Prove that the perimeters of (i) Triangle ABE is 177/
20
cm (ii) the rectangle BCDE is 47/
26
Answers
Answer:
= AB + BE +AE
= \frac{5}{2}+\ 2\ \frac{3}{4}\ +\ 3\ \frac{3}{5}
2
5
+ 2
4
3
+ 3
5
3
= \left[\frac{5}{2}\ +\ \frac{11}{4\ }\ +\ \frac{18}{5}\right][
2
5
+
4
11
+
5
18
]
= \left[\frac{5\times10+11\times5+18\times4}{20}\right]\ =\ \left[\frac{50+55+72}{20}\right][
20
5×10+11×5+18×4
] = [
20
50+55+72
]
= \frac{177}{20}\ =\ 8\ \frac{17}{20}
20
177
= 8
20
17
Hence, the perimeter of \triangle ABE\ \ =\ 8\ \frac{17}{20}cm△ABE = 8
20
17
cm
ii) Perimeter of rectangle BCDE
= 2 x length + breadth
= 2\ \times\ \left[2\ \frac{3}{4}+\frac{7}{6}\right]2 × [2
4
3
+
6
7
]
= 2\ \times\frac{11}{4}\ +\ \frac{7}{6}2 ×
4
11
+
6
7
= 2\ \times\ \frac{11\times3+7\times2}{12}2 ×
12
11×3+7×2
= 2\times\ \left[\ \frac{33+14}{12}\right]\ =\ 2\times\ \frac{47}{12}2× [
12
33+14
] = 2×
12
47
= \frac{47}{6}=\ 7\ \frac{5}{6}
6
47
= 7
6
5
Hence, the required perimeter is = 7\ \frac{7}{6}7
6
7
Since 8\ \frac{17}{20\ }>7\frac{5}{6}8
20
17
>7
6
5
Thus, perimeter of triangle ABE is greater than the perimeter of the rectangle BCDE