7. Find the perimeters of ODABE id the rectangle BCDE in this figure. Which figure has greater perimeted and by how much? 34 zona fcmܝ c 0 1 2 3 cm E
Answers
Answer:
Perimeter of ΔABE = AB+BE+AE
= 3
3
1
+2
5
1
+4
3
2
=
3
10
+
5
11
+
3
14
=
15
50+33+70
=
15
153
=10
15
3
cm
Perimeter of Rectangle BCDE = 2(l+b)
=2(2
5
1
+1
3
2
)
=2(
5
11
+
3
5
)
=2(
15
33+25
)
=2(
15
58
)
=
15
116
=7
15
11
cm
So,ΔABE has greater perimeter
Difference of perimeter =
15
153
−
15
116
=
15
37
=2
15
7
cm
Hence ΔABE has greater perimeter by =2
15
7
cm