Page 180 - Ganit Kaushal

Hall: (23 ft × 15 ft); Area = 345 sq ft

Toilet: (5 ft × 10 ft); Area = 50 sq ft.
Entrance: (15 ft × 7 ft); Area = 105 sq ft.
(b) Area of the house = (180 + 50 + 70 + 180 + 120 + 345 + 105) sq ft
= 1050 sq ft (or 25 ft × 42ft = 1050 sq ft.)
(c) Perimeter of Sharan’s house = 2 × (25 ft + 42 ft) = 134 ft.
Students are advised to prepare the comparison table themselves, using the above obtained
measurements in this example 3 and example 4 before Exercise 6.4.
Ex 4. Look at the tangram pieces given in the adjoining figure.
P Q
(a) Explore and figure out how many pieces have the same area.

(b) How many times bigger is Shape D as compared to Shape C? What B E x
is the relationship between Shapes C, D and E? a a

(c) Which shape has more area: Shape D or F? Give reasons for your 2x A D
answer. C a

(d) Which shape has more area: Shape F or G? Give reasons for your F x
answer. G

(e) What is the area of Shape A as compared to Shape G? Is it twice as S x x R
big? Four times as big?

(f) Can you now figure out the area of the big square formed with all seven pieces in terms of the area
of Shape C?

(g) Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the
area of Shape C? Give reasons for your answer.

(h) Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same?
Explain your answer. (NCERT)

Sol. PQRS is a square, let its side be 2x. D is a square, let its side be a (as shown in the figure).
2
2
\ Area of square PQRS = 4x and Area of square D = a .
(a) In the tangram pieces, by overlapping the shapes over each other, we can find out that shapes A and
B have the same area, Shapes C and E are isosceles right-angled triangles, both having an equal
side as ‘a’, so they have the same area. \ A = B; C = E.
(b) We also figured out that Shapes C and E are triangles with the same base and height as square D.
1
\ Area of DC = Area of DE = (Area of Square D) [ Triangle-square relation]
2
1
\ D = 2C = 2E ⇒ Ar. of C = Ar. of E = a 2
2
(c) It can be easily observed that F can be overlapped by C and E (Imagine C sliding down along the
diagonal PR and E sliding down along the side QR; F will be completely covered by C and E).
\ Area of DF = Area of DC + Area of DE i.e., F = C + E
⇒ F = 2C [ E = C from part (a)]
⇒ Area of F = Area of D [ D = 2C from part (b)]

⇒ Shape F and D are equal in area.
1
We have: Area of shape F = x 2 [ of triangle square relation]
2

178 Mathematics-6