Playing with Constructions — NCERT Solutions, AI Tutor & Practice

For accurate constructions we mainly use:

1. Ruler (scale): Used to draw straight line segments and to measure lengths in centimetres and millimetres.
2. Compass: Used to draw circles and arcs, and to mark off equal lengths by keeping the same opening.

Other helpful tools include the divider (to compare lengths) and the set-squares and protractor (to draw and measure angles). In this chapter, most curved shapes and equal-length markings are made using the compass, while straight parts are drawn with the ruler.

Steps:

1. Place the ruler on the paper and mark a point $A$ at the $0\ \text{cm}$ line.
2. Move along the ruler to the $6\ \text{cm}$ mark and make a point $B$.
3. Join $A$ and $B$ with a sharp pencil along the edge of the ruler.

The drawn segment is $AB = 6\ \text{cm}$.

$A \bullet\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\! \bullet B \quad (6\ \text{cm})$

Steps:

1. Open the compass so that the distance between the pointer and the pencil tip is exactly $4\ \text{cm}$ (check against a ruler).
2. Mark a centre point $O$ on the paper.
3. Place the compass pointer at $O$ and, holding the top, rotate the pencil all the way around to draw the circle.

Diameter: $\text{Diameter} = 2 \times \text{radius} = 2 \times 4 = 8\ \text{cm}$

Properties used: All four sides are equal ($5\ \text{cm}$) and each angle is $90^\circ$.

Steps:

1. Draw $PQ = 5\ \text{cm}$ with a ruler.
2. At $P$, use a set-square to draw a line perpendicular to $PQ$ and mark $S$ on it so that $PS = 5\ \text{cm}$.
3. At $Q$, draw a perpendicular to $PQ$ and mark $R$ so that $QR = 5\ \text{cm}$.
4. Join $S$ and $R$.

Check: $SR = 5\ \text{cm}$ and all angles are $90^\circ$, so $PQRS$ is a square.

Steps:

1. Draw a circle with centre $A$ and radius $r$ (say $3\ \text{cm}$).
2. Mark a point $B$ on this circle.
3. Without changing the compass opening, place the pointer at $B$ and draw a second circle of the same radius $r$.

Observation: The two circles cut each other at two points, and the overlapping (common) region is a leaf-like shape called a vesica. Also, since $AB = r$ and both crossing points are at distance $r$ from $A$ and from $B$, joining these points to $A$ and $B$ gives equilateral triangles. This is the starting idea for many symmetric designs.

Steps:

1. Draw $AB = 6\ \text{cm}$.
2. At $A$ and $B$, draw perpendiculars to $AB$ using a set-square.
3. On these perpendiculars mark $D$ and $C$ so that $AD = BC = 4\ \text{cm}$.
4. Join $D$ and $C$; then $DC = 6\ \text{cm}$.

$ABCD$ is the required rectangle.

Difference from a square: In a square all four sides are equal, but in a rectangle only the opposite sides are equal (length $= 6\ \text{cm}$, breadth $= 4\ \text{cm}$). Both have all angles equal to $90^\circ$.

Steps:

1. Draw $AB = 5\ \text{cm}$ using the ruler.
2. Open the compass to $5\ \text{cm}$. With pointer at $A$, draw an arc above $AB$.
3. Keeping the same $5\ \text{cm}$ opening, put the pointer at $B$ and draw another arc that cuts the first arc. Call the crossing point $C$.
4. Join $AC$ and $BC$.

Since $AB = AC = BC = 5\ \text{cm}$, triangle $ABC$ is equilateral.

Steps:

1. Take the given segment $AB$.
2. Open the compass to a radius more than half of $AB$. With pointer at $A$, draw arcs above and below the segment.
3. With the same radius and pointer at $B$, draw arcs that cut the first arcs at two points, $P$ (above) and $Q$ (below).
4. Join $P$ and $Q$ with a ruler. The line $PQ$ crosses $AB$ at point $M$.

$M$ is the midpoint, so $AM = MB$. The points $P$ and $Q$ are equidistant from $A$ and $B$, so the line $PQ$ passes exactly through the middle of $AB$.

Steps:

1. Draw a circle with centre $O$ and radius $r$ (say $3\ \text{cm}$).
2. Keep the same radius $r$. Place the compass pointer at any point $A$ on the circle and mark an arc on the circle; label the crossing point.
3. Move the pointer to that new mark and again cut an arc of radius $r$ on the circle. Repeat all the way round.
4. Since a radius steps around a circle exactly six times, you get $6$ equally spaced points.
5. With each of these $6$ points as a centre, draw a circle of radius $r$.

The six outer circles pass through $O$ and overlap to form a beautiful six-petal flower (rose) pattern.

Steps:

1. Construct a square $ABCD$ of side $4\ \text{cm}$ (draw $AB = 4\ \text{cm}$, raise perpendiculars at $A$ and $B$, mark $D$ and $C$ at $4\ \text{cm}$, and join $DC$). This is the wall of the house.
2. The top side $DC$ becomes the base of the roof, with $DC = 4\ \text{cm}$.
3. Open the compass to $4\ \text{cm}$. With pointer at $D$, draw an arc above $DC$; with the same radius and pointer at $C$, draw another arc crossing it at $E$.
4. Join $DE$ and $CE$.

Triangle $DEC$ is equilateral ($DE = EC = DC = 4\ \text{cm}$) and sits neatly on the square as the roof, completing the house.

Steps:

1. On line $\ell$, with $P$ as centre and a convenient radius, cut two arcs on the line, one on each side of $P$. Call these points $X$ and $Y$ (so $PX = PY$).
2. Now increase the compass radius. With pointer at $X$, draw an arc above the line.
3. With the same radius and pointer at $Y$, draw another arc cutting the first at a point $Z$.
4. Join $P$ and $Z$.

The line $PZ$ is perpendicular to $\ell$ at $P$, i.e. $\angle ZP X = 90^\circ$, because $Z$ is equidistant from the equal points $X$ and $Y$.

Finding the radius:

The circle touches all four sides, so its centre is the centre of the square. The distance from the centre to each side is half the side length.

$\text{Radius} = \frac{\text{side}}{2} = \frac{6}{2} = 3\ \text{cm}$

Steps to draw:

1. Construct the square $ABCD$ of side $6\ \text{cm}$.
2. Find its centre $O$ by joining the diagonals $AC$ and $BD$; they meet at $O$.
3. Open the compass to $3\ \text{cm}$, place the pointer at $O$, and draw the circle.

The circle will just touch the midpoint of each side of the square.