Page 181 - Ganit Kaushal

1
2
2
\ x = a (area of D)
2
2
2
⇒ x = 2a , this gives a relation between x and a.
(d) Area of shape G = 2(Area of shape C)
1
2
2
= Area of shape D = a = x (from part (c))
2
⇒ Area of shape F = Area of shape G
1 1 1 D C
(e) Area of shape A = × (Area of ∆PQS) = × ( Area of Sq. PQRS)
2 2 2
1 1 G
= × × (2x) 2 E
2 2 F
1
2
2
⇒ Area of shape A = × 4x = x = 2 × Area of shape G (From part (d))
4
2
2
(f) Area of square PQRS = 4x = 8a = 8 × Area of D = 8(2Area of C) B
= 16 × Area of C
A
(g) Seven pieces have been arranged to form a rectangle, as shown in the given
figure. The area will remain the same i.e., 16 × (Area of part C).
Are there more ways to arrange these seven pieces to form a rectangle?
Yes, certainly, there are. Students are advised to explore more arrangements.

(h) The perimeter will change. Perimeter of the square = 8x.
Perimeter of the rectangle = (2a + 2a) + (4a + 4a) = 12a = 12  x  = 62x


2 

So, the perimeter will increase. ( 62 = 8.48 > 8)
Ex 5. Match the shapes (each side measures 2 cm) in column I Column I Column II
with the corresponding perimeters in column II.
Sol. (a) → (iv) (a) (i) 16 cm
(b) → (i)
(c) → (ii) (b) (ii) 20 cm

(d) → (iii)

(c) (iii) 24 cm

(d) (iv) 28 cm

Ex 6. State True or False for each of the following statements.

(a) If the length of a rectangle is halved and the breadth is doubled, then the area of the rectangle
obtained remains the same.
(b) The area of a square is doubled if the side of the square is doubled.
(c) The perimeter of a regular octagon of side 6 cm is 36 cm.
(d) To find the cost of a frame of a picture, we need to find the perimeter of the picture.

Perimeter and Area 179