This unit facilitates you in,

stating Pythagoras theorem

proving Pythagoras theorem logically.

stating converse of Pythagoras

theorem.

proving converse of Pythagoras

theorem logically.

explaining the meaning of Pythagorean

triplets.

reasoning deductively and prove the

riders based on Pythagoras theorem

and its converse.

analysing and solving real life problems

based on pythagoras theorem.

Pythagoras Theorem

Pythagoras Theorem.

Converse of Pythagorastheorem.

Pythagorean triplets

Proof of Pythagoras theoremand its converse

Problems and riders based onPythagoras theorem.

11

Pythagoras

(569-479 B.C, Greece)

Pythagoras, a pupil of Thales is

perhaps best known for the

theorem on right angled triangle

which bears his name. He helped

to devlop the study of geometry in

a logical way by using undefined

terms, definitions, postualtes and

logical deductions. The influence

of his school in the fields of

mathematics, science, music,

religion and philosophy is

apparent even today.

There is geometry in the humming of the strings,

there is music in the spacing of spheres.

-Pythagoras

270 UNIT-11

BC

P

E

F

A

D

G

5

Let us recall some of the properties of a right angled triangle which we have already

learnt.

1. The side opposite to right angle is called hypotenuse.

2. In a right angled triangle, hypotenuse is the longest side.

3. In an isosceles right angled triangle, each acute angle is 45°.

4. Area of a right angled triangle is half the product of the sides

containing the right angle.

5. The perpendicular drawn from the right angled vertex to the

hypotenuse divides the triangle into two similar triangles which are similar to the

given right angled triangle.

Now, let us learn one more very interesting property about right angled triangles.

To understand the property, do the following activity.

Construct a ABC with AB = 4cm, BC = 3cm, AC = 5cm.

Observe that ABC will be exactly 90°.

Construct squares on all the three sides and divide

them into squares of sides 1cm each.

Now, count the number of small squares on all the

three sides. They are 9, 16 and 25. If we add 9 and

16, we get 25. Repeat this activity for right angled

triangles with sides, {6cm, 8cm and 10cm}, {5cm,

12cm, 13cm} etc.

What do we infer from this? We can infer that,

"In a right angled triangle, the square on the hypotenuse is equal to the sum of

the squares on the other two sides".

This statement which gives the relationship between areas of sides of a right angled

triangle is called Pythagoras theorem, named after the Greek mathematician

Pythagoras, who lived around 500BC. This theorem which is used in many branches

of mathematics has attracted the attention of many mathematicians and today we

have hundreds of varieties of proofs for it.

The statement of Pythagoras theorem can be proved practically by another way.

This was given by Henry Perigal (1830). Hence, it is called Perigal

dissection.

Study the activity given below and do it in groups.

Construct a ABC, right angled at C on a card boardas shown in the figure.

Draw squares on AB, BC and CA. In these squares, dothe constructions as shown in the figure and stepsmentioned below.

Mark the middle point (P) of the square drawn on thelongest side containing the right angle, i.e. BC. (Thiscan be obtained by joining the diagonals of the square)

Through P draw DE || AB and FG DE .

A

B C

Hypotenuse

3 cm

3 cm

3cm

3cm

4cm

4

4 cm

4 cm

cm

5cm

m

5cm

5cm

5c

A

B C

Pythagoras Theorem 271

Mark the quadrilaterals formed as 1,2,3 and 4, and the square on AC as 5. (Observethe figure)

Cut the quadrilaterals 1,2,3,4, and 5.

Arrange them on the square drawn on the hypotenuse AB. (observe the figure)

What conclusion we can draw about the areas of the squares on sides of the right

angled triangle?

Pythagoras Theorem

In a right angled triangle, the square on the hypotenuse is equal to the sum of

the squares on the other two sides.

Data : In ABC, ABC = 90°

To Prove : AB2 + BC2 = CA2

Construction : Draw BD AC.

Proof : Statement Reason

Compare ABC and ADB,

ABC ADB 90 (Data and construction)

BAD is common.

ABC ADB (Equiangular triangles)

AB

AD =

AC

AB(A A similarity criteria)

AB2 = AC.AD .....(1)

Compare ABC and BDC,

ABC BDC 90 (Data and construction)

ACB is common

ABC BDC [Equiangular Triangles]

BC AC

DC BC[AA similarity criteria]

BC2 = AC.DC .....(2)

By adding (1) and (2) we get

AB2 + BC2 = (AC. AD) + (AC. DC)

AB2 + BC2 = AC (AD + DC)

AB2 + BC2 = AC. AC = AC2 [AD + DC = AC)

AB2+ BC2 = AC2 QED

A

B C

D

272 UNIT-11

Alternate proof for Pythagoras theorem using area of trapezium

Data : In ABC, ABC 90

To prove : a2 + c2 = b2

Construction : Extend BC to D such that,

CD = AB.

At D, draw a

and mark a point E on it such that,

ED = BC.

Join E, C and E,A.

Proof : Statement Reason

Compare ABC and CDE

AB = CD (Construction)

ABC = CDE 90 (Data and construction)

BC = DE (Construction)

ABC CDE (SAS)

1) AC = CE [CPCT]

2) BAC = DCE and

3) ACB = CED

In ABC, BAC + ACB = 90° [ BAC = ECD ]

DCE + ACB = 90° [ adjacent angles of linear pair]

ACE = 90°

A

E

B DC

a

cb

a c

b b

l

l

d

Know this!

Baudhayana, an ancient Indian mathematician inhis Shulva sutras composed during 800 BC statesthat, "The diagonal of a rectangle produces both areasof which its length and breadth produce separately".

d2 = l2 + b2

For this reason, this theorem is also referred to asthe 'Baudhayana Theorem'.

Pythagoras Theorem 273

Area of ABDE = Area of ABC + Area of ACE + Area of CDE

½(a + c)(a + c) = ½ac + ½b2 + ½ac

(a + c)2 = 2ac + b2 [by cancelling 1

2 throughout]

a2 + c2 + 2ac = 2ac + b2

a2 + c2 = b2 QED

Know this

To-day we have varieties of proofs for Pythagoras Theorem. It is said that there aremore than 300 proofs, some of them are proved based on trigonometry, co-ordinategeometry, vectors etc.

Try to collect some of them and discuss in groups.

ILLUSTRATIVE EXAMPLES

Numerical problems based on Pythagoras theorem

Example 1:

In a right angled ABC, B = 90° , AC = 17cm and AB = 8cm, find BC.

Sol. Given, in ABC, B 90

AC2 = AB2 + BC2 Pythagoras theorem]

BC2 = AC2 - AB2

BC2 = 172 - 82

BC2 = 289 - 64 = 225

BC = 225 15cm

Example 2 :

In ABC, ABC = 45° , AM BC, AM = 4cm and BC = 7cm. Find the length of AC.

Sol. In AMB, AMB 90 , ABM 45 BAM 45

AMB is an isosceles right angled triangle

MA = MB = 4cm

MC = BC – MB = 7 – 4 = 3cm

MC = 3cm

In AMC, AC2= AM2 + MC2 [ Pythagoras theorem]

AC2 = 42 + 32

AC2 = 16 + 9

AC2 = 25

AC = 25 5cm

?

A

B

M C

4 c

m

7cm

45°A

B C

17cm

8cm

?

274 UNIT-11

Example 3 :

In the rectangle WXYZ, XY + YZ = 17cm and

XZ + YW = 26 cm,

calculate the length and breadth of the rectangle.

Sol. XZ + YW = 26 cm

d1 + d

2= 26 cm

2d = 26cm [d1 = d

2]

d = 13 cm

XZ = YW = 13cm

Let length = XY = x cm breadth = XW = (17 – x)cm

In WXY, WX2 + XY2 = WY2 [pythagoras theorem]

(17 – x)2 + x2 = 132

(289 – 34x + x2) + x2 = 169

(2x2 – 34x + 120 = 0) 2

x2 – 17x + 60 = 0

x2 – 12x – 5x + 60 = 0

x(x – 12) –5(x – 12) = 0

(x – 12)(x – 5) = 0

x – 12 = 0 or x – 5 = 0

x = 12 or x = 5

Length = 12cm, breadth = 5cm

Example 4 : In the ABC, the hypotenuse is greater than one of the other side by 2units and it is twice greater than the another side by 1 unit. Find the measure

of the sides.

Sol. Let AB = x and BC = y

AC = x + 2 or AC = 2y + 1

x + 2 = 2y + 1

x + 1 = 2y

1

2

x= y

AB2 + BC2 = AC2 Pythagoras theorem)

x2 + y 2 = (2y + 1)2

2

2 1

2

xx = 4y2 + 4y + 1

2

2 2 1

4

x xx =

21 1

4 4 12 2

x x

W

X Y

Z

A

B C

x

y

(+ 2)

x

or(2

+ 1)

y

Pythagoras Theorem 275

A

B

6 =x

=8-x xC D

2 24 2 1

4

x x x=

24( 2 1) 4( 1)1

4 2

x x x

4x2 + x2 + 2x + 1 = 4(x2 + 2x + 1) + 4(2x + 2) + 4

24x + x2 + 2x + 1 = 24x + 8x + 4 + 8x + 8 + 4

x2 + 2x – 16x + 1 – 16 = 0

x2 – 14x – 15 = 0

(x – 15)(x + 1) = 0

x =15 or x = – 1

If x = 15, x + 2 = 17 and

y = 1 16

82 2

x

AB = 15 units, BC = 8 units, AC = 17 units.

Example 5 :

An insect 8 m away from the foot of a lamp post which is 6m tall, crawls towards

it. After moving through a distance, its distance from the top of the lamp post is

equal to the distance it has moved. How far is the insect away from the foot of

the lamp post? [Bhaskaracharya's Leelavathi]

Sol. Distance between the insect and

the foot of the lamp post = BD = 8m.

The height of the lamp post = AB = 6m.

After moving a distance, let the insect be at C,

Let AC = CD = x m.

BC = (8 - x) m.

In ABC, B 90

AC2 = AB2+BC2 (Pythagoras theorem)

x2 = 62 + ( 8- x)2

2x = 36 + 64 - 16x + 2x

16x = 100

x = 100

6.2516

BC = 8–x = (8–6.25) = 1.75 m

The insect is 1.75 m. away from the foot of the lamp post.

Note : We can also consider BC = x, then CD = AC = (8 – x)

276 UNIT-11

Riders based on Pythagoras Theorem.

Example 6 : In the given figure, AD BC, Prove that AB2 + CD2 = BD2 + AC2

Sol. In ADC, ADC 90

AC2 = AD2 + CD2 (Pythagoras theorem) .......(1)

In DBA ADB 90

AB2 = AD2 + BD2 (Pythagoras theorem) .......(2)

Subtracting (1) from (2), we get

AB2 – AC2 = AD2 + BD2 - AD

2- CD2

AB2 – AC2 = BD2 - CD2

AB2 + CD2 = BD2 + AC2

Example 7 :

In ABC, (i) ACB = 60° (ii) AD BC.

Derive an expression for AB in terms of AC and BC.

Sol. Const: Draw DE such that EDC 60

DEC is an equilateral [ Data and

construction]

DE = DC = EC

DEA is an isosceles AED 120 and ADE 30 DAE

DE = EA

DC = EA = EC [ Axiom - 1]

AC = 2DC

Now in ADB, AB2 = AD2 + BD2 [ pythagoras theorem]

AB2 = (AC2 – DC2) + (BC – DC)2 [ pythagoras theorem]

AB2 = 4DC2 – DC2 + BC2 – 2BC . DC + DC2 [AC = 2DC]

AB2 = 4DC2 + BC2 – 2BC.DC

AB2 = 4DC2 – 2BC.DC + BC2

AB2 = 2DC (2DC – BC) + BC2

AB2 = AC (AC – BC) + BC2 [ 2DC = AC)

AB2 = AC2 – AC.BC + BC2

or AB2 = AC2 + BC2 – AC.BC

By this we have expressed AB interms of AC and BC only.

Example 8 :

Derive the formula for height and area of an equilateral triangle.

Sol. In the equilateral ABC, let AB = BC = CA = 'a' units.

Draw AD BC Let AD = 'h' units

B

AC

D

A

B

CD

E

60°

Pythagoras Theorem 277

BD = DC [ RHS theorem]

BD = DC = 2

a

Now there are two right angled triangles, ADB and ADC.

In ADB, ADB = 90° [ AD BC]

AB2= AD2 + BD2 [ Pythagoras theorem]

a2= h2 +

2

2

a

a2 = h2 + 2

4

a

2 2

1 4

a a= h2

2 2

4

4a a= h2

23

4

a= h2

Take square root on either sides.

23

4

a = 2h

3.

2

a= h

h= 3

2

a

Height of an equilateral triangle of side a units, h =3

2

a

Let us now find the area of equialteral ABC.

Area of ABC = 1

base height2

A = 1 3

2 2

aa

A = 2 3

4

a

Area of an equilateral triangle, A = 2 3

4

a

A

B CD

a a

2

a

2

a

278 UNIT-11

Example 9 :

Equilateral triangles are drawn on the sides of a right

angled triangle. Show that the area of the triangle

on the hypotenuse is equal to sum of the area of

triangles on the other two sides.

Sol.

Data : In ABC, ABC 90

Equilateral les PAB, QBC and RAC are drawnon sides AB, BC and CA respectively.

To prove : Area (PAB) + Area (QBC) = Area (RAC)

Proof : PAB QBC RAC [ Equiangular les are similar]

ConsiderArea ( PAB) Area( QBC)

Area ( RAC) Area ( RAC)=

2 2

2 2

AB BC

AC AC[ Theorem]

Area( PAB) Area( QBC)

Area( RAC)=

2 2

2

AB BC

AC

2

2

Area ( PAB) Area( QBC) AC

Area( RAC) AC

[pythagoras theoremAC2 = AB2 + BC2]

Area ( PAB) Area( QBC)

Area( RAC)= 1

Area (PBA) + Area (QBC) = Area (RAC)

Alternate Proof : APB + BQC = 2 2 2 23 3 3

4 4 4x y x y

RAC = 2

2 2 2 23 3

4 4x y x y

APB + BQC = RAC

Example 10 :

In ABD, C is a point on BD such that BC : CD = 1 : 2 and ABC is an equilateral

triangle Prove that AD2 = 7AC2

Sol. Data: In ABD, BC : CD = 1 : 2

In ABC, AB = BC = CA

To Prove: AD2 = 7AC2

Construction: Draw AE BC

Proof: In ABC,

BE = EC = 2

a and

AE = 3

2

a

2 2

3

4x yA

B C

R

P

Q

x

y

23

4

x

2

3

4y

xy2

2

+

23area of equilateral le=

4side

A

B C DE

a a

2a

3

2

a

2

a

2

a

Pythagoras Theorem 279

In ADE, AED 90 [ Construction]

AD2 = AE2 + ED2 [ Pythagoras theorem]

AD2 =

2 23

22 2

a aa

AD2 =

223 5

4 2

a a

AD2 = 2 23 25

4 4

a a

AD2 = 2 23 25

4

a a

AD2 = 228

4

a

AD2 = 7a2

AD2 = 7AC2

EXERCISE 11.1

Numerical problems based on Pythagoras theorem.

1. The sides of a right angled triangle containing the right angle are 5cm and 12cm,

find its hypotenuse.

2. Find the length of the diagonal of a square of side 12cm.

3. The length of the diagonal of a rectangular playground is 125m and the length of

one side is 75m. Find the length of the other side.

4. In LAW, LAW 90 , LNA 90 and LW = 26cm,

LN = 6cm and AN = 8 cm. Calculate the length of WA.

5. A door of width 6 meter has an arch above it having a

height of 2 meter. Find the radius of the arch.

6. The sides of a right angled triangle are in an arithmetic progression. Show that the

sides are in the ratio. 3 : 4 : 5.

7. A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole

situated just below the pillar from a distance of 27 feet away from the pillar will fly to

catch it. If both posess the same speed, how far from the pillar they are going to

meet?

W A

L

26 cmN 8cm

6cm

h = 2m

6m

280 UNIT-11

Riders based on Pythagoras theorem.

8. In MGN, MP GN. If MG = a units, MN = b units,

GP = c units and PN = d units.

Prove that (a + b)(a – b) = (c + d)(c – d).

9. In LAB, ALB 90 and LM AB

Prove that 2

2

LA AM

MBLB

10. In ABC, ABC = 900, BD AC. If AB = 'c' units,

BC = 'a' units, BD = 'p' units , CA = 'b' units.

Prove that 2 2 2

1 1 1

a c p.

Pythagorean triplets

You have learnt that the measures of three sides of a right angled triangle are

related in a special way.

If the three numbers, which are the measures of three sides of a right angled

triangle are natural numbers then they are called Pythagorean triplets.

Some of them are listed in the table given below. study them.

A B

3, 4, 5 6, 8, 10

5, 12, 13 15, 36, 39

7, 24, 25 14, 48, 50

11, 60, 61 44, 240, 244

Compare each of the Pythagorean triplets in column A with the corresponding tripletin column B.

We can conclude that; if (x , y, z) is a pythagorean triplet, then

(kx, ky, kz) is also a Pythagorean triplet where k N.

Know this!

Pythagorean triplets can be found using the following general form.

For natural number: 2n, (n2 – 1), (n2 + 1) Here 'n' may be even or odd.

For odd natural numbers: n, 21

( 1),2

n 21( 1),

2n Here n is odd

where n N.

From the above general forms any number of pythagorean triplets can be generated bygiving values to 'n'.

M

G NP

a

d

b

c

L

MA B

A

B C

D

c

a

bp

Pythagoras Theorem 281

We have learnt that a square number can be equal to sum of two square numbers.

Does this type of relationship apply to cubes or other powers?

Know this!

Fermat's last theorem [Pierre de Fermat (1601-1665), Frenchmathematician]

"It is impossible to write

- a cube as the sum of two cubes.

- a fourth power as the sum of fourth powers"

or

In general, "it is impossible to write any power beyond the second as thesum of two similar powers."

There are no a, b, c, N for which an + bn = cn where n N and n > 2

Andrew Wiles (Born in cambridge, England - 1963) saw this problem, whenhe was ten years old.

Thirty years later and with seven years of intense work Andrew wiles,proved Fermat's last theorem.

Converse of Pythagoras Theorem

Let us now learn the converse of pythagoras theorem. For this, consider the two

examples.

Example 2

• Let PQR be another triangle with QR =

5cm, PQ = 4cm and PR = 4.3 cm.

• Imagine squares on all the three sides

and find their areas.

• Here, the longest side is QR and its

length is 5 cm. The area of the square

on the longest side QR is 25 sq. cm.

• The sum of the squares on the other

two sides PQ and PR will be (16 + 18.49)

sq cm = 34.49 sq. cm which is not

equal to the area of the square on the

longest side QR.

• Now measure QPR which is

opposite to the longest side QR, you will

find that QPR=74° but not 90°.

• So, PQR is not a right angled triangle.

Example 1

• Let ABC be a triangle with

BC = 5cm, AB = 4cm and AC = 3cm.

• Imagine squares on all the three sides

and find their areas.

• Here, the longest side is BC and its

length is 5cm. The area of the square

on the longest side BC is 25 sq. cm

• The sum of the squares on the other

two sides AB and AC will be (16 + 9) sq

cm which is also 25 sq. cm.

• Now measure BAC which is opposite

to the longest side BC. you will find

that BAC = 90°.

• So, ABC is a right angled triangle,

right angled at the vertex A.

A

B C5 cm

4 cm

3 cm

P

Q R5 cm

4 cm

4.3 cm

282 UNIT-11

So, what is the basic condition for a triangle to be a right angled triangle? Compare

the areas on the sides of two triangles in the above examples and try to figure out the

necessary condtion.

The condition is: "If the square on the longest side of a triangle is equal to the

sum of the squares on the other two sides then those two sides contain a right angle."

This is converse of Pythagoras theorem. It was first mentioned and proved by Euclid.

Now let us prove the converse of Pythagoras theorem logically.

Converse of pythagoras theorem

"If the square on the longest side of a triangle is equal to the sum of the squares

on the other two sides, then those two sides contain a right angle."

A

B CD

Data : In ABC, AC2 = AB2 + BC2

To prove : ABC = 90°

Construction : Draw a perpendicular on AB at B. Select a point D

on it such that, DB = BC. Join 'A' and 'D'.

Statement Reason

Proof : In ABD,

ABD = 90° (Construction)

AD2 = AB2 + BD2(Pythagoras theorem)

But in ABC,

AC2 = AB2 + BC2(Data)

AD2 = AC2

AD = AC

Compare ABD and ABC

AD = AC (Proved)

BD = BC (Construction)

AB is common

ABD ABC (SSS)

ABD = ABC (CPCT)

ABD = ABC = 90° QED

Pythagoras Theorem 283

Comparision of Pythagoras theorem and its converse.

Now observe the following table. Study the relationship between the areas of three

sides of a triangle.

Discuss

The Pythagoras theorem has two fundamental aspects, where one is about areas and the

other is about lengths. This landmark theorem connects two main branches of

mathematics - Geometry and Algebra.

Pythagoras theorem

"In a right angled triangle, the square

on the hypotenuse is equal to the sum

of the squares on the other two sides.

A

B C

Data: In ABC, ABC = 90°

To prove: AC2 = AB2 + BC2

Converse of Pythagoras theorem

"If the square on the longest side of a

triangle is equal to the sum of the

squares on the other two sides, then

those two sides contain a right angle."

A

B C

Data: In ABC, AC2 = AB2 + BC2

To prove: ABC = 90°

A

B C a

bc

In ABC,

ABC = 90°

AC2 = AB2 + BC2

b2 = c2 + a2

If b2= c2 + a2

then ABC = 90°

A

B C

bc

a*

In ABC,

ABC < 90°

AC2< AB2 + BC2

b 2< c2 + a2

If b2< c2 + a2

then ABC < 90°

*

A

B C a

bc

In ABC,

ABC > 90°

AC2> AB2 + BC2

b2> c2 + a2

If b2> c2 + a2

then ABC > 90°

Converse

284 UNIT-11

ILLUSTRATIVE EXAMPLES

Example1 : Verify whether the following measures represent the sides of a right angled

triangle.

(a) 6, 8, 10

Sol. Sides are : 6, 8, 10

Consider the areas of square on the sides : 62, 82, 102

i.e., 36, 64, 100

Consider the sum of areas of squares on the two smaller sides : 36 + 64 = 100

62 + 82 = 102

We observe that, square on the longest side of the triangle is equal to the sum of

squares on the other two sides.

By converse of Pythagoras theorem, those two smaller sides must contain a right

angle.

Conclusion: The sides 6, 8 and 10 form the sides of a right angled triangle with

hypotenuse 10 units and 6 and 8 units as the sides containing the right angle.

Note: Without actually constructing the triangle for the given measurements of

sides it is now possible to say whether the sides represent the sides of a right

angled triangle using converse of Pythagoras theorem.

(b) 1, 2, 3

Sides are : 1, 3 , 2

Consider the areas of

squares on the sides : 12, 2

3 , 22

i.e., : 1, 3, 4

Consider the sum of the areas of squares on the two smaller sides

: 1 + 3 = 4

12 + 2

3 = 22

We observe that the square on the longest side (2 units) is equal to the sum of the

squares on the other two sides.

By converse of pythagoras Theorem, these two smaller sides must contain a right

angle.

Conclusion: 1, 2, 3 forms the sides of a right angled triangle with 2 units as

hypotenuse and 1 and 3 units as sides containing the right angle.

(c) 4, 5, 6

Sides are : 4, 5, 6

Areas of squares on the sides : 42, 52, 62

i.e., : 16, 25, 36

Pythagoras Theorem 285

Sum of areas of squares on

the two smaller sides : 16 + 25 = 41

42 + 52 62

We observe that the square on the longest side of the triangle is not equal to the

sum of the squares on the other two sides.

By converse of Pythagoras theorem, these two sides cannot contain a right angle.

Hence, 4, 5, and 6 cannot form the sides of right angled triangle.

Example 2 : In the quadrilateral ABCD, ABC 90 and

AD2 = (AB2 + BC2 + CD2). Prove that : ACD 90

A

B C

D

Proof: In ABC, ABC 90 [ Data]

AC2 = AB2 + BC2 [ Pythagoras theorem]

But AD2 = (AB2 + BC2)+CD2 [ Data]

AD2 = AC2 + CD2 [by data AB2 + BC2 = AC2]

ACD = 90° [ converse of Pythagoras theorem]

EXERCISE 11.2

1. Verify whether the following measures represent the sides of a right angled triangle.

(a) 2, 3, 5 (b) 6 3, 12, 6

(c) , 1, 2 1n n n (d) x2 – 1, 2x , x2 + 1

(e)1 1

, ,2 2

x xx (f) m2 – n2, 2mn, m2 + n2

2. In ABC, a + b = 18 units, b + c = 25 units and c + a = 17units. What type of triangle is ABC ? Give reason.

A

B Ca

bc

286 UNIT-11

EXERCISE 11.1

1] 13 cm 2]12 2 cm 3] 100m 4] 24cm 5] 3.25m 6] 3: 4: 5 7] 12 ft

ANSWERS

3. In ABC, CD AB, CA = 2AD and BD = 3AD, Prove that BCA 90 .

4. The shortest distance AP from a point 'A' to QR is 12 cm.

Q and R are respectively 15 cm and 20cm from 'A' and on

opposite sides of AP. Prove that QAR 90

5. In the isosceles ABC, AB = AC, BC = 18 cm, AD BC,

AD = 12 cm, BC is produced to 'E' and AE = 20cm.

Prove that BAE 90

6. In the quadrilateral ABCD, ADC 90 , AB = 9cm,

BC = AD = 6cm and CD = 3cm, Prove that 0ACB 90

7. ABCD is a rectangle. 'P' is any point outside it such that

PA2 + PC2 = BA2 + AD2. Prove that APC 90

R

A

P

20 cm

15 c

m

12 c

m

Pythagoras theorem

Statement of Pythagoras

theorem

Converse of Pythagoras

theorem

Pythagorean

triplets

Logical proof Riders

A

B CD E18 cm

20cm

12

cm

A

B C

D

PA

B

CD

9 cm

3 cm

6cm

6 cm