The document defines various types of quadrilaterals and their properties: quadrilateral, parallelogram, rectangle, rhombus, square. It also lists several important theorems about areas of triangles and parallelograms, including: parallelograms on the same base and between the same parallels have equal areas; the area of a triangle is half the area of a parallelogram on the same base and between the same parallels. Several questions and proofs are provided applying these theorems to calculate and compare areas of figures.

2. Quadrilateral
I have exactly four sides.
A quadrilateral is a
2-dimensional closed shape
With four straight sides. Rectangle
I have all of the properties of
The llgm PLUS:
Parallelogram (+) One angle is of 90 or 4 right
I have:
-2 pairs of opposite Angles
-Sides are equal and ll (+) Diagonals are equal, congruent
-Sides. Rhombus
-Opposite pairs of angles
-Are equal(congruent) I have all of the properties
-Diagonals bisect each other Of the llgm plus:
-Consecutive angles supplementary (+) Adjacent sides are equal
Or co-interior are 180 Or 4 congruent sides
Diagonals form 2 congruent Triangles (+) Diagonals are
perpendicular.
Square
Hey look at me!
I have all the properties of the
Parallelogram and the Rectangle and the Rhombus!
I have it all!!

3. Important Theorems
1.A median divides a Triangle in 2 parts of equal area.
2. A diagonal divides a Parallelogram in 2 parts of equal area.
3. Area of parallelogram is the product of base and adjacent side
4. Area of Triangle is half the product of base and height.
5. Parallelogram lying between same base and between same parallels are equal in area
6. Triangle lying on same base and between same parallels are equal in area.
7. If a Triangle and Parallelogram lie on same base and between same parallels then the
Area of Triangle is half the area of parallelogram.
8. Two triangles having the same base and equal area lie between the same parallels
9. Triangles having same base and equal area must have equal altitude

4. Theorem 5
llgms lying on same base and between same parallels are equal in
area.
Pf- In triangle ADE and CBF
Angle A = Angle B ( Corresponding Ang.)
Angle E= Angle F ( “ “ “)
AD=BC ( Opp. Sides of llgm)
Adding Area of CDEB to both Triangles we get:
Tri. ADE + area of EDCB= Tri. BCF+ area of EDCB
= llgm ABCD= llgm DEFC

5. Question 1 on Theorem 7
If E,F,G and H are respectively the mid-points of the sides of a
llgm ABCD.
Show that ar(EFGH)=1/2 ar(ABCD)
Solution-
Given-ABCD is a llgm.E,F,G and H are its mid- points. So EFGH is a llgm.
To Prove-ar(EFGH)=1/2 ar(ABCD)
Const.-Join H to F such that 2 llgms DHFC and HFBA are formed.
Pf-ar(GHF)=1/2 ar(DHFC){if a triangle and a llgm lie on same base and
b/w same lls then ar of triangle=1/2 the ar of llgm}
ar(HEF)=1/2 ar(HFBA){“ “ “ “ “}
Ar(GHF+HEF)=1/2 ar(DHFC+HFBA)
So ar(EFGH)=1/2 ar(ABCD),hence proved.

6. Question 2 on Theorems 5 and 7
In figure,PQRS and ABRS are llgms and X is any point on side BR.Show
that-(i)ar(PQRS)=ar(ABRS),(ii)ar(AXS)=1/2 ar(PQRS).
Solution-
Given-PQRS and ABRS are two llgms.
To Prove-(i)ar(PQRS)=ar(ABRS),(ii)ar(AXS)=1/2 ar (PQRS)
Pf-i)ar(PQRS)=ar(ABRS){llgms lying on same base and b/w same lls
are equal in area)
ii)ar(AXS)=1/2 ar(ABRS){if a triangle and llgm lie on same base and
b/w same lls then ar of triangle=1/2 ar of llgm}
Ar(ABRS)=Ar(PQRS){Proved above}
So ar(AXS)=1/2 ar (PQRS).

7. Question 3
Show that the diagonals of a llgm divide it into 4 triangles of equal
areas.
Solution-
Given-ABCD is a llgm
To Prove-diagonals of a llgm divide it into 4 triangles of equal areas.
Pf-ar(ADB)=ar(ABC){triangles lying on same base and b/w same lls are
equal in area}
subtract AOB from both sides
=ar(AOD)=ar(BOC)
ar(ABC)=ar(DBC){“ “ “ “ “}
subtract COB from both sides
=ar(AOB)=ar(COD)
So ar(AOD)=ar(BOC)=ar(AOB)=ar(COD),hence proved.

8. Question 4
P and Q are any two points lying on the sides DC and AD respectively
of a llgm ABCD.Show that ar(APB)=ar(BQC).
Solution-
Given-ABCD is a llgm and APB and BQC are two
triangles in it.
To Prove-ar(APB)=ar(BQC)
Pf-ar(APB)=1/2 ar (ABCD){if a tringle and a llgm are on same base and
b/w same lls then ar of triangle=1/2 ar llgm}
ar(BQC)=1/2 ar (ABCD){“ “ “ “}
From above,
ar(APB)=ar(BCQ),hence proved.

9. Given: XY ll AC, ABCD= trapezium
To prove: Area of Triangle AXD= Trian.ACY
Pf : 1. Area of triangle AXD= Area of Tri. AXC
(Same base AX and between same lls AX and
DC)
2. Area of Tri.AXC= Triangle AYC ( Same base AC
and same lls AC and XY)
From 1 and 2
3. Area of Triangle. AXD= Area of Tri. ACY

10. Given: BF ll AC
To prove: Area . AEDF= Area. ABCDE
Pf. 1. Area of ACB= Area. ACF
( same base AC and between same lls
AC and BF)
2. Area ACB + Quadri.ACDE= Area ACF + Quadri. ACDE
= Area of AEDF= Area of ABCDE

11. Given: Triangle ABC, D,E,F mid –points of Sides BC, AC and AB respectively.
To Proove:1. BDEF is a llgm and whose area =1/2=Tri. ABC
2. DEF=1/2 ABC
Pf. 1. E and F are mid-points and A line segment Joining two pint F and E
Is parallel to third side and FE=1/2 Of BC
2. But BD=DC, FE=1/2(BD + DC)
FE=BD and FE ll BD Hence a Quadrilateral whose opposite sides
are equal and parallel is a parallelogram.
Triangles. FBD=DEF, DEF=AFE,DEF=EDC( Diagonal of a llgm Divides a llgm in 2 Parts of
Equal area)
Hence we can say Area. Of BDEF=1/2 ABC
2. Triangles. FBD=DEF, DEF=AFE,DEF=EDC( Diagonal of a llgm Divides a llgm in 2 Parts of
Equal area) Hence area of DEF=1/4 ABC