Math, asked by chandrajini89, 4 hours ago

5. Find the perimeters of (i) AABE (ii) the rectangle BCDE in this
figure. Whose perimeter is greater?
3
ide​

Answers

Answered by kritikavashishth58
2

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

Step-by-step explanation:

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Answered by sukantipanda779
0

Answer:

tere to nahi ata tu kyun answer de raha hai

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