Pythagoras Theorem

Given: Three sides of a triangle are 8 cm, 15 cm and 17 cm.

To check if the triangle is a right-angled triangle, we verify by Pythagoras theorem.

Taking the longest side to be hypotenuse and other two sides as base and perpendicular, we have,

Hypotenuse, h = 17 cm

Perpendicular, p = 8 cm

Base, b = 15 cm

According to Pythagoras theorem,

h² = p² + b²

⇒ 17² = 8² + 15²

⇒ 289 = 64 + 225

⇒ 289 = 289

Hence, verified that the triangle is a right- angled triangle.

Given: Three sides of the triangle are 9 cm, 11 cm and 6 cm.

To check if the triangle is a right-angled triangle, we verify by Pythagoras theorem.

Taking the longest side to be hypotenuse and other two sides as base and perpendicular, we have,

Hypotenuse, h = 11 cm

Perpendicular, p = 6 cm

Base, b = 9 cm

According to Pythagoras theorem,

h² = p² + b²

⇒ 11² = 6² + 9²

⇒ 121 = 36 + 81

⇒ 121 = 117

Given: Length of ladder, l = 15 m

Now, the situation can be shown in a diagram as:

As the wall is always perpendicular to the floor or road, so ΔAFD and ΔBFC are right angled triangle.

Here b_road represents breadth of road.

To find b_road first we need to find the lengths AF and FB.

In right ΔAFD,

h = 15 m

p = 12 m

b = AF

By applying Pythagoras theorem, we have,

h² = p² + b²

⇒ 15² = 12² + AF²

⇒ 225 = 144 + AF²

⇒ AF²= 225 – 144

⇒ AF² = 121

⇒ AF = √121

⇒ AF = 11 m …………………… (1)

In right ΔBFC,

h = 15 m

p = 9 m

b = FB

By applying Pythagoras theorem, we have,

h² = p² + b²

⇒ 15² = 9² + FB²

⇒ 225 = 81 + FB²

⇒ FB² = 225 – 81

⇒ FB² = 144

⇒ FB = √144

⇒ FB = 12 m …………………… (2)

Now,

b_road = AF + FB

Substituting from eqn. (1) and (2), we have,

b_road = 11 + 12 m

b_road = 13 m

Therefore, the breath of road is 13 m.

Given: Side of rhombus = 10 cm

Length of one diagonal = 12 cm

The figure for the question is:

We know that diagonals a rhombus are perpendicular bisector to each other.

So, the ΔAOB is a right angled triangles with ∠AOB as right angle.

Now, in ΔAOB,

h = 10 cm

p = 6 cm

b = AO

By applying Pythagoras Theorem we have,

h² = p² + b²

10² = 6² + AO²

⇒ 100 = 36 + AO²

⇒ AO² = 100 – 36

⇒ AO² = 64

⇒ AO = √64 = 8 cm

Also,

OD = AO = 1/2AD [∵ Diagonals of rhombus bisect each other]

⇒ AD = 2 × AO

⇒ AD = 2 × 8

⇒ AD = 16 cm

Thus the length of the other diagonal is 16 cm.

Given: ∠Q is a right angle.

To prove: PS² + QR²= PR² + QS²

Proof:

It can be seen both the triangles ΔPQS and ΔPQR are right angled triangles.

Applying Pythagoras Theorem to ΔPQS gives,

PS² = PQ² + QS² [∵ H² = P² + B²]

⇒ PQ² = PS² - QS² ……………… (1)

Applying Pythagoras theorem to ΔPQR gives,

PR² = PQ² + QR² [∵ H² = P² + B²]

Substituting PQ² from (1) gives,

PR² = (PS² - QS²) + QR²

⇒ PR² = PS² - QS² + QR²

⇒ PR² + QS² = PS² + QR²

⇒ PS² + QR²= PR² + QS²

Hence proved.

Given: ABCD is a rhombus.

To prove: AB² + BC² + CD² + DA² = AC² + BD²

Proof:

We know that diagonals of a rhombus are perpendicular bisector to each other.

So,

AO = OC = 1/2AC

BO = OD = 1/2BD

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Also all sides of rhombus are equal,

AB = BC = CD = DA ………………… (1)

By applying Pythagoras Theorem to ΔAOB, we get,

AB² = AO² + BO² [∵ H² = P² + B²]

⇒ AB² = (1/2AC)² + (1/2BD)²

⇒ AB² = 1/4AC² + 1/4BD²

⇒ AB² = 1/4(AC² + BD²)

⇒ 4AB² = AC² + BD²

⇒ AB² + BC² + CD² + DA² = AC² + BD² [By using eqn. (1)]

Hence proved.

Given: AB = BC = CA = x

∠ADC = ∠ADB = 90°

To prove: AB² + BC² + CA² = 4AD²

Proof:

We know that in an equilateral triangle perpendicular from any vertex on opposite side bisects it.

So, BD = DC = 1/2BC ……………… (1)

By applying Pythagoras theorem in ΔABD, we get,

AB² = AD² + BD² [∵ H² = P² + B²]

By substituting BD from eqn. (1) we get,

AB² = AD² + (1/2BC)²

⇒ AB² = AD² + 1/4BC²

⇒ 4AB² = 4AD² + BC²

⇒ 4AB² - BC² = 4AD²

⇒ 4AB² - AB² = 4AD² [Given: AB = BC]

⇒ 3AB² = 4AD²

⇒ AB² + BC² + CA² = 4AD² [∵ AB = BC = CA]

Hence proved.

Given: ABC is a right angled triangle with ∠A = 90°

To prove: BQ² + PC² = BC² + PQ²

Proof:

By applying Pythagoras theorem in ΔAPQ, we get,

PQ² = AP² + AQ² ……………… (1)

By applying Pythagoras theorem in ΔABQ, we get,

BQ² = AB² + AQ² ……………… (2)

By applying Pythagoras theorem in ΔAPC, we get,

PC² = AP² + AC² ………………… (3)

By applying Pythagoras theorem in ΔABC, we get,

BC² = AB² + AC² ……………… (4)

By adding (1) and (2) we get,

BQ² + PC² = AB² + AQ² + AP² + AC²

⇒ BQ² + PC² = AB² + AC² + AQ² + AP²

Substituting from (1) and (4) we get,

BQ² + PC² = BC² + PQ²

Hence proved.

Given: AC ⊥ BD

To prove: AB² + CD² = BC² + DA²

Proof:

By applying Pythagoras theorem in ΔAOB, we get,

AB² = AO² + BO² ……………… (1)

By applying Pythagoras theorem in ΔBOC, we get,

BC² = BO² + CO² ……………… (2)

By applying Pythagoras theorem in ΔCOD, we get,

CD² = CO² + DO² ……………… (3)

By applying Pythagoras theorem in ΔDOA, we get,

DA² = AO² + DO² ……………… (4)

By adding eqn. (1) and (3), we get,

AB² + CD² = AO² + BO² + CO² + DO²

⇒ AB² + CD² = (BO² + CO²) + (AO² + DO²)

Substituting from eqn. (2) and (4) gives,

AB² + CD² = BC² + DA²

Hence proved.

Given: In ΔABC, AB>AC

Height, h = AD

To prove: AB² - AC² = BD² - CD²

Proof:

By applying Pythagoras theorem in ΔACD, we have,

AC² = AD² + CD² ……………… (1)

By applying Pythagoras theorem in ΔABD, we have,

AB² = AD² + BD² ……………… (2)

Subtracting eqn. (1) from (2), we get,

AB² - AC² = AD² + BD² - (AD² + CD²)

⇒ AB² - AC² = AD² + BD² - AD² - CD²

⇒ AB² - AC² = BD² - CD²

Hence proved.

Given: In ΔABC, AC>AB

CE ⊥ AB

BD ⊥ AC

To prove: AC² + BP² = AB² + CP²

Proof:

Given: ABC is an isosceles triangle. So, AC = BC

∠C = right angle

To prove: AD² + DB² = 2CD²

Proof:
By Pythagoras Theroem,

In ΔABC, We have

AB² = AC² + BC² [1]

In ΔADC and ΔADB, we have

AC = BC [Given]

CD = CD [Common]

AD = DB [D is the mid point]

⇒ ΔADC ≅ ΔADB [By SSS Congruency Criterion]

⇒ ∠ADC = ∠ADB [Corresponding parts of congruent triangles are equal]

Also, ∠ADC + ∠ADB = 180° [Linear Pair]

⇒ ∠ADC + ∠ADC = 180°

⇒ ∠ADC = ∠ADB = 90°

Hence, ADC and ADB are right triangles, ∴ By Pythagoras theorem

AC² = CD² + AD² [2]

BC² = CD² + BD² [3]

Adding [2] and [3], we have

AC² + BC² = AD² + BD² + 2CD²

⇒ AB² = AD² + BD² + 2CD²

⇒ (2AD)² = AD² + AD² + 2CD²

[∵ D is mid-point of AB, AD = BD and AB = 2AD = 2BD]

⇒ 4AD² = 2AD² + 2CD²

⇒ 2AD² = 2CD²

⇒ AD² + BD² = 2CD² [∵ AD = BD]

Hence, Proved!

Given: ∠A is right angle.

CD is median.

To prove: BC² = CD² + 3AD²

Proof:

We know that median divides the side into two halves.

So,

AB = 2AD = 2BD ………… (1)

By applying Pythagoras theorem in ΔADC, we have,

CD² = AD² + AC²

⇒ AC² = CD² - AD² ……… (2)

By applying Pythagoras theorem in ΔABC, we have,

BC² = AB² + AC² ………… (3)

Substituting from eqn. (1) and (2) into (3) gives,

BC² = (2AD)² + (CD² - AD²)

⇒ BC² = 4AD² + CD² - AD²

⇒ BC² = CD² + 3AD²

Hence proved.

By applying Pythagoras theorem in ΔAOZ, we get,

AO² = AZ² + ZO²

⇒ AZ² = AO² - ZO² ……………… (1)

By applying Pythagoras theorem in ΔBOX, we get,

BO² = BX² + XO²

⇒ BX² = BO² - XO² ………………… (2)

By applying Pythagoras theorem in ΔCOY, we get,

CO² = CY² + YO²

⇒ CY² = CO² - YO² ………………… (3)

By adding eqn. (1), (2) and (3), we have,

AZ² + BX² + CY² = AO² - ZO² + BO² - XO² + CO² - YO²

⇒ AZ² + BX² + CY² = AO² + BO² + CO² - XO² - YO² - ZO² …… (i)

By applying Pythagoras theorem in ΔAOY, we get,

AO² = AY² + YO²

⇒ AY² = AO² - YO² ………… (4)

By applying Pythagoras theorem in ΔBOZ, we get,

BO² = BZ² + ZO²

⇒ BZ² = BO² - ZO² …………… (5)

By applying Pythagoras theorem in ΔCOX, we get,

CO² = CX² + XO²

⇒ CX² = CO² - XO² …………… (6)

By adding eqn. (4), (5) and (6), we have,

AY² + BZ² + CX² = AO² - YO² + BO² - ZO² + CO² - XO²

⇒ AY² + CX² + BZ² = AO² + BO² + CO² - XO² - YO² - ZO² …… (ii)

By comparing eqn. (i) and (ii) we get,

AZ² + BX² + CY² = AY² + CX² + BZ²

Hence proved.

Given: ΔRST is a right angled triangle with ∠S as right angle.

X is midpoint of RS.

Y is midpoint of ST.

To prove: RY² + XT² = 5XY²

Proof:

In ΔRSY,

RY² = RS² + SY² [∵ H² = P² + B²]

⇒ RY² = (2XS)² + SY² [∵ X is midpoint of RS]

⇒ RY² = 4XS² + SY² ……………… (1)

In ΔXST,

XT² = XS² + ST² [∵ H² = P² + B²]

⇒ XT² = XS² + (2SY)² [∵ Y is midpoint of ST]

⇒ XT² = XS² + 4SY² ……………… (2)

In ΔXSY,

XY² = XS² + SY² …………… (3) [∵ H² = P² + B²]

Now, adding eqn. (1) and (2) gives,

RY² + XT² = 4XS² + SY² + XS² + 4SY²

⇒ RY² + XT² = 5XS² + 5SY²

⇒ RY² + XT² = 5(XS² + SY²)

Substituting from (3) gives,

RY² + XT² = 5XY²

Hence proved.

Let the person start his journey from a point O.

The figure given above traverses his journey. Length of BO gives his distance from starting point.

It can be seen that ΔAOB is a right angled triangle because the direction axis are always perpendicular to each other.

By applying Pythagoras theorem to ΔAOB we get,

h² = p² + b²

⇒ BO² = 10² + 24²

⇒ BO² = 100 + 576

⇒ BO² = 676

⇒ BO = √676 = 26 m

Thus, he is 26 m away from starting point.

The given triangle is shown below:

By applying Pythagoras theorem in right angled triangle ADC, we get,

AC² = AD² + DC² [∵ h² = p² + b²]

⇒ AD² = AC² – DC²

⇒ AD² = BC² – DC² [AC = BC, sides of equilateral Δ]

⇒ AD² = (2DC) ² - DC² [BC = 2DC, AD bisects BC]

⇒ AD² = 4DC² - DC²

⇒ AD² = 3DC²

Given: AB² = 2AC²

⇒ AB² = AC² + AC²

⇒ AB² = AC² + BC² [∵ Given: AC = BC]

The expression given above is equivalent to expression of Pythagoras theorem, h² = p² + b², which corresponds to a right angled triangle.

Also, h = AB and angle opposite to hypotenuse is right angle.

Thus ∠C = 90°.

The situation can be depicted by figure given below:

Here AD and BC are two rods.

Distance between top CD can be given by Pythagoras theorem.

By applying Pythagoras theorem in ΔCED, we get,

h² = p² + b²

⇒ CD² = 6² + 8²

⇒ CD² = 36 + 64

⇒ CD² = 100

⇒ CD = √100 = 10 cm

We know that the diagonals of a rhombus are perpendicular bisector to each other.

It can be solved using figure given below:

By applying Pythagoras theorem to ΔAOB, we get,

AB² = 5² + 12²

⇒ AB² = 25 + 144

⇒ AB² = 169

⇒ AB = √169 = 13 cm

Perimeter of rhombus = 4 × side

= 4 × AB

= 4 × 13 cm

= 52 cm

(i) True

Let the sides of triangle be 3x, 4x and 5x.

By applying Pythagoras theorem to this triangle, we get,

(5x) ² = (3x) ² + (4x) ²

⇒ 25x² = 9x² + 16x²

⇒ 25x² = (9 + 16)x²

⇒ 25x² = 25x²

Thus, all triangles having sides in ratio 3:4:5 will form right angled triangle.

(ii) False

Given: Length of chord = 10 cm

Since, the chord subtends right angle at the centre, then the triangle is a right angled triangle.

Let the radius of circle be x.

By applying Pythagoras theorem to this triangle we get,

10² = x² + x²

⇒ 100 = 2x²

⇒ 2x² = 100

⇒ x² = 50

⇒ x = √50 = 2√5 cm

It is found that the radius of the circle must be 2√5 cm which mismatch from the radius given in question as 5 cm. Hence, it is false statement.

(i) In a right angled triangle, the area of square drawn on the hypotenuse is equal to the SUM of the areas of the squares drawn on the other two sides.

Explanation:

It is inferred from expression of Pythagoras theorem i.e.

H² = P² + B²

(ii) In an isosceles right-angled triangle if the length of each of two equal sides is 42 cm, then the length of the hypotenuse will be 42√2 cm.

Explanation:

By Pythagoras theorem we get,

h² = 42² + 42²

⇒ h² = 1764 + 1764

⇒ h² = 3528

⇒ h = √3528 = 42√2 cm

(iii) In a rectangular figure ABCD, the two diagonals AC and BD intersect each other at point O, if AB = 12 cm, AO = 6.5 cm, then the length of BC is 5 cm.

Explanation:

We know that, diagonals of a rectangle bisect each other.

So, AC = 2 × AO

⇒ AC = 2 × 6.5

⇒ AC = 13 cm

Now, applying Pythagoras theorem to ΔABC gives,

13² = 12² + BC²

⇒ 169 = 144 + BC²

⇒ BC² = 169-144 = 25

⇒ BC = √25 = 5 cm

Let’s consider that the triangle is a right angled triangle,

By applying Pythagoras theorem we get,

(2a + 1)² = (2√2a)² + (2a-1)²

⇒ (2a)² + 1² + 2×2a×1 = 4×2a + (2a)² + 1² - 2×2a×1

⇒ 4a² + 1 + 4a = 8a + 4a² + 1 - 4a

⇒ 4a² + 1 + 4a = 4a² + 1 + 4a

Since, the sides of the triangle satisfy Pythagoras theorem, thus the triangle is a right angled triangle with angle opposite to hypotenuse as right angle.

Here,

H = 2a + 1

P = 2a-1

B = 2√2a

Thus, ∠BAC is a right angle with measure 90°.

Given: PQR and POQ are right angled triangles.

∠QPR = 90°

∠POQ = 90°

OP = 6 cm

OQ = 8 cm

PR = 24 cm

By applying Pythagoras theorem to ΔPOQ, we get,

PQ² = OP² + OQ²

⇒ PQ² = 6² + 8²

⇒ PQ² = 36 + 64

⇒ PQ² = 100 = 10²

⇒ PQ = 10 cm

Now, by applying Pythagoras Theorem to ΔPQR, we get,

QR² = PQ² + PR²

⇒ QR² = 10² + 24²

⇒ QR² = 100 + 576

⇒ QR² = 676 = 26²

⇒ QR = 26 cm

Thus, QR is 26 cm long.

Given: In a rectangle ABCD with an interior point O,

OA = 5 cm

OB = 6 cm

OD = 8 cm

We know in a rectangle ABCD with an interior point O,

OA² + OC² = OB² + OD²

⇒ 5² + OC² = 6² + 8²

⇒ 25 + OC² = 36 + 64

⇒ OC² = 100-25 = 75

⇒ OC = √75 = 5√3 cm

Thus OC is 5√3 cm long.

Given: BD = 8 cm

DC = 2 cm

AD = 4 cm

By applying Pythagoras theorem to ΔACD we get,

AC² = AD² + CD²

AC² = 4² + 2²

AC² = 16 + 4

AC² = 20 ……………………… (1)

By applying Pythagoras theorem to ΔABD we get,

AB² = AD² + BD²

AB² = 4² + 8²

AB² = 16 + 64

AB² = 80 ……………………… (2)

Now,

BC = BD + DC

BC = 8 + 2

BC = 10

BC² = 10² = 100 ……………… (3)

By adding eqns. (1) and (2) and equating with (3) we get,

BC² = AB² + AC²

⇒ h² = p² + b²

OR,

∠BAC is a right angle.