB² + BC²

So by converse of Pythagoras Theorem, ΔABC is right angled and right angle at B.

∴ Option (c) is correct.

Let us assume,

⇒ AB = x, AC = 2x and BC = √3x

As we see that, AC² = AB² + BC²

So by converse of Pythagoras Theorem, ΔABC is right angled and right angle at B.

Let ∠C = θ,

⇒ θ = 30°

∴ Option (b) is correct.

Let ∠X = A,

⇒ ∠Y = 2A and ∠Z = 3A

As sum of all angles in a triangle is equal to 180°

∠X + ∠Y + ∠Z = 180

⇒ 6A = 180

⇒ A = 30

⇒ ∠Z = 90°

⇒ YZ = 7.5

∴ Option (d) is correct.

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

BD² = AD.CD

⇒ 5² = 5.CD

⇒ CD = 5

∴ Option (c) is correct.

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB² + AC² = 2(AD² + BD²)

⇒ 130 = 2(7² + BD²)

⇒ 65 = 49 + BD²

⇒ 16 = BD²

⇒ BD = 4

∴ Option (a) is correct.

Let the side of square be a.

Diagonal = √2a

⇒ 5√2 = √2a

⇒ a = 5

∴ Option (b) is correct.

Let the other side of rectangle be x

By Pythagoras Theorem,

AD² = AC² + CD²

13² = AC² + 5²

⇒ 169 – 25 = AC²

⇒ 144 = AC²

⇒ AC² = 12²

⇒ AC = 12

Perimeter = 2(AC + CD)

⇒ Perimeter = 2(12 + 5)

⇒ Perimeter = 34

∴ Option (b) is correct.

Median of an equilateral triangle is also a perpendicular to the base.

Let the side be a

⇒ a = 2

∴ Option (c) is correct.

Let the length of altitude be x and side be a.

⇒ 3a = 6

⇒ a = 2

⇒ x = √3

∴ Option (d) is correct.

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB² + AC² = 2(AD² + BD²)

⇒ BC² = 2 (36 + BD²)

⇒ 4BD² = 72 + 2BD²

⇒ 2BD² = 72

⇒ BD² = 36

⇒ BD = 6

So, BC = 12 as BC = 2BD

By Pythagoras Theorem in ΔABC,

AC² = BC² – AB²

⇒ AC² = 144 – 100

⇒ AC = 2√11

∴ Option (d) is correct.

Since ∠Q = 90 and PQ = QR , therefore ∠QPR = ∠QRP = θ

As sum of all of triangle = 180°

⇒ 2θ + 90 = 180

⇒ θ = 45°

⇒ PQ = 2√2

∴ Option (b) is correct.

By Pythagoras Theorem in ΔABD,

AB² = AD² + BD² ...... (1)

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

AD² = BD.CD ...... (2)

From equation (1) and (2)

AB² = BD.CD + BD²

⇒ AB² = BD(CD + BD)

⇒ AB² = BD.BC

∴ Option (a) is correct.

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

⇒ AD² = BD.DC

∴ Option (d) is correct.