Page 168 - Ganit Kaushal

Sol. To find the starting points for the two runners, we need to measure 150 m B
350 metres backwards from the common finishing line. Since 25 is a
common factor of 100, 150, and 350, we can mark every 25 metres on 100 m
both tracks (as shown by the black dots in the figure). These markings 25 m
help us count the distance easily in steps of 25 metres.
Now, since 14 × 25 = 350, we need to count 14 such marks backwards
from the finishing line to find the starting points, ‘A’ and ‘B’, for the A
two runners. Remember, the corners also count as a mark. The starting
points ‘A’ and ‘B’ are shown as black round dots in the figure. Common Finishing Line
Ex 5. Look at the adjoining figure, Akshi says that the perimeter of this triangle shape is
9 units. Toshi says it can’t be 9 units and the perimeter will be more than 9 units.

What do you think? (NCERT)
Sol. The figure uses two types of unit lengths. One: straight-line units (red coloured -
horizontal or vertical) and the other: diagonal-line units (blue coloured). We will
write 1 unit of straight line as (1s) and 1 unit of diagonal line as (1d). By actual measurement, or even
by the naked eye, it can be observed that:

1 diagonal-line unit > 1 straight-line unit i.e., 1d > 1s.

The perimeter of the triangle consists of 6 straight-line units + 3 diagonal-line units. We can write this
in a short form as: (6s + 3d) units.
Now, (6s + 3d) > (6s + 3s) ( 1d > 1s). ⇒ 6s + 3d > 9s.

Therefore, Toshi is right that the perimeter will be more than 9 units.
Ex 6. Write the perimeters of the figures in
terms of straight and diagonal units
as explained in Example 5 above.
(NCERT)
Sol. Here 1 unit of straight line = 1s and
1 unit of diagonal line = 1d. (i) (ii) (iii) (iv)

(i) (8s + 2d) units (ii) (4s + 6d) units (iii) (12s + 6d) units (iv) (18s + 6d) units
Ex 7. A rectangle having side lengths of 5 cm a nd 3 cm is made using a piece of wire. If the wire is
straightened and then bent to form a square, what will be the length of a side of the square? (NCERT)
Sol. Length of a straightened wire = Perimeter of the rectangle
5 cm
= 2 × (5 cm + 3 cm) [ P = 2 × (l × b)]

= 16 cm 3 cm
Let ‘a’ be the length of a side of the square. Then the perimeter of the square
should be equal to the length of the wire.
\ 4 × a = 16 cm

⇒ a = (16 ÷ 4) cm = 4 cm

166 Mathematics-6