Find the perimeter of
•∆ ABE
•the rectangle BCDE
in this figure. Whose perimeter is greater ?
plz answer the questions with solution
Answers
Answer:
the answer is here
Step-by-step explanation:
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
\begin{gathered}Difference\: of\: the\: perimeter \\= \frac{153}{15}-\frac{116}{15}\\=\frac{153-116}{15}\\=\frac{37}{15}\end{gathered}Differenceoftheperimeter=15153−15116=15153−116=1537
Therefore ,
Perimeter of ∆ABE > Perimeter of BCDE
Perimeter of ∆ABC (37/15) cm greater than perimeter of BCDE.
••••
Given :
- AB = 5/2 cm
- BE = 2 3/4 cm
- AE = 3 3/5 cm
- EB = 7/6 cm
To find :
- perimeter of ∆ ABE
- the perimeter of rectangle BCDE
- whose perimeter is greater
Solution :
⇒ Let's find the Perimeter of ∆ ABE.
Perimeter of Triangle = Sum of all sides
= a + b + c
= 5/2 + 2 3/4 + 3 3/5
= 5/2 + 11/4 + 18/5 [LCM of 2,4 and 5 is 20]
= 5 × 10/2 × 10 + 11 × 5/4 × 5 + 18 × 4/5 × 4
= 50/20 + 11/20 + 72/20
= 50 + 11 + 72/20
= 133/20
- Therefore, the perimeter of ∆ ABE is 133/20 cm.
⇒ Let's find the Perimeter of the Rectangle.
= 2 (l + b)
= 2 (2 3/4 + 7/6)
= 2 (11/4 + 7/6) [LCM of 4 and 6 is 12]
= 2 {(11 × 3/4 × 3) + (7 × 2/6 × 2)}
= 2 (33/12) + (14/12)
= 2 (33 + 14/12)
= 2 × (47/12)
= 2/1 × (47/12)
= 2 × 47/1 × 12
= 94/12
- Therefore, the perimeter of Rectangle BCDE is 94/12 cm.
⇒ Whose perimeter is greater?
- To compare them, they must have the same denominators.
⇒ 133/20
⇒ 94/12
LCM of 20 and 12 = 60
⇒ 133 × 3/20 × 3
= 399/60
⇒ 94/12
= 94 × 5/12 × 5
= 470/60
→ 470/60 ≥ 399/60
→ 94/12 ≥ 133/20
- Therefore, the perimeter of the rectangle is greater.