These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.5

Question 1.

Which of the following are models for perpendicular lines:

(a) The adjacent edges of a table top.

(b) The lines of a railway track.

(c) The line segments forming the letter ‘L’.

(d) The letter V.

Answer:

(a) The adjacent edges of a table top.

.’. They form perpendicular lines,

(b) The lines of a railway track

Here, the two adjacent sides don’t meet.

χ They do not form perpendicular lines

(c) The line segments forming the letter ‘L’.

Here, the line segments form a right angle

∴ They form perpendicular lines.

(d) Here, the angle between the lines is not a right angle.

∴ They do not form perpendicular lines.

Question 2.

Let \(\overline{\mathrm{PQ}}\) be the perpendicular to the line segment \(\overline{\mathrm{XY}}\). Let \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{XY}}\) intersect in the point A. What is the measure of ∠PAY?

Answer:

Since, \(\overline{\mathrm{XY}}\) ⊥ PQ

Angle between them is a right angle.

.’. ∠PAY = 90°

Question 3.

There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?

Answer:

The angle are 90°, 45°, 45°

The angles are 90°, 60°, 30°

So, Angle 90° is common between them

Question 4.

Study the diagram. The line l is perpendicular to line m.

(a) Is CE = EG?

(b) Does PE bisect CG?

(c) Identify any two line segments for which PE is the perpendicular bisector.

(d) Are these true?

(i) AC > FG (ii) CD = GH (iii) BC < EH

Answer:

(a) CE = CD + DE = 1 + 1=2 EG = EF + FG = 1 + 1 = 2

(b) PE & CG Intersect at point E & CE = EG

.’. PE is the bisector of CG

(c) PE is perpendicular bisector for

\(\overline{\mathrm{CG}}\)

As CE = EG = 2

& \(\overline{\mathrm{CE}}\) = EG = 2

\(\overline{\mathrm{XY}}\) ⊥ \(\overline{\mathrm{CG}}\)

\(\overline{\mathrm{BH}}\)

As \(\overline{\mathrm{BE}}=\overlline{\mathrm{BE}}\) = 3

\(\overline{\mathrm{PE}}=\overlline{\mathrm{BH}}\)

(d) (i) True

AC = AB + BC = 1 + 1 = 2

FG = 1

∴ AC > FG

(ii) True

CD = 1

GH = 1

CD > GH

(iii) True

BC = 1

EH = EF + FG + GH = 1 + 1 + 1 = 3

∴ CD > GH