Question

7.

Find the perimeters of (1) AABE (ii) the rectangle BCDE in
this figure. Which figure has greater perimetre and by how
much?
​

Answer 1

Answer:

1)AABE 2) the rectangle BCDE in this figure

Answer 2

Answer:

( i ) :

Sides of Δ ABE : -

3 \dfrac{1}{3}3

3

1

cm

2 \dfrac{1}{5}2

5

1

cm

4 \dfrac{2}{3}4

3

2

cm

Perimeter of ΔABE = sum of all sides

Perimeter of ΔABE = 3 \dfrac{1}{3}\: cm + 2 \dfrac{1}{5}\: cm +4 \dfrac{2}{3}3

3

1

cm+2

5

1

cm+4

3

2

cm

Perimeter of ΔABE = 10 / 3 cm + 11 / 5 cm + 14 / 3 cm

Perimeter of ΔABE = 50 / 15 + 33 / 15 + 70 / 15 cm

Perimeter of ΔABE = ( 50 + 33 + 70 ) / 15 cm

Perimeter of ΔABE = 153 / 15 cm

( ii ) :

Perimeter of rectangle = 2( length + breadth )

Perimeter of rectangle = 2( 2 \dfrac{1}{5}\:cm+ 1\dfrac{2}{3}2

5

1

cm+1

3

2

)

Perimeter of rectangle = 2( 11 / 5 cm + 5 / 3 cm )

Perimeter of rectangle = 2( 33 / 15 cm + 25 / 15 cm )

Perimeter of rectangle = 2( 58 / 15 )

Perimeter of rectangle = 116 / 15 cm

As 153 / 15 > 1116 / 15. Triangular figure has greater perimeter and by ( 153 - 116 ) / 15 ; 37 / 15