Construction of Specific Basic Triangles
Construction of Equilateral Triangle
An equilateral triangle is a special type of triangle characterized by having all three sides of equal length. A direct consequence of this property is that all three interior angles are also equal, and since the sum of angles in a triangle is $180^\circ$, each angle in an equilateral triangle measures $180^\circ / 3 = 60^\circ$.
Constructing an equilateral triangle is one of the most basic compass-and-straightedge constructions.
Construction Steps (Given Side Length)
Given: The desired length of each side, say $s$ units.
Tools: Ruler (for drawing the first side), Compass (for transferring the length and finding the third vertex), Straightedge (for drawing the sides).
Goal: Construct an equilateral triangle with side length $s$.
Steps:
- Draw the Base Side: Use the ruler to draw a line segment, say $AB$, with length exactly equal to the given side length $s$. This will be one side of the equilateral triangle.
- Draw Arc from One Endpoint: Place the pointed end of the compass on vertex $A$. Set the compass width precisely equal to the side length $s$ (i.e., the length of $AB$). Draw an arc above (or below) the line segment $AB$. The third vertex of the triangle, $C$, must be located somewhere on this arc, as its distance from $A$ must be $s$.
- Draw Arc from the Other Endpoint: Keeping the same compass width $s$ (do not change the compass setting), place the pointed end of the compass on vertex $B$. Draw another arc that intersects the arc drawn in Step 2. Label the point of intersection of the two arcs as $C$. This point $C$ is $s$ units away from $B$.
- Join the Vertices: Use the straightedge to draw straight line segments connecting point $A$ to point $C$ and point $B$ to point $C$.
The resulting figure is $\triangle ABC$.
Example
Example 1. Construct an equilateral triangle with a side length of 5 cm.
Answer:
Given: Side length $s = 5$ cm.
To Construct: An equilateral triangle with side 5 cm.
Construction Steps:
- Draw a line segment $XY$ of length 5 cm using a ruler.
- With $X$ as center and compass width 5 cm, draw an arc above $XY$.
- With $Y$ as center and the same compass width 5 cm, draw an arc intersecting the previous arc. Label the intersection point $Z$.
- Join $X$ to $Z$ and $Y$ to $Z$ using a straightedge.
The resulting triangle $\triangle XYZ$ is equilateral with $XY=YZ=ZX=5$ cm. All angles $\angle X, \angle Y, \angle Z$ will measure $60^\circ$.
Competitive Exam Note:
Constructing an equilateral triangle given its side is a fundamental skill. It's also directly related to constructing a $60^\circ$ angle, as seen in the construction of angles section (I1). The same principle of using equal radii from two points defines the third vertex of the equilateral triangle. This construction is often a preliminary step in more complex geometric problems.
Justification of Equilateral Triangle Construction
The justification proves that the triangle $\triangle ABC$ constructed using the steps outlined in I1, with a given side length $s$, is indeed an equilateral triangle. By definition, this means we need to show that all three sides of $\triangle ABC$ are equal in length ($AB = BC = AC$).
Proof
Given: A line segment $AB$ of length $s$. Point $C$ is constructed by drawing an arc from $A$ with radius $s$ and an arc from $B$ with radius $s$, and $C$ is the intersection of these arcs.
To Prove: $\triangle ABC$ is an equilateral triangle, i.e., $AB = BC = AC$.
Proof:
Let's examine the lengths of the sides of $\triangle ABC$ based on the construction steps:
- Side $AB$: By construction (Step 1), the line segment $AB$ was drawn with the given length $s$.
$\text{Length of } AB = s$
(By construction)
- Side $AC$: Point $C$ was obtained by the intersection of an arc drawn from center $A$. The radius of this arc was set equal to $s$ (Step 2). Therefore, the distance from $A$ to any point on this arc, including point $C$, is $s$.
$\text{Length of } AC = s$
(Radius of arc from A)
- Side $BC$: Point $C$ was also obtained by the intersection of an arc drawn from center $B$. The radius of this arc was set equal to the same length $s$ (Step 3). Therefore, the distance from $B$ to any point on this arc, including point $C$, is $s$.
$\text{Length of } BC = s$
(Radius of arc from B)
Combining the lengths of the three sides:
$AB = s$
... (i)
$AC = s$
... (ii)
$BC = s$
... (iii)
From (i), (ii), and (iii), we can see that:
$\therefore AB = AC = BC$
(All sides are equal to s)
By the definition of an equilateral triangle, a triangle with all three sides of equal length is equilateral.
Therefore, $\triangle ABC$ is an equilateral triangle. Q.E.D.
As a property of equilateral triangles, it also follows that all interior angles are equal, i.e., $\angle A = \angle B = \angle C = 60^\circ$. The construction of a $60^\circ$ angle (I1) is indeed based on forming an equilateral triangle implicitly.
Competitive Exam Note:
The justification for the equilateral triangle construction is straightforward and directly follows from the definition of the construction steps and the definition of an equilateral triangle. It reinforces the concept that setting the compass to a specific radius and drawing intersecting arcs ensures that the point of intersection is equidistant from the centers of the arcs, which in this case establishes the equality of all three sides.
Construction of an Isosceles Triangle (Given base and equal sides or base and base angles)
An isosceles triangle is defined as a triangle that has at least two sides of equal length. A key property of isosceles triangles is that the angles opposite the two equal sides, called the base angles, are also equal. The side between the two equal angles is called the base.
There are typically two common scenarios for constructing an isosceles triangle using compass and straightedge/ruler.
Construction 1: Given Base and Length of Equal Sides
This construction is very similar to the equilateral triangle construction, but the lengths of the base and the equal sides might be different.
Given: The length of the base, say $b$, and the length of the two equal sides, say $s$.
Condition for Construction: For the two arcs to intersect and form a triangle, the sum of the lengths of the two equal sides must be greater than the length of the base. By the Triangle Inequality Theorem, $s + s > b$, which simplifies to $2s > b$. If $2s \le b$, the arcs will not intersect or will only touch at a single point, and no triangle can be formed.
Tools: Ruler, Compass, Straightedge.
Goal: Construct an isosceles triangle $ABC$ with base $BC=b$ and equal sides $AB=AC=s$.
Steps:
- Draw the Base: Use the ruler to draw the base line segment $BC$ with length equal to $b$.
- Draw Arc from One Endpoint: Place the pointed end of the compass on vertex $B$. Set the compass width equal to the length of the equal sides, $s$. Draw an arc above (or below) the base line segment $BC$. The third vertex, $A$, must lie somewhere on this arc, as its distance from $B$ must be $s$.
- Draw Arc from the Other Endpoint: Keeping the same compass width $s$ (do not change the compass setting from Step 2), place the pointed end of the compass on vertex $C$. Draw another arc that intersects the arc drawn in Step 2. Label the point of intersection of the two arcs as $A$. This point $A$ is $s$ units away from $C$.
- Join the Vertices: Use the straightedge to draw straight line segments connecting point $A$ to point $B$ and point $A$ to point $C$.
The resulting figure is $\triangle ABC$. By construction, $BC=b$, $AB=s$ (radius from $B$), and $AC=s$ (radius from $C$). Since $AB=AC$, $\triangle ABC$ is an isosceles triangle with base $BC$ and equal sides $AB$ and $AC$.
Construction 2: Given Base and Measure of Base Angles
This construction is based on the ASA congruence criterion and the property that base angles of an isosceles triangle are equal.
Given: The length of the base, say $b$, and the measure of the two equal base angles, say $\beta$.
Condition for Construction: The sum of the two base angles must be less than $180^\circ$ for a triangle to be possible. $2\beta < 180^\circ$, which means $\beta < 90^\circ$. If $\beta \ge 90^\circ$, the rays will not intersect on the required side of the base.
Tools: Ruler, Compass, Straightedge, Protractor (needed if $\beta$ is not a standard constructible angle like $30^\circ, 45^\circ, 60^\circ, \dots$).
Goal: Construct an isosceles triangle $ABC$ with base $BC=b$ and base angles $\angle ABC = \angle ACB = \beta$.
Steps:
- Draw the Base: Use the ruler to draw the base line segment $BC$ with length equal to $b$.
- Construct Angle at One Endpoint: At vertex $B$ (one endpoint of the base), construct an angle with measure $\beta$. Draw a ray $BX$ starting from $B$ such that $\angle CBX = \beta$. This ray $BX$ will contain one of the equal sides of the isosceles triangle ($AB$).
- Construct Angle at the Other Endpoint: At vertex $C$ (the other endpoint of the base), construct an angle with the same measure $\beta$. Draw a ray $CY$ starting from $C$ such that $\angle BCY = \beta$. Ensure you draw this ray $CY$ on the same side of the base $BC$ as ray $BX$. This ray $CY$ will contain the other equal side ($AC$).
- Locate the Third Vertex: The third vertex of the triangle, $A$, must lie on both ray $BX$ and ray $CY$. Therefore, the point $A$ is the point of intersection of rays $BX$ and $CY$. Extend the rays $BX$ and $CY$ until they intersect. Label the point of intersection as $A$.
The resulting figure is $\triangle ABC$. By construction, $BC=b$, $\angle ABC = \beta$, and $\angle ACB = \beta$. Since the base angles $\angle ABC$ and $\angle ACB$ are equal, the sides opposite to them must be equal in length, i.e., $AC = AB$. Thus, $\triangle ABC$ is an isosceles triangle with base $BC$.
Examples
Example 1. Construct an isosceles triangle $ABC$ with base $BC = 5$ cm and equal sides $AB = AC = 7$ cm.
Answer:
Given: Base $b=5$ cm, Equal sides $s=7$ cm.
To Construct: Isosceles $\triangle ABC$ with $BC=5$, $AB=AC=7$.
Check Condition: $2s > b \Rightarrow 2 \times 7 > 5 \Rightarrow 14 > 5$. True. Construction is possible.
Construction Steps:
- Draw line segment $BC$ of length 5 cm.
- With $B$ as center and compass width 7 cm, draw an arc above $BC$.
- With $C$ as center and the same compass width 7 cm, draw an arc intersecting the previous arc. Label the intersection point $A$.
- Join $A$ to $B$ and $A$ to $C$.
$\triangle ABC$ is the required isosceles triangle.
Example 2. Construct an isosceles triangle $PQR$ with base $QR = 6$ cm and base angles $\angle Q = \angle R = 50^\circ$.
Answer:
Given: Base $b=6$ cm, Base angles $\beta = 50^\circ$.
To Construct: Isosceles $\triangle PQR$ with $QR=6$, $\angle Q = \angle R = 50^\circ$.
Check Condition: $2\beta < 180^\circ \Rightarrow 2 \times 50^\circ = 100^\circ < 180^\circ$. True. Construction is possible.
Construction Steps:
- Draw line segment $QR$ of length 6 cm.
- At point $Q$, construct an angle of $50^\circ$ using a protractor. Draw ray $QX$.
- At point $R$, construct an angle of $50^\circ$ using a protractor. Draw ray $RY$ on the same side of $QR$ as $QX$.
- Extend rays $QX$ and $RY$ to intersect at point $P$.
$\triangle PQR$ is the required isosceles triangle with $QR=6$ cm and $\angle Q = \angle R = 50^\circ$. Consequently, $PQ=PR$. The third angle $\angle P = 180^\circ - (50^\circ + 50^\circ) = 80^\circ$.
Competitive Exam Note:
Isosceles triangle constructions are variants of SSS and ASA. When given base and equal sides, it's SSS with two sides being equal. When given base and base angles, it's ASA with two angles being equal. The conditions $2s > b$ and $\beta < 90^\circ$ are important sanity checks. Remember that constructing an isosceles triangle based on base angles automatically makes the opposite sides equal, and vice-versa.
Construction of a Right Angled Triangle when Hypotenuse and Length of one Side is Given (RHS Criterion)
Constructing a right-angled triangle using the lengths of the hypotenuse and one leg is a common scenario, particularly useful in contexts involving coordinates or distances. This construction is based on the RHS (Right angle-Hypotenuse-Side) congruence criterion, which states that if the hypotenuse and a leg of one right-angled triangle are congruent to the hypotenuse and a leg of another right-angled triangle, then the triangles are congruent. This implies the unique determination of a right triangle by these two lengths and the presence of a right angle.
Condition for Construction
In a right-angled triangle, the hypotenuse is always the longest side. Therefore, for the construction to be possible, the given length of the hypotenuse must be strictly greater than the given length of the leg. If the given leg length is $a$ and the hypotenuse length is $c$, the condition is $c > a$. If $c \le a$, the arc from $C$ will not intersect the perpendicular ray (or will only touch it at $B$), and no triangle (specifically, no right triangle) can be formed.
Construction Steps
Given: The length of one leg, say $a$, and the length of the hypotenuse, say $c$. We aim to construct $\triangle ABC$ where $\angle B = 90^\circ$, the leg $BC = a$, and the hypotenuse $AC = c$.
Tools: Ruler (for measuring length), Compass (for drawing arcs), Straightedge (for drawing line segments and rays).
Goal: Construct right-angled $\triangle ABC$ with leg $a$ and hypotenuse $c$.
Steps:
- Draw One of the Legs: Use the ruler to draw the line segment corresponding to the given leg. Draw the line segment $BC$ with length equal to $a$. This will be one leg of the right-angled triangle.
- Construct the Right Angle: At vertex $B$ (one endpoint of the segment drawn in Step 1), construct a perpendicular line (a $90^\circ$ angle). Draw a ray $BX$ starting from $B$ such that $\angle CBX = 90^\circ$. Use the standard compass and straightedge method for constructing a $90^\circ$ angle at a point on a line (as described in I1, Construction of Perpendicular from a Point on the Line). The third vertex, $A$, must lie somewhere on this ray $BX$.
- Mark the Hypotenuse Length: Place the pointed end of the compass on vertex $C$ (the other endpoint of the base leg). Set the compass width equal to the length of the hypotenuse, $c$. Draw an arc that intersects the perpendicular ray $BX$ drawn in Step 2. The point of intersection is vertex $A$, as it must be $c$ units away from $C$ and must lie on the perpendicular line from $B$. Label the point of intersection as $A$.
- Join the Vertices: Use the straightedge to draw the line segment connecting point $A$ to point $C$. This segment $AC$ is the hypotenuse.
The resulting figure is $\triangle ABC$. By construction, $BC = a$, $\angle ABC = 90^\circ$, and $AC = c$. Thus, this is the required right-angled triangle.
Justification:
The construction directly creates a triangle with a right angle at $B$, one leg $BC$ of length $a$, and the vertex $A$ located on the perpendicular ray $BX$ at a distance $c$ from $C$. This means the side opposite the right angle, $AC$, has length $c$, which is the hypotenuse. Thus, the triangle $ABC$ satisfies the given conditions: a right angle at $B$, a leg $BC=a$, and hypotenuse $AC=c$. By the RHS congruence criterion, a right triangle is uniquely determined by these properties, confirming the construction's validity.
If the other leg $AB$ was given instead of $BC$, you would draw $AB$ first, construct the right angle at $B$, and then draw an arc from $A$ with radius $c$ to intersect the perpendicular ray from $B$ at $C$.
Example
Example 1. Construct a right-angled triangle $LMN$, right-angled at $M$, with $MN = 4$ cm and hypotenuse $LN = 5$ cm.
Answer:
Given: Leg $MN = 4$ cm, Hypotenuse $LN = 5$ cm, $\angle M = 90^\circ$.
To Construct: Right-angled $\triangle LMN$.
Check Condition: Hypotenuse > Leg $\Rightarrow 5$ cm $> 4$ cm. True. Construction is possible.
Construction Steps:
- Draw the line segment $MN$ of length 4 cm using a ruler.
- At point $M$, construct a perpendicular line (a $90^\circ$ angle). Draw a ray $MX$ such that $\angle NMX = 90^\circ$.
- With $N$ as center (the other endpoint of the leg) and compass width 5 cm (length of the hypotenuse), draw an arc intersecting the perpendicular ray $MX$. Label the intersection point $L$.
- Join $L$ to $N$ using a straightedge.
The resulting triangle $\triangle LMN$ is the required right-angled triangle with $\angle M = 90^\circ$, $MN = 4$ cm, and $LN = 5$ cm. The length of the other leg, $LM$, can be calculated using the Pythagorean theorem ($LM^2 + MN^2 = LN^2 \Rightarrow LM^2 + 4^2 = 5^2 \Rightarrow LM^2 + 16 = 25 \Rightarrow LM^2 = 9 \Rightarrow LM = 3$ cm).
Competitive Exam Note:
The RHS construction is specific to right-angled triangles. It combines the construction of a $90^\circ$ angle with using a compass to locate the third vertex based on the hypotenuse length. Always draw the given leg first, construct the right angle at one of its endpoints, and then use the hypotenuse length from the *other* endpoint of the leg. Remember the condition that the hypotenuse must be longer than the given leg.