Transcript PowerPoint of lesson study

Triangle
Ratio of the area of triangles
D
A
l2
A2
l1
A1
B
C
Theorem 1
E
A1 l1 2
( )
A2 l 2
F
Example
D
In the figure, BC// DE, AC= 3 cm and CE= 4 cm.
Find
area of ABC
area of ADE
B
A
3 cm
area of ABC  AC 


area of ADE  AE 
3
 
7
9

49
2
2
C
4 cm
E
Class work
In the figure, PSQ, QXR and RYP are straight lines.
If the area of PQR is 225 cm 2 , find the area of the
parallelogram SXRY.
P
S
Q
20 cm X
Y
30 cm
R
Identify the similar triangles first.
PSY ~ PQR
area of PSY  SY 

 
area of PQR  QR 
P
2
81
area of PSY  30 
 
225
 50 
area of PSY  81
2
S
Q
20 cm X
Y
30 cm
R
SQX ~ PQR
area of SQX  QX 

 
area of PQR  QR 
2
area of SQX  20 
 
225
 50 
2
area of SQX  36
P
81
S
Y
36
Q
20 cm X
Area of parallelogram = 225 – 81 - 36
= 108
Area of parallelogram is 108 cm 2
30 cm
R
Triangle
C
A
B
Triangle
C
height
A
base
B
Triangle
C
A
B
Triangle
C
base
height
A
B
Triangle
C
A
B
Triangle
C
height
A
base
B
Area of Triangle
C
height
A
Area
base
B
1
  base  height
2
What is the relationship between the heights of
ABD and ADC ?
A
B
D
C
Triangles have common height h
A
h
B
D
C
For triangles with common height,
find
area of ABD
area of ADC
1
 BD  h
 2
1
 DC  h
2
h
B

BD
DC
A
D
C
For triangles with common height,
Theorem 2
area of ABD b1

area of ADC b2
Eg.3) Given that BC : DC = 5: 1
area of ABD
Find
area of ADC
A
B
D
C
Eg.3) Given that BC : DC = 5: 1
area of ABD
Find
area of ADC
ABD and ADC
have common height,
A
area of ABD BD

area of ADC DC
B
D
C
Eg.3) Given that BC : DC = 5: 1
area of ABD
Find
area of ADC
ABD and ADC
have common height,
A
area of ABD BD

area of ADC DC
4

1
=4
B
D
C
Eg.4) Given that AD = 3 cm and CD = 1 cm.
area of ABD
Find
area of BDC
A
3 cm
D
1 cm
B
C
Eg.4) Given that AD = 3 cm and CD = 1 cm.
area of ABD
Find
area of BDC
ABD and ADC
have common height,
area of ABD AD

area of BDC CD
A
3 cm
3

1
D
1 cm
3
B
C
Class work
5.) In the figure, find the area of
PRX
: area of
QRX .
R
5 cm
P
x
4 cm
X
x
Q
Class work
5.) In the figure, find the area of
PRX
PRX and RXQ have the common height
: area of
R
5 cm
P
x
QRX .
4 cm
X
x
Q
Class work
5.) In the figure, find the area of
PRX
: area of
QRX .
R
area of PRX
area of RXQ
5 cm
4 cm
x

x
P
=1
x
X
x
Q
6.) In the figure, if the area of PQR is 44 cm
given that QX:XR = 5:6 and PY:YR =5:3
calculate the area of (a) PXR
(b) RXY
P
Y
Q
X
R
2
6.) In the figure, if the area of PQR is 44 cm
given that QX:XR = 5:6 and PY:YR =5:3
2
PXR , PQX and PQR have the common height
area of PXR
6x

area of PQR
5x  6x

6x
11x

6
11
P
Y
Q
5x
X
6x
R
6.) In the figure, if the area of PQR is 44 cm
given that QX:XR = 5:6 and PY:YR =5:3
2
PXR , PQX and PQR have the common height
area of PXR 
6
 44  24
11
P
Y
Q
5x
X
6x
R
6.) In the figure, if the area of PQR is 44 cm
given that QX:XR = 5:6 and PY:YR =5:3
PXR and RXY have the common height
3y
area of RXY

5x  3 y
area of PXR
3y

8y

3
8
P
5y
Y
3y
Q
5x
X
6x
R
2
6.) In the figure, if the area of PQR is 44 cm
given that QX:XR = 5:6 and PY:YR =5:3
3
area of RXY   24
8
9
P
5y
Y
3y
Q
5x
X
6x
R
2
7) In the figure, PQRS is a rectangle.
M is a midpoint of QR. PR and MS intersects at N.
Find the area of NRS : area of PQMN.
P
S
N
Q
M
R
In the figure, PQRS is a rectangle.
M is a midpoint of QR. PR and MS intersects at N.
Find the area of NRS : area of PQMN.
P
S
N
Q
M
R
In the figure, PQRS is a rectangle.
M is a midpoint of QR. PR and MS intersects at N.
Find the area of NRS : area of PQMN.
P
S
N
Q
x
M
x
R
Find the area of NRS : area of PQMN.
2x
P
S
4A
N
A
Q
x
x
M
R
PNS ~ RNM
area of PNS  PS 
2x 


 
 =4
area of RNM  RM 
 x 
2
Let area of RNM  A
2
Then, area of PNS  4 A
Find the area of NRS : area of PQMN.
x2
S
P
4A
2y
2A
N
y
R
x
A
M
Considering MNR and RNS
x
Q
They have the common height.
area of MNR
y
1


2y
area of RNS
2
Let area of MNR  A
then area of RNS  2 A
Find the area of NRS : area of PQMN.
x2
S
P
4A
2y
2A
N
y
R
x
A
M
x
Q
Considerin g PRS , area of PSR  6 A
Since PQR  RSP , area of PQR  6 A
Hence,
area of PQMN  6 A  A  5 A
area of NRS : area of PQMN = 2A : 5A
=2:5
8.) In the figure, PX:XQ = 1: 2, PY:YR = 3:2.
Area of QXY : Area of PQR = ?
Q
X
P
Y
R
In the figure, PX:XQ = 1: 2, PY:YR = 3:2.
Area of QXY : Area of PQR = ?
PXY and XYQ have the common height
area of PXY 1

area of XYQ 2
Q
2x
X
x
P
Y
R
In the figure, PX:XQ = 1: 2, PY:YR = 3:2.
Area of QXY : Area of PQR = ?
PXY and XYQ have the common height
area of PXY 1

area of XYQ 2
Q
2x
X
let area of PXY  A
then area of XYQ  2 A
x
P
2A
A
Y
R
In the figure, PX:XQ = 1: 2, PY:YR = 3:2.
Area of QXY : Area of PQR = ?
PYQ and QYR have the common height
Q
area of PYQ 3

area of QYR 2
2x
X
area of PYQ  3 A
then area of QYR  2 A
x
2A
A
P
3y
Area of QXY : area of PQR = 2A : 5A
=2:5
Y
2y
R
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
P
Q
S
Y
R
10.) In the figure, if
Find
area of RSX 1 ,

area of QRX 3
area of RSX
area of trapezium PQRS
S
R
X
P
Q
10.) In the figure, if
area of RSX 1 ,

area of QRX 3
RSX and QRX have the common height
S
R
A
3A
X
P
Q
10.) In the figure, if
area of RSX 1 ,

area of QRX 3
SX 1

XQ 3
x

3x
S
R
x
A
3A
X
3x
P
Q
area of RSX 1
10.) In the figure, if
 ,
area of QRX 3
Since RSX and PQX are similar ,
area of RSX  SX 

 
area of PQX  QX 
S
x
2
 x 
 
 3x 
2
1

9
R
A
3A
X
9A
P
3x
Q
area of RSX 1
10.) In the figure, if
 ,
area of QRX 3
Since SPQ and RPQ have the same base and height ,
SPQ and RPQ have equal area
area of SPQ  3 A
S
R
x
3A
A
3A
X
9A
P
3x
Q
area of RSX 1
10.) In the figure, if
 ,
area of QRX 3
area of RSX
A
A
1



area of trapezium PQRS A  3 A  9 A  3 A 16 A
16
S
R
x
3A
A
3A
X
9A
P
3x
Q
Ratio of the area of
D
similar triangles
A
l2
A2
l1
A1
B
C
Theorem 1
E
A1
l1 2
( )
A2
l2
F
For triangles with common height,
Theorem 2
b1
area of ABD

area of ADC
b2