How will you construct and angles?

Understand the Question A angle is constructed by bisecting a angle. A angle is constructed by bisecting a angle. Therefore, the construction follows three main stages: 1. Construct a angle using an arc of equal radius. 2. Bisect the angle to obtain . 3. Bisect the angle to obtain . Step 1 · Construct a Angle 1. Draw a ray . 2. With as centre, draw an arc of any radius cutting at point . 3. With the same radius and centre , draw an arc cutting the first arc at point . 4. Draw ray passing through . Step 2 · Bisect to Construct a Angle 1. With as centre, draw an arc in the interior of . 2. With the same radius and centre , draw another arc intersecting the previous arc at point . 3. Draw ray passing through . Step 3 · Bisect to Construct a Angle 1. With as centre, draw an arc in the interior of . 2. With the same radius and centre , draw another arc intersecting the previous arc at point . 3. Draw ray passing through . Answer > Common Mistakes Changing Compass Radius: Changing the compass setting between drawing the base arc and marking the intersection point will distort the initial angle. Incorrect Centres for Bisection: When bisecting an angle, always place the compass point at the intersection points on the two arms of that specific angle (e.g., using points and to bisect ).

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Question 31

How will you construct $30^\circ$ and $15^\circ$ angles?

Solution

Understand the Question

A $30^\circ$ angle is constructed by bisecting a $60^\circ$ angle.
A $15^\circ$ angle is constructed by bisecting a $30^\circ$ angle.
Therefore, the construction follows three main stages:
1. Construct a $60^\circ$ angle using an arc of equal radius.
2. Bisect the $60^\circ$ angle to obtain $30^\circ$ .
3. Bisect the $30^\circ$ angle to obtain $15^\circ$ .

Step 1 · Construct a $60^\circ$ Angle

Draw a ray $\text{OA}$ .
With $\text{O}$ as centre, draw an arc of any radius cutting $\text{OA}$ at point $\text{P}$ .
With the same radius and centre $\text{P}$ , draw an arc cutting the first arc at point $\text{Q}$ .
Draw ray $\text{OB}$ passing through $\text{Q}$ .

$\text{Angle AOB} = 60^\circ$

Step 2 · Bisect $60^\circ$ to Construct a $30^\circ$ Angle

With $\text{P}$ as centre, draw an arc in the interior of $\angle \text{AOB}$ .
With the same radius and centre $\text{Q}$ , draw another arc intersecting the previous arc at point $\text{R}$ .
Draw ray $\text{OC}$ passing through $\text{R}$ .

\begin{aligned} \text{Angle AOC} &= \dfrac{\text{Angle AOB}}{2} \\[0.6em] &= \dfrac{60^\circ}{2} \\[0.6em] &= 30^\circ \end{aligned}

Step 3 · Bisect $30^\circ$ to Construct a $15^\circ$ Angle

With $\text{P}$ as centre, draw an arc in the interior of $\angle \text{AOC}$ .
With the same radius and centre $\text{R}$ , draw another arc intersecting the previous arc at point $\text{S}$ .
Draw ray $\text{OD}$ passing through $\text{S}$ .

\begin{aligned} \text{Angle AOD} &= \dfrac{\text{Angle AOC}}{2} \\[0.6em] &= \dfrac{30^\circ}{2} \\[0.6em] &= 15^\circ \end{aligned}

Answer

$\text{Angle AOC} = 30^\circ \quad \text{and} \quad \text{Angle AOD} = 15^\circ$

Common Mistakes

Changing Compass Radius: Changing the compass setting between drawing the base arc and marking the intersection point $\text{Q}$ will distort the initial $60^\circ$ angle.
Incorrect Centres for Bisection: When bisecting an angle, always place the compass point at the intersection points on the two arms of that specific angle (e.g., using points $\text{P}$ and $\text{R}$ to bisect $\angle \text{AOC}$ ).

More questions in IT

How do we find such $A$ and $B$ ?

From $X$ and $Y$ , draw arcs above and below $XY$ , with the same radii. The two points at which the arcs meet, above and below $XY$ , give us $A$ and $B$ , respectively.

Use this to construct an eye.

In Fig. 6.1, join $A$ and $B$ with a line. Where does $AB$ intersect $XY$ , and what is the angle formed between them?

Will the line joining the two points at which the arcs meet, above and below $XY$ , always be the perpendicular bisector of $XY$ , i.e., when $XY$ is of any length, and the arcs are drawn using a radius of any length?

Which two triangles should be congruent for $AB$ to be the perpendicular bisector of $XY$ (that is, $O$ is the midpoint of $XY$ and $AB$ is perpendicular to $XY$ )?

How do we get these different shapes? Try!

Will $C$ and $D$ lie on the perpendicular bisector $\text{AB}$ ?

Justify the following statement using the facts that we have established.

Any point that has the same distance from $X$ and $Y$ lies on the perpendicular bisector of $\text{XY}$ .

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

Can we extend the method of constructing the perpendicular bisector to construct a $90^\circ$ angle at any point on a line? Draw a line and mark a point $O$ on it. Construct a $90^\circ$ angle at point $O$ .

Q10

Find a segment of this line for which $O$ is the midpoint.

Q11

How do we construct this figure?

Q12

What is the angle between two adjacent lines?

Q13

How do we construct a $45^\circ$ angle using only a ruler and a compass?

Q14

Construct the following figure.

Q15

Draw an angle. Create a copy of this angle using only a ruler and compass.

Q16

How do we implement this idea using a ruler and a compass?

Q17

How did they make these arches?

Q18

Construct this arch shape on a piece of paper.

Let us think about the support lines this figure will need.

For symmetry, we should have $AB = CD$ , and $\angle BAD = \angle CDA$ . How would you construct these support lines?

Q19

Use these support lines to construct an arch. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

Q20

How do we construct this shape?

What supporting lines will you use to draw this arch?

Remember 'Wavy Wave' from the Grade 6 Textbook?

The supporting lines are just two line segments of equal length.

Q21

If their midpoints are marked, will you be able to construct a pointed arch?

Q22

How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)? To begin with, try to construct a pentagon and hexagon with equal sidelengths.

Q23

Can we break a regular hexagon into smaller pieces that can be constructed?

Q24

Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will it result in a regular hexagon?

Q25

Consider this figure. Will the $70^\circ$ angle fit into the gap? What is the gap angle $\angle AOI$ ?

We have, $40^\circ + 60^\circ + 50^\circ + 30^\circ + 40^\circ + 90^\circ + \text{gap angle} = 360^\circ$

Use this to determine whether the $70^\circ$ angle fits the gap.

Q26

In Fig. 6.12 can you explain why $AOD$ , $BOE$ and $COF$ are straight lines?

Q27

Construct a regular hexagon with a sidelength $4\text{ cm}$ using a ruler and a compass.

Q28

Context: We can construct a regular hexagon more directly if we can construct a $120^\circ$ angle using a ruler and a compass.

Q. How do we do it?

Q29

Why is $\angle \text{CAX} = 60^\circ$ ? Is there an equilateral triangle here?

Q30

Construct a regular hexagon of sidelength $5\text{ cm}$ .

Q31

How will you construct $30^\circ$ and $15^\circ$ angles?

Q32

Construct the following 6-pointed star. Note that it has a rotational symmetry.

Q33

Are the six triangles forming the 6 points of the star — $\Delta \text{AGH}$ , $\Delta \text{BHI}$ , $\Delta \text{CIJ}$ , $\Delta \text{DJK}$ , $\Delta \text{ELK}$ , $\Delta \text{FLG}$ — equilateral? Why?

[Hint: Find the angles.]

Q34

Can a $4 \times 6$ grid be tiled using multiple copies of $2 \times 1$ tiles? We are allowed to rotate a $2 \times 1$ tile and use it.

Q35

Can a $4 \times 7$ grid be tiled using $2 \times 1$ tiles?

Q36

What about a $5 \times 7$ grid?

Complete the justification.

Q38

Is an $m \times n$ grid tileable with $2 \times 1$ tiles, if both $m$ and $n$ are even? If yes, come up with a general strategy to tile it.

Q39

Is an $m \times n$ grid tileable with $2 \times 1$ tiles, if one of $m$ and $n$ is even and the other is odd? If yes, come up with a general strategy to tile it.

Q40

Is an $m \times n$ grid tileable with $2 \times 1$ tiles, if both $m$ and $n$ are odd? Give reasons.

Q41

Here is a $5 \times 3$ grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with $2 \times 1$ tiles?

Q42

Is the following region tileable with $2 \times 1$ tiles?

Q43

Context: We are considering whether a given region can be tiled using $2 \times 1$ tiles.

Q. What about this one?

Q44

Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a $5 \times 3$ grid, makes it non-tileable?

Q45

If the plain grid is tileable, is the black-and-white-grid tileable?

Q46

If the black-and-white grid is tileable, is the plain grid tileable?

Q47

Use this idea to find another unit square that, when removed from a $5 \times 3$ grid, makes it non-tileable?