Find the perimeters of 1 ∆ ABE 2the rectangular BCDE in this figure . Whose perimeter is greater
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Perimeter of ∆ABE
= AB + BE + EA
= 3 1/3 cm + 2 1/5 cm + 4 2/3 cm
= 10/3 + 11/5 + 14/3
= ( 50 + 33 + 70 )/15
=153/15 --------( 1 )
ii ) Perimeter of BCDE
= BC + CD + DE + EB
= DE + BE + DE + BE
[ Since Opposite sides are equal ]
= 2( DE + BE )
= 2 [ 1 2/3 + 2 1/5 ]
= 2 [ 5/3 + 11/5 ]
= 2[ ( 25 + 33 )/15 ]
= 2 × 58/15
= 116/15 ------------( 2 )
From ( 1 ) and ( 2 ) , we observe that
153/15 > 116/15
Therefore ,
Perimeter of ∆ABE > Perimeter of BCDE
Perimeter of ∆ABC (37/15) cm greater than perimeter of BCDE.
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