Page 174 - Ganit Kaushal

 1 1  1  1 1  1  1 1   1

(b) Perimeter of cross =  2 × 2 + 2 +   2 × 2  +  2 × 2 + 2 +  2 × 2 +   2 × 2   + 2 + 2 ×   2









+ 2 ×  1  + 1 + 2 ×  1   2




2
 2
 1 1 1 1 
11+
= 1+ 2 + 11++ 2 ++ 2 + 11++ 2 + 1 m = 10 m




(c) Since 10 m > 6 m, the second arrangement (cross) has a greater perimeter.
(d) We will have the greatest perimeter by arranging them 1 m
horizontally (as explained above in example 2). 2
This arrangement (adjoining figure) is in the form of a 9 m
9 1 2
rectangle with length and breadth as m and m respectively.
2 2
 9 1 10
\ Perimeter = 2 ×  2 + 2 =× 2 = 10 m, which is the same as that of the cross.
2


So, we conclude that there is no other way of getting a perimeter more than 10 m.
Exercise 6.2
1. Find the area of the triangle shown in the figures
alongside, drawn on square grid paper, using the
conventions C1–C4.
(a) (b)
2. Using conventions C1–C4, find the area of shapes (a) and (b)
shown alongside, drawn on a square grid paper.

(a) (b)
Area of a Triangle and Its Relation with a Rectangle
Draw a rectangle on a piece of paper and draw one of its diagonals. Cut the rectangle along
that diagonal to get two triangles.
Do the two triangles overlap each other exactly? Do they have the same area?

Yes, they overlap, and they have the same area.
Try this with more rectangles having different dimensions. Flip
You can check this with a square as well. 1

Can you draw any inferences from this 2 2 1
exercise?
Yes, we can infer the following:

“We can cut rectangles (or squares) along their diagonals into two overlapping right triangles
1
/
s
having the same area. Area of each triangle = ( Area of rectangle quare) ”
2
The area of the blue rectangle will be the same as that of the yellow triangle, because as we see in the figures,
part 1 and part 2 of the blue rectangle are the same as part 1 (after flipping) and part 2 of the yellow triangle.

172 Mathematics-6