practice sums on parallelograms

SOLVED EXAMPLES

1. In figure 7-15, ABCD is a parallelogram and A  50o . Find the other angles. Sol:



In parallelogram ABCD,

A  B  180o



(Since Consecutive interior angles are supplementary) Thus, 50o  B  180o or B  180o  50o  130o Also A  C  50o and B  D  130o Therefore B  130o , C  50o and D  130o 2. In figure 7-16, ABCD is a parallelogram. Find all the angles of the parallelogram (opposite angles of a parallelogram are equal)



Sol:



We know that the opposite angles of a parallelogram are equal, Therefore A  C  7 y o Now, in  ABD , A  B  C  180o or 7 y o  5 y o  6 y o  180o or 18 y o  180o 180o or y   10o 18

o



(Angle sum property of triangle)



We know that the opposite sides of a parallelogram are parallel to each other

Therefore DC parallel to AB and DB is the transversal.



Now, DBC  ABD or DBC  6 y o



(Alternate interior angles are equal)



Now, in parallelogram ABCD, B  ABD  DBC



B  5 y o  6 y o  11 y o  1110  110o



( Since y o  10o )



Since opposite angles of a parallelogram are equal

Therefore D  B



or D  110o



( Since B  110o )



Also, C  7 y o  7  10  70o Therefore A  70o (Opposite angles of a parallelogram are equal)



Therefore A  70o , B  110o , C  70o and D  110o 3. One of the adjacent angles of a parallelogram is twice the other. Find all the angles of the parallelogram. Sol:



We know that the adjacent angles of a parallelogram are supplementary Therefore A  B  180o or x o  2 x o  180o



or 3 x o  180o 180o or x   60o 3

o



A  60o and B  2  60o  120o



As opposite angles of a parallelogram are equal,

A  C  60o and B  D  120o



Hence A  60o , B  120o , C  60o and D  120o 4. In a Triangle ABC , D, E and F are respectively the mid points of sides AB, BC and CA, and C  60o and DEF  50o . Find all the angles of the Triangle DEF .



Sol: Since F and E are respectively the mid-points of sides CA, and CB of Triangle ABC

Therefore CF CE  FA EB



(Basic proportionality theorem)



Thus, FE parallel to AB



(Converse of Basic proportionality theorem)



… (1)



Also, E and D are the mid-points of sides BC and AB respectively

Therefore BE BD  EC DA



(Basic proportionality theorem) (Converse of Basic proportionality theorem)



So, ED parallel to CA … (2)



Now, from (1) and (2), we find that in quadrilateral

AD parallel to FE and AF parallel to ED ,



Therefore Quadrilateral ADEF is a parallelogram (By definition) So, A  50o (Since Opposite angles of a parallelogram are equal) Again in  BAC , since D is the mid point of AB and F is the mid point of AC,

Therefore AD AF  DB FC



By converse of Basic proportionality theorem, we have

FD parallel to BC



… 93)



Now, in quadrilateral DECF, using (2) and (3), we have

CE parallel to FD and ED parallel to CF



Therefore Quadrilateral DECF is a parallelogram (By definition) Since opposite angle of a parallelogram are equal Therefore FDE  60o Now, in Triangle DEF , FDE  DEF  EFD  180o (Angle sum property of triangle) or 60o  50o  EFD  180o



or 110o  EFD  180o or EFD  180o  110o  70o

B  70o



(Opposite angles of a parallelogram are equal)

EFD  70o and in



Therefore In Triangle DEF , FDE  60o , DEF  50o and Triangle ABC , A  50o , B  70o and C  60o



5. Given Triangle ABC . Lines are drawn through A, B and C parallel respectively to the sides BC, CA and AB forming Triangle PQR . Show that 1 BC  PR 2 Sol: In the given figure 7-19, PA parallel to BC and PB parallel to AC



Therefore PBCA is a parallelogram (by definition) Thus, BC=PA (Opposite sides of a parallelogram are equal) … (1)



Again, AR parallel to BC and RC parallel to AB



Therefore ABCR is a parallelogram (by definition) Thus, BC=QR (Opposite sides of a parallelogram are equal) Adding (1) and (2), we get BC+BC=PA+AR or 2BC=PR or BC 

1 PR 2



… (2)



( Since PA  AR  PR )



6. ABCD is a parallelogram and X, Y are the mid-points of the sides AB and DC respectively. Show that AXCY is a parallelogram. Sol: Since ABCD is a parallelogram

Therefore AB parallel to DC and AB=DC(in a parallelogram opposite sides are parallel and equal to each other)



Thus,



1 1 AB  DC (halves of equals are equal) 2 2



or AX=YC



(Since X is the mid point of AB and Y is the mid point of DC)



and AB parallel to DC implies that AX parallel to YC Hence, AXCY is a parallelogram (If a quadrilateral opposite sides are parallel and equal, then the quadrilateral is a parallelogram) In figure 7-21, Triangle ABC is an isosceles triangle in which AB=AC. AP is the bisector of exterior angles CAD and CP parallel to BA . Prove that PAC  BCA and ABCP is a parallelogram Sol: In figure isosceles triangle ABC, it is given that AB=AC

Therefore ABC  ACB (In an isosceles triangles, angle opposite equal sides are equal)



… (1)



exterior CAD  ABC  ACB

CAD  ABC  ABC CAD  2ABC



( Since ABC  ACB ) … (2)



Since AP bisects CAD

Therefore CAD  2DAP



… (3)



Now from (2) and (3), we have

DAP  ABC



But DAP and ABC are corresponding angles. Hence, AP parallel to BC , CP paralell to BA (Given)



Therefore Quadrilateral ABCP is a parallelogram (By definition) 8. In figure 7-22, ABCD is a parallelogram, Find the values x and y Sol: Since ABCD is a parallelogram

Therefore AB parallel to DC and AD parallel to BC



(By definition)



Thus,

BAC  DCA (Alternate interior angles are equal)



and DAC  ACB (Alternate interior angles are equal)



Therefore 10 y  40



and and

x



4 x  32



or y 



40 4 10



32 8 4



Hence x  8 and y  4 9. Prove that the line segment joining the mid points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. Sol:



Given: A trapezium ABCD in which AB parallel to DC , P and Q are the mid points of its diagonals AC and BD respectively. To prove: 1. PQ parallel to AB or DC 2. PQ 

1 ( AB  DC ) 2



Construction: Join DP and produce DP to meet AB in R. Proof: Since AB parallel to DC and transversal AC intersects them at A and C respectively

Therefore 1  2



… (1) (Alternate interior angles are equal)



Now, in Triangle APR and Triangle DPC , we find that

1  2



(Proved above) (Since P is the mid point of AC)



AP=CP



3  4



(Vertically opposite angles are equal)



Therefore By ASA criterion of congruence

Triangle APR  Triangle DPC . This implies that AR=DC and PR=DP … (2) (cpctc)



Now, in Triangle DRB , P and Q are mid points of DR and DB respectively

Therefore PQ parallel to RB



or PQ parallel to AB



(Since RB is the part of AB)



or PQ paralell to AB or DC( Since AB parallel to DC given) Proves (1) part Further, P and Q are the mid points of DR and DB respectively, in  DRB

Therefore PQ  1 1 1 RB  ( AB  AR )  ( AB  DC ) 2 2 2



( Since AR  DC )



10. In figure 7-24, ABCD is a parallelogram, X and Y are points on the diagonal BD such that DX=BY. Prove that AXCY is a parallelogram. Sol:



Given: ABCD is a parallelogram. X, Y are points on the diagonal BD such that DX=BY. To prove: AXCY is a parallelogram. Construction: Join AC to meet DB in O. Proof: As the diagonals of a parallelogram bisect each other, therefore, AC and BD bisect each other in O.

Therefore OB  OD



(Since O is the mid point of bisection) (Given)



and BY=DX



Therefore OB  DY  OD  DX



Thus OY=OX Now, in quadrilateral AYCX, OX=OY and OA=OC. Therefore, its diagonals bisect each other at O. A quadrilateral, whose diagonals bisect each other, is parallelogram. Hence AXCY is a parallelogram.