Constructions and Tilings - NCERT Solutions Class 7 Maths (Ganita Prakash)

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Constructions and Tilings – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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Constructions and Tilings – NCERT Solutions

Q.1: Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass?

Solution:

Steps:

Taking some fixed radius, from X and then Y, construct two sufficiently long arcs above XY. Name the point where the arcs meet as A.
Using the same radius, from X and then Y, construct two sufficiently long arcs below XY. Name the point where the arcs meet as B.
AB is the required perpendicular bisector.

Q.2: Construct at least 4 different angles. Draw their bisectors.

Solution:

Steps for Angle Bisection:

Mark points A and B such that OA = OB.
Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C.
OC bisects {tex}\angle {\text{AOB}}{/tex}.

So, a 45^o angle can be constructed by first constructing a 90^o angle and then bisecting it.

Similarly, you can draw next 3 angles and bisect them.

Q.3: How do we construct a 45^o angle using only a ruler and a compass?

Solution:

Steps for Angle Bisection:

Mark points A and B such that OA = OB.
Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C.
OC bisects {tex}\angle {\text{AOB}}{/tex}.

So, a 45^o angle can be constructed by first constructing a 90^o angle and then bisecting it.

Q.4: How do we construct a 60^o angle?

Solution:

We get a 60^o angle if we construct an equilateral triangle! We can use the following steps for this.
Step 1:

Construct an arc with centre A and any radius.
Step 2:

With the same radius, cut another arc from B that meets the first arc. Let C be the point at which the arcs meet.
We have {tex}\angle {\text {CAX}} = 60^o{/tex}.

Q.5: How will you construct 30^o and 15^o angles?

Solution:

Constructing a 30^o Angle
Steps:

Draw a ray.
At its endpoint, construct a {tex}{6 0}^{\boldsymbol{\circ}}{/tex} angle using an equilateral triangle construction.
Bisect this {tex}60^{\circ}{/tex} angle using a compass (draw arcs from both rays and join their intersection to the vertex).
The bisected angle is {tex}{3 0}^{\circ}{/tex}.
So {tex} 60^{\circ} \div 2=30^{\circ} {/tex}

Constructing a {tex}15^{\circ}{/tex} Angle
Steps:

First construct a {tex}{3 0}^{\boldsymbol{\circ}}{/tex} angle (as above).
Bisect this {tex}30^{\circ}{/tex} angle again by the same arc-intersection method.
The resulting angle is {tex}{1 5}^{\boldsymbol{\circ}}{/tex}.
So {tex} 30^{\circ} \div 2={1 5}^{\circ} {/tex}