Question

Find the perimeters of
(i) ΔABE
(ii) the rectangle BCDE in this figure. Which figure has greater perimeter and by how much?
5/2 3/2 3(3/5) are the sides

Answer 1

bhai please provide the figure although i will try to answer your questions

(i) perimeter of triangle ABE is side +side +side ......ie...sum of all sides

(ii) the side of the triangle are given but sides of rectangle are not given so i can't find it

but i can help u by a simple trick of finding the perimeter of any

....

Yes any figure of the world u can find the perimeter of it by this trick

here is the trick

just add the length of all the sides of that figure u can get the perimeter of that figure

JAI SHREE KRISHNA

Answer 2

the perimeter of triangle ABE is

the perimeter of triangle is l*b.....

Solution :

From the figure given above ,

i ) 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.