1.
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 rightI 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 equalOr co-interior are 180 Or 4 congruent sidesDiagonals 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!!
2.
Important Theorems1.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 side4. Area of Triangle is half the product of base and height.5. Parallelogram lying between same base and between same parallels are equal in area6. 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 theArea of Triangle is half the area of parallelogram.8. Two triangles having the same base and equal area lie between the same parallels9. Triangles having same base and equal area must have equal altitude
3.
Theorem 5llgms lying on same base and between same parallels are equal inarea.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
4.
Question 1 on Theorem 7If E,F,G and H are respectively the mid-points of the sides of allgm 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.
5.
Question 2 on Theorems 5 and 7In figure,PQRS and ABRS are llgms and X is any point on side BR.Showthat-(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).
6.
Question 3Show that the diagonals of a llgm divide it into 4 triangles of equalareas.Solution-Given-ABCD is a llgmTo 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.
7.
Question 4P and Q are any two points lying on the sides DC and AD respectivelyof a llgm ABCD.Show that ar(APB)=ar(BQC).Solution-Given-ABCD is a llgm and APB and BQC are twotriangles 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.
8.
Given: XY ll AC, ABCD= trapeziumTo prove: Area of Triangle AXD= Trian.ACYPf : 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 23. Area of Triangle. AXD= Area of Tri. ACY
9.
Given: BF ll ACTo prove: Area . AEDF= Area. ABCDEPf. 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
10.
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. ABC2. DEF=1/2 ABCPf. 1. E and F are mid-points and A line segment Joining two pint F and EIs parallel to third side and FE=1/2 Of BC2. 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 ofEqual area) Hence we can say Area. Of BDEF=1/2 ABC2. Triangles. FBD=DEF, DEF=AFE,DEF=EDC( Diagonal of a llgm Divides a llgm in 2 Parts ofEqual area) Hence area of DEF=1/4 ABC
A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later.
Be the first to comment