Please solve question no.19.c.
Dear Student,
Please find below the solution to the asked query:
Given : ABCD is a square of side 7 cm. If AE = FC = CG = HA = 3 cm, So
BE = BF = DG = DH = 4 ( We get from : 7 - 3 = 4 ) , So
AEH , BEF , CGF and DGH are isosceles right angle triangle ( As two sides equal and all have one angle at 90 from vertex of square ABCD )
i ) In AEH from angle sum property we get
AEH + AHE + HAE = 180
AEH + AEH + 90 = 180 ( Here AEH = AHE from base angle theorem as we know AE = HA and HAE = 90 vertex angle of square ABCD )
2 AEH = 90 ,
AEH = 45 --- ( 1 )
And in BEF from angle sum property we get
BEF + BFE + EBF = 180
BEF + BEF + 90 = 180 ( Here BEF = BFE from base angle theorem as we know BE = BF and EBF = 90 vertex angle of square ABCD )
2 BEF = 90 ,
BEF = 45 --- ( 2 )
And
AEH + HEF + BEF = 180 ( Linear pair angles )
45 + HEF + 45 =180 ( From equation 1 and 2 we get )
HEF = 90 , Similarly we can show : EFG = 90 , FGH = 90 and GHE = 90
Therefore,
EFGH is a rectangle . ( Hence proved )
ii ) From Pythagoras theorem we get in AEH :
HE2 = AE2 + HA2 , Substitute values we get
HE2 = 32 + 32 ,
HE2 = 9 + 9 ,
HE2 = 18 ,
HE = 3 cm and FG = 3 ( Opposite sides of rectangle EFGH as we proved in previous part )
And now we apply Pythagoras theorem and get in BEF :
EF2 = BE2 + BF2 , Substitute values we get
EF2 = 42 + 42 ,
EF2 = 16 + 16 ,
EF2 = 32 ,
EF = 4 cm and GH = 4 ( Opposite sides of rectangle EFGH as we proved in previous part )
Then,
We know area of rectangle = Length Breadth and Perimeter of rectangle = 2 ( Length + Breadth ) , So
Area of EFGH = 3 4 = 24 cm2 ( Ans )
Perimeter of EFGH = 2 ( 3 + 4 ) = 2 ( 7 ) = 14 cm ( Ans )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards,
Please find below the solution to the asked query:
Given : ABCD is a square of side 7 cm. If AE = FC = CG = HA = 3 cm, So
BE = BF = DG = DH = 4 ( We get from : 7 - 3 = 4 ) , So
AEH , BEF , CGF and DGH are isosceles right angle triangle ( As two sides equal and all have one angle at 90 from vertex of square ABCD )
i ) In AEH from angle sum property we get
AEH + AHE + HAE = 180
AEH + AEH + 90 = 180 ( Here AEH = AHE from base angle theorem as we know AE = HA and HAE = 90 vertex angle of square ABCD )
2 AEH = 90 ,
AEH = 45 --- ( 1 )
And in BEF from angle sum property we get
BEF + BFE + EBF = 180
BEF + BEF + 90 = 180 ( Here BEF = BFE from base angle theorem as we know BE = BF and EBF = 90 vertex angle of square ABCD )
2 BEF = 90 ,
BEF = 45 --- ( 2 )
And
AEH + HEF + BEF = 180 ( Linear pair angles )
45 + HEF + 45 =180 ( From equation 1 and 2 we get )
HEF = 90 , Similarly we can show : EFG = 90 , FGH = 90 and GHE = 90
Therefore,
EFGH is a rectangle . ( Hence proved )
ii ) From Pythagoras theorem we get in AEH :
HE2 = AE2 + HA2 , Substitute values we get
HE2 = 32 + 32 ,
HE2 = 9 + 9 ,
HE2 = 18 ,
HE = 3 cm and FG = 3 ( Opposite sides of rectangle EFGH as we proved in previous part )
And now we apply Pythagoras theorem and get in BEF :
EF2 = BE2 + BF2 , Substitute values we get
EF2 = 42 + 42 ,
EF2 = 16 + 16 ,
EF2 = 32 ,
EF = 4 cm and GH = 4 ( Opposite sides of rectangle EFGH as we proved in previous part )
Then,
We know area of rectangle = Length Breadth and Perimeter of rectangle = 2 ( Length + Breadth ) , So
Area of EFGH = 3 4 = 24 cm2 ( Ans )
Perimeter of EFGH = 2 ( 3 + 4 ) = 2 ( 7 ) = 14 cm ( Ans )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards,