Triangle

FE

D

CB

A

Ratio of the area of triangles

1l2l

1A2A

2

1

A

A 2

2

1 )(l

lTheorem 1

Example

In the figure, BC// DE, AC= 3 cm and CE= 4 cm.

FindADEofarea

ABCofarea

ADEofarea

ABCofarea

2

7

3

49

9

D

4 cm3 cmEC

B

A

2

AE

AC

Class work

In the figure, PSQ, QXR and RYP are straight lines.

If the area of is , find the area of the parallelogram SXRY.

PQR 2225 cm

Y

X

S

RQ

P

30 cm20 cm

Y

X

S

RQ

P

30 cm20 cm

Identify the similar triangles first.PQRPSY ~

2

QR

SY

PQRofarea

PSYofarea

2

50

30

225

PSYofarea

81PSYofarea

81

Y

X

S

RQ

P

30 cm20 cm

81

PQRSQX ~

2

QR

QX

PQRofarea

SQXofarea

2

50

20

225

SQXofarea

36SQXofarea 36

Area of parallelogram = 225 – 81 - 36

= 1082108 cmArea of parallelogram is

TriangleC

BA

Triangle

baseA B

C

height

C

BA

Triangle

C

BA

base

height

Triangle

C

BA

Triangle

C

BA

Triangle

baseheight

Area of Triangle

baseA B

C

height

heightbase2

1Area

D CB

A

What is the relationship between the heights of

and ?ABD ADC

h

D CB

A

Triangles have common height h

ADCofarea

ABDofarea

h

D CB

A

For triangles with common height,

find

hDC

hBD

2

12

1

DC

BD

For triangles with common height,

D CB

A

ADCofarea

ABDofarea

2

1

b

bTheorem 2

D CB

A

Eg.3) Given that BC : DC = 5: 1

ADCofarea

ABDofarea

Find

D CB

A

Eg.3) Given that BC : DC = 5: 1

ADCofarea

ABDofarea

Find

ABD ADCand have common height,

ADCofarea

ABDofarea

DC

BD

D CB

A

Eg.3) Given that BC : DC = 5: 1

ADCofarea

ABDofarea

Find

ABD ADCand have common height,

ADCofarea

ABDofarea

DC

BD

1

4

= 4

D1 cm

3 cm

CB

A

Eg.4) Given that AD = 3 cm and CD = 1 cm.

Find BDCofarea

ABDofarea

D1 cm

3 cm

CB

A

Eg.4) Given that AD = 3 cm and CD = 1 cm.

Find BDCofarea

ABDofarea

ABD ADCand have common height,

BDCofarea

ABDofarea

CD

AD

1

3

3

Class work5.) In the figure, find the area of : area of .PRX QRX

xx

4 cm5 cm

X

R

QP

Class work5.) In the figure, find the area of : area of .PRX QRX

xx

4 cm5 cm

X

R

QP

heightcommonthehaveRXQandPRX

Class work5.) In the figure, find the area of : area of .PRX QRX

RXQofarea

PRXofarea

x

x

= 1xx

4 cm5 cm

X

R

QP

6.) In the figure, given that QX:XR = 5:6 and PY:YR =5:3

calculate the area of (a) (b)

PXR

RXY

Y

X RQ

P

244 cmisPQRofareatheif

6.) In the figure, given that QX:XR = 5:6 and PY:YR =5:3

Y

X RQ

P

244 cmisPQRofareatheif

5x 6x

heightcommonthehavePQRandPQXPXR ,

PQRofarea

PXRofarea

xx

x

65

6

x

x

11

6

11

6

6.) In the figure, given that QX:XR = 5:6 and PY:YR =5:3

Y

X RQ

P

244 cmisPQRofareatheif

5x 6x

heightcommonthehavePQRandPQXPXR ,

PXRofarea 4411

6 24

6.) In the figure, given that QX:XR = 5:6 and PY:YR =5:3

Y

X RQ

P

244 cmisPQRofareatheif

5x 6x

heightcommonthehaveRXYandPXR

PXRofarea

RXYofarea

yx

y

35

3

y

y

8

3

8

3

5y

3y

6.) In the figure, given that QX:XR = 5:6 and PY:YR =5:3

Y

X RQ

P

244 cmisPQRofareatheif

5x 6x

RXYofarea 248

3 9

5y

3y

N

M

S

RQ

P

7) In the figure, PQRS is a rectangle.

M is a midpoint of QR. PR and MS intersects at N.

Find the area of : area of PQMN.NRS

In the figure, PQRS is a rectangle.

N

M

S

RQ

P

Find the area of : area of PQMN.

M is a midpoint of QR. PR and MS intersects at N.

NRS

M is a midpoint of QR. PR and MS intersects at N.

In the figure, PQRS is a rectangle.

xx

N

M

S

RQ

PFind the area of : area of PQMN.NRS

2x

xx

N

M

S

RQ

P

Find the area of : area of PQMN.NRS

RNMPNS ~2

RM

PS

RNMofarea

PNSofarea2

2

x

x= 4

ARNMofareaLet APNSofareaThen 4,

A

4A

Find the area of : area of PQMN.NRS

4A

Ay

2y

Considering RNSandMNR They have the common height.

RNSofarea

MNRofarea

AMNRofareaLet ARNSofareathen 2

2A

y

y

2

2

1

Find the area of : area of PQMN.NRS

4A

Ay

2y

APSRofarea 6

AA 6

2A

PQRofarea

Hence, A5

= 2A : 5A = 2 : 5

PQMNofareaNRSofarea :

,PRSgConsiderin ,RSPPQRSince

PQMNofarea

A6

Y

X

R

Q

P

8.) In the figure, PX:XQ = 1: 2, PY:YR = 3:2.Area of : Area of = ?QXY PQR

In the figure, PX:XQ = 1: 2, PY:YR = 3:2.Area of : Area of = ?QXY PQR

PXY XYQand have the common height

2x

x

Y

X

R

Q

P

XYQofarea

PXYofarea

2

1

In the figure, PX:XQ = 1: 2, PY:YR = 3:2.Area of : Area of = ?QXY PQR

PXY XYQand have the common height

APXYofarealet 2A

A

2x

x

Y

X

R

Q

P

XYQofareathen

2

1

XYQofarea

PXYofarea

A2

In the figure, PX:XQ = 1: 2, PY:YR = 3:2.Area of : Area of = ?QXY PQR

PYQ QYRand have the common height

APYQofarea 3

QYRofareathen

QYRofarea

PYQofarea

2y3y

2A

A

2x

x

Y

X

R

Q

P

= 2A : 5A

= 2 : 5

2

3

A2

PQRofareaQXYofArea :

Y

X

S

RQ

P

9.) In the figure, PQRS is a rectangle. RSX is a straight line and PX// QS. If the area of PQRS is 24 and Y is a point on QR such that QY :YR = 3:1, find the area of . SXY

X

S R

QP

3

1

QRXofarea

RSXofarea10.) In the figure, if ,

Find PQRStrapeziumofarea

RSXofarea

X

S R

QP

3

1

QRXofarea

RSXofarea10.) In the figure, if ,

heightcommonthehaveQRXandRSX

A

3A

X

S R

QP

3

1

QRXofarea

RSXofarea10.) In the figure, if ,

3

1

XQ

SX

x

3x

x

x

3

A

3A

X

S R

QP

3

1

QRXofarea

RSXofarea10.) In the figure, if ,

x

3x

,similararePQXandRSXSince

PQXofarea

RSXofarea

2

QX

SX2

3

x

x

9

1

A

3A

9A

X

S R

QP

3

1

QRXofarea

RSXofarea10.) In the figure, if ,

x

3x

,heightandbasesamethehaveRPQandSPQSince

A

3A

9A

areaequalhaveRPQandSPQ

SPQofarea A3

3A

X

S R

QP

3

1

QRXofarea

RSXofarea10.) In the figure, if ,

x

3x

A

3A

9A

3A

PQRStrapeziumofarea

RSXofarea AAAA

A

393

A

A

16

16

1

FE

D

CB

A

Ratio of the area of

similar triangles

1l2l

1A2A

2

1

A

A 2

2

1 )(l

l

Theorem 1

For triangles with common height,

D CB

A

ADCofarea

ABDofarea

2

1

b

b

Theorem 2