Triangle

FE

D

CB

A

Ratio of the area of triangles

1l2l

1A2A

2

1

AA 2

2

1 )(ll

Theorem 1

Example

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

Find ADEofareaABCofarea

ADEofareaABCofarea

2

73

499

D

4 cm3 cmEC

B

A

2

AEAC

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

QRSY

PQRofareaPSYofarea

2

5030

225

PSYofarea

81PSYofarea

81

Y

X

S

RQ

P

30 cm20 cm

81

PQRSQX ~

2

QRQX

PQRofareaSQXofarea

2

5020

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

heightbase21Area

D CB

A

What is the relationship between the heights of

and ?ABD ADC

h

D CB

A

Triangles have common height h

ADCofareaABDofarea

h

D CB

A

For triangles with common height,

find

hDC

hBD

2121

DCBD

For triangles with common height,

D CB

A

ADCofareaABDofarea

2

1

bb

Theorem 2

D CB

A

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

ADCofareaABDofarea

Find

D CB

A

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

ADCofareaABDofarea

Find

ABD ADCand have common height,

ADCofareaABDofarea

DCBD

D CB

A

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

ADCofareaABDofarea

Find

ABD ADCand have common height,

ADCofareaABDofarea

DCBD

14

= 4

D1 cm

3 cm

CB

A

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

Find BDCofareaABDofarea

D1 cm

3 cm

CB

A

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

Find BDCofareaABDofarea

ABD ADCand have common height,

BDCofareaABDofarea

CDAD

13

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

RXQofareaPRXofarea

xx

= 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 ,

PQRofareaPXRofarea

xxx65

6

xx

116

116

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 44116

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

PXRofareaRXYofarea

yxy35

3

yy

83

83

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 2483 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

RMPS

RNMofareaPNSofarea 22

xx

= 4

ARNMofareaLet APNSofareaThen 4,

A

4A

Find the area of : area of PQMN.NRS

4A

Ay

2y

Considering RNSandMNR They have the common height.

RNSofareaMNRofarea

AMNRofareaLet ARNSofareathen 2

2A

yy

2

21

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

XYQofareaPXYofarea

21

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

PXYQofareathen

21

XYQofareaPXYofarea

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

QYRofareaPYQofarea

2y3y

2A

A

2x

x

Y

X

R

Q

P

= 2A : 5A

= 2 : 5

23

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

31

QRXofareaRSXofarea10.) In the figure, if ,

Find PQRStrapeziumofarea

RSXofarea

X

S R

QP

31

QRXofareaRSXofarea10.) In the figure, if ,

heightcommonthehaveQRXandRSX

A

3A

X

S R

QP

31

QRXofareaRSXofarea10.) In the figure, if ,

31

XQSX

x

3x

xx

3

A

3A

X

S R

QP

31

QRXofareaRSXofarea

10.) In the figure, if ,

x

3x

,similararePQXandRSXSince

PQXofareaRSXofarea

2

QXSX 2

3

xx

91

A

3A

9A

X

S R

QP

31

QRXofareaRSXofarea

10.) In the figure, if ,

x

3x

,heightandbasesamethehaveRPQandSPQSince

A

3A

9A

areaequalhaveRPQandSPQ

SPQofarea A3

3A

X

S R

QP

31

QRXofareaRSXofarea

10.) In the figure, if ,

x

3x

A

3A

9A

3A

PQRStrapeziumofareaRSXofarea

AAAAA

393

AA

16

161

FE

D

CB

A

Ratio of the area of

similar triangles

1l2l

1A2A

2

1

AA 2

2

1 )(ll

Theorem 1

For triangles with common height,

D CB

A

ADCofareaABDofarea

2

1

bb

Theorem 2