Question

find the perimeters of (i)∆ABE (ii) rectangle BCDE in this figure. which figure has greater perimeter and by how much​

Answer 1

Perimeter of the triangle = 11/5 + 10/3 + 14/3

= (33+50+70)/15

= 153/15

Perimeter of the rectangle= 5/3 + 5/3 + 5/3 + 5/3

=20/3

So, Perimeter of triangle = 10.2 cm

And, Perimeter of rectangle= 6.66 cm

So, the triangle has a greater Perimeter and by 3.54 cm.

Hope this helps you :)

Answer 2

Answer:

$\red{\bold{\underline{\underline{ \mathcal{Answer:}}}}} \\ \blue{(i) \: Perimeter \: of \: {\Delta}ABE \: = AB + BE + AE.} \\ \blue{ \tt:{ \implies{Perimeter \: of \: {\Delta}ABE \: = 3 \frac{1}{3} + 2\frac{1}{5} + 4 \frac{2}{3}.}}} \\ \blue{ \tt:{ \implies{Perimeter \: of \: {\Delta}ABE \: = \frac{10}{3} + \frac{11}{5} + \frac{14}{3}.}}} \\ \blue{ \tt:{ \implies{Perimeter \: of \: {\Delta}ABE \: = \frac{10 \times 5 + 11 \times 3 + 14 \times 5}{15}.}}} \\ \blue{ \tt:{ \implies{Perimeter \: of \: {\Delta}ABE \: = \frac{50 + 33 + 70}{15}.}}} \\ \blue{ \tt:{ \implies{Perimeter \: of \: {\Delta}ABE \: = \frac{153}{15} = \frac{51}{5} = 10.2cm.}}} \\ \red{And} \\ \green{(ii) \: Perimeter \: of \: rectangle \: BCDE = 2(DE + BE)} \\ \green{ \tt:{ \implies {( Perimeter \: of \: rectangle \: BCDE) = 2(1 \frac{2}{3} + 2 \frac{1}{5}).}}} \\ \green{ \tt:{ \implies { Perimeter \: of \: rectangle \: BCDE = 2( \frac{5}{3} + \frac{11}{5}).}}} \\ \green{ \tt:{ \implies { Perimeter \: of \: rectangle \: BCDE =2( \frac{5 \times 5 + 11 \times 3}{15}).}}} \\ \green{ \tt:{ \implies { Perimeter \: of \: rectangle \: BCDE =2( \frac{25 + 33}{15}).}}} \\ \green{ \tt:{ \implies { Perimeter \: of \: rectangle \: BCDE =2( \frac{58}{15}).}}} \\ \green{ \tt:{ \implies { Perimeter \: of \: rectangle \: BCDE =7.733cm.}}} \\ \orange{We \: can \: see \: perimeter \: of \: {\Delta}ABE \: is \: greater \: than \: the \: perimeter \: of \: BCDE.}$