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Answers
Answer:
From the fig,
AB = (5/2) cm
AE =
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 13= 18/5 cm
BE =
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 14= 11/4 cm
ED = 7/6 cm
(i) We know that,
Perimeter of the triangle = Sum of all sides
Then,
Perimeter of triangle ABE = AB + BE + EA
= (5/2) + (11/4) + (18/5)
The LCM of 2, 4, 5 = 20
Now, let us change each of the given fraction into an equivalent fraction having 20 as the denominator.
= {[(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]}
= (50/20) + (55/20) + (72/20)
= (50 + 55 + 72)/20
= 177/20
=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 15cm
(ii) Now, we have to find the perimeter of the rectangle,
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
Then,
Perimeter of rectangle BCDE = 2 × (BE + ED)
= 2 × [(11/4) + (7/6)]
The LCM of 4, 6 = 12
Now, let us change each of the given fraction into an equivalent fraction having 20 as the denominator
= 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]}
= 2 × [(33/12) + (14/12)]
= 2 × [(33 + 14)/12]
= 2 × (47/12)
= 47/6
= NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 16
Finally, we have find which one is having greater perimeter.
Perimeter of triangle ABE = (177/20)
Perimeter of rectangle BCDE = (47/6)
The two perimeters are in the form of unlike fraction.
Changing perimeters into like fractions we have,
(177/20) = (177/20) × (3/3) = 531/60
(43/6) = (43/6) × (10/10) = 430/60
Clearly, (531/60) > (430/60)
Hence, (177/20) > (43/6)
∴ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)