Page 169 - ICSE Math 6

By a divider and a ruler: Let’s measure the given line segment AB.

(a) Put one end of the divider at A and open the divider so
that the other end reaches B.

(b) Without disturbing the opening of the divider, put it on
the ruler in such a way that one of its ends is at ‘0’.

(c) Read the marking on the ruler where the other end of
the divider reaches.

0
(d) This measure is the length of line segment AB. Clearly
AB measures 4 cm.

Example 5: Why is it better to use a divider and a ruler than a ruler only, while measuring the length
of a line segment?

Solution: It is better to use a divider and a ruler than a ruler only to avoid positioning error. This
error happens due to angular viewing.

Example 6: Draw any line segment, say AB. Take any point C lying between A and B. Measure
the lengths of AB, AC and CB. Is AB = AC + CB? What inference can you draw from
this?
Solution: Yes, AB = AC + CB. A C B

If AB is a line segment and C is a point in between, then measure of the line segment
is equal to the sum of the two smaller segments. Three points are collinear if measure
of the line segment is equal to the sum of the two smaller segments.

Measurement of an Angle

The unit for measuring an angle is degree which is denoted by ‘°’. The
angle made by a ray, say OA, while completing a rotation around the
vertex O is 360º (read as 360 degrees). When we divide 360° into 360 O A
equal parts, then each part is 1°. If we further divide 1° into 60 equal
parts, then each part is known as a minute denoted by a single prime ‘′’. If 1′ (read as 1 minute) is
again divided into 60 equal parts, then each part is known as a second denoted by two primes ‘″’. So,
1 complete rotation (turn) = 360°

1° = 60′
1′ = 60″
Example 7: Find the angle whose magnitude is 10° 15′ 30″ smaller than the magnitude of the angle
120° 10′ 50″.

Solution: Required angle = (120° 10′ 50″) – (10° 15′ 30″)

120° 10′ 50″ (Q 1° = 60′)
– 10° 15′ 30″
109° 55′ 20″

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