Constructions of Basic Triangles

A triangle is a simple closed curve or polygon created by three line-segments. 

The basic elements of the triangle are sides, angles, and vertices.

Basic Constructions

Why do we learn constructions? Suppose a building needs to be constructed. One needs to know the accurate map of the building. Construction means to draw the lines and angles accurately. Also to draw the road maps, learning the basic constructions is of great importance. Let us now learn about the basic constructions in geometry.

Construct perpendicular bisector of a line segment

 

Basic Constructions

1. Construction of a bisector of a given angle

Construction:

1. With A as a centre and using compasses, draw an arc that cuts both rays of A.
2. Label the points of intersection as B and C.
3.  Now with B as a centre, draw (in the interior of A) an arc whose radius is more than half the length BC.
4. With the same radius and with C as a centre, draw another arc in the interior of A.
5. Let the two arcs intersect at D
6. So we get AD as the required bisector of A.

2. Construction of the perpendicular bisector of a given line segment.

Construction:

1. Draw a line segment AB
2. Now take  A as a centre, using compasses, draw a circle.
3. The radius of your circle should be more than half the length of AB.
4. With the same radius and with B as a centre, draw another circle using compasses.
5. Let it cut the previous circle at C and D.
6. Join CD so it cuts AB at O.
7. O is the midpoint of AB. Also, COA and COB are right angles.
8. Therefore, the CD is the perpendicular bisector of AB.

3.  Construction of 60° Angle

Construction:

1. Take a ruler and draw a line l and make a point O on it. Take a compass and put its one end at point O and draw an arc with any convenient radius.
2. We take the centre O and draw an arc. We get the point A.
3. Now take the compass and with the same orientation taking the point A as the centre draw an arc that passes through O. Draw an arc such that it intersects the existing arc.
4. Here OA and OB are nothing but the radius which is equal in length.
5. So actually we are trying to draw an equilateral triangle which means the angle we see in the above construction is 60 degree.
6. We get ∠BOA which measures 60 degrees.

Solved Examples for You on Basic Constructions

Question 1: To construct a perpendicular to a line L from a point P outside the line, steps are given in the jumbled form. Identify the second step from the following.

1)Draw line PQ
2)Draw a line L and consider point P outside the line.
3)Take  P as a centre, draw 2 arcs on line L  and name it as points A and B respectively.
4)Taking A and B  as a centre one by one and keeping the same distance in compass, draw the arcs on another side of the plane.The point where these arcs intersect name that point as Q

1. 4
2. 3
3. 2
4. 1

Solution: B is the correct option. 3 is the second step amongst all.

Question 2: While constructing a parallel line to a given line, we ______.

1. Copy a segment
2. Bisect a segment
3. Copy an angle
4. Construct a perpendicular

Solution: C is the correct option. While constructing a parallel line to a given line, we copy an angle.

How to Draw Different Types of a Triangle?

Construction of  a Triangle

Triangles are the three-sided polygon geometric figure that comprises three edges and three vertices. The most significant property of a triangle is that the sum of all the interior angles of a triangle is 180°. Pythagoras theorem and trigonometry concepts are dependent on the properties of a triangle. It should be noted that the measurement of the three sides and angles of a triangle may or may not be equal in the construction of a triangle. Through ruler, compasses, protractor, and pencil, we can draw different types of a triangle. The construction of a triangle is possible if, 

• All the three sides of a triangle are mentioned

• Two sides of a triangle one included angle is given.

• Two angles of a triangle and one included side is given.

• The measurement of the hypotenuse side and one other side is given.

Properties of a Triangle

Although different triangles have their own different properties, there are some common properties that are similar for all triangles. Some of the common properties that are similar for all the triangles are given below:

• The sum of all three interior angles of a triangle is equal to 180 degrees.

• The measurement of the exterior angle of a triangle is equivalent to the addition of the opposite interior angles.

• The sum of the length of any two sides of a triangle is always greater than the third side of a triangle.

• In a right-angle triangle, the sum of the square of the length of the hypotenuse side which is opposite to the right angle is equal to the sum of the square of the length of the other two sides of a triangle.

How to Construct a Triangle?

Triangle and the other geometrical shapes can be accurately constructed using a protractor, ruler and a pair of compasses.

Following Three Properties are needed to Construct a Triangle:

Construction of Triangles : Three parameters define a triangle and the following possible combinations uniquely define a triangle

 •  Side-Side-Side (SSS)

 •  Side-Angle-Side (SAS)

 •  Angle-Side-Angle (ASA)

 •  right angle-hypotenuse-side (RHS)

 •  side-angle-altitude (SAL)

The geometrical instruments are used to construct the triangles of given parameters.

Criteria for Construction of Triangle

For the construction of the Triangle, all its dimension and angles need not be required.

Any of the following set of measurement is required to construct a triangle:

1. Construction for SSS criteria

All three sides are given (SSS)

2. Construction of triangle when perimeter and two angles are given

The perimeter of the Triangle and two base angle

3. Constructing Triangles using ASA criteria

Two angles and side between them (ASA)

4. Constructing Triangles using SAS criteria

Two sides and the angle between them (SAS)

5. Constructing Triangles using RHS criteria

Right-angled Triangle given Hypotenuse and a Side.

The three sides of the triangle are given. SSS criterion

Two sides and the included angle are given. SAS criterion

One side and any two angles are given. AAS criterion

Two angles and including side is given. ASA criterion

POSSIBILITY OF CONSTRUCTION

A triangle cannot be constructed if:

1. 3 angles are given. Since, the length of the sides can vary.

2. Two sides and non-included angle is given.

3. Sum of two sides should always be greater than third side.

SSS CRITERION

Draw a rough sketch of the triangle ABC.

Draw line segment AB = 6 cm.

With A as centre and radius = AC = 9 cm. Draw an arc.

With B as the centre and radius = BC = 5 cm. Draw another arc to intersect the previous arc at point C.

With B as the centre and radius = BC = 5 cm. Draw another arc to intersect the previous arc at point C.

Join A to C and B to C. Triangle ABC is the required triangle.

SAS CRITERION

Draw a rough sketch of the triangle PQR.

Draw line segment AB = 7 cm.

Using a protractor or a compass, construct angle of 60° at the point P.

With P as the centre and radius = PR = 6 cm. Draw an arc to intersect XP at a point R.

Join RQ. Triangle PQR is the required triangle.

ASA CRITERION

Draw a rough sketch of the triangle ABC.

Draw line segment BC = 8 cm.

At B draw ∠PBC = 60° using a compass.

At C draw an ∠QCB = 70° using a protractor.

The point of intersection of PB and QC is the vertex A.

Triangle ABC is the required triangle.

RIGHT TRIANGLE

Draw a rough sketch of the triangle XYZ.

Draw line segment XY = 3 cm.

At X draw an ∠AXY = 90° using compass.

With Y as the centre and radius 5 cm. Draw an arc to intersect AX at Z.

Join YZ. Triangle XYZ is the required triangle.

Criteria for the Construction of Triangles 

All the dimensions and angles are not required to construct a triangle. A triangle can be constructed if any of the below-given criteria meets.

Below are the following sets of measurements that are required for the construction of a triangle.

Criteria for the Construction of SSS Triangles.

Construction for the SSS triangle is possible when all the three sides of a triangle are given.

Criteria for the construction of the triangle when the perimeter and the two angles are given

Below, a given triangle can be constructed when the perimeter and two angles of the triangle are given.

Criteria for the construction of ASA triangles.

Construction for the ASA triangle is possible when any of the two angles and one side of a triangle is given.

Criteria for the Construction of SAS Triangles.

Construction for the SAS triangle is possible when any of the two sides and one angle of a triangle are given.

Criteria for the Construction of RHS Triangles.

Construction for the RHS triangle is possible when the hypotenuse side and one other side of a triangle are given.

Solved examples

1. How to construct a triangle PQR with the length of PQ=5 cm, PR= 6 cm and QR = 4.5 cm.

Solution: 

Given

PQ =5 cm

PR = 6 cm

QR = 4.5 cm

Steps of Construction

1. Draw a line of length PQ = 5cm and mark the points P and Q in it.

2. To draw the line segment PR, extend the two arms of the compass 6 cm away, place the one endpoint of the compass at point P and draw an arc with the pencil end.

3. To draw the line segment QR, fix the compass to 4.5 cm. Place the one end point of the compass at point Q and mark an arc with the pencil end. Draw an arc that intersects the previous arc drawn in step 2.Mark the intersection point of two arcs as R.

4. Join the line segments PR and QR.

Now , PQR is the required triangle 

 2. How to construct a triangle PQR with the length of PQ=4 CM, PR= 6.5 CM and ∠PQR = 60°

Solution : 

Given- PQ = 4cm

PR = 6.5 cm

∠PQR = 60°

Steps Of Construction

1. Draw a line segment QR of length 6.5cm 

2. Using a protractor and placing it at point Q, draw a line segment QX making an angle of 60° with QR.

3. Taking Q as center, draw an arc of radius 6 cm that cut the line QX at P.

4. Join PR.

Now, PQR is the required triangle 

Quiz Time

1. While constructing a triangle whose both perimeter and base angles are given, the first step is to

1. To draw the  base of any length

2. To draw the base angle from any random paint

3. Draw base of length= 1/2* perimeter

4. Draw a base of any length= perimeter

2. If the construction of a triangle PQR in which PQ= 6cm, ∠P=70° and ∠40° is possible, then find the measurement of ∠C.

1. 40° 

2. 70° 

3. 80° 

4. 50° 

3. To draw the perpendicular bisector of line segment AB, we open the compass

1. More than ½ AB

2. Less than ½ AB

3. Equal to ½ AB

4. None of these

Constructions of Triangles

1. Construct a scalene triangle when the base, one base angle and the sum of the lengths of the other two sides are given.

Constructions of Triangles

Construction:

Draw the base BC and at the point, B makes an angle, say XBC equal to the given angle.

Cut a line segment BD equal to AB + AC from the ray BX.

Join DC and make an angle DCY equal to BDC.

Let CY intersect BX at A

ABC is the required triangle.

In triangle ACD, ∠ACD = ∠ ADC

So, AB = BD – AD = BD – AC

AB + AC = BD

2. Construct a scalene triangle when the base, one base angle and the difference between the lengths of the other two sides are given.

Constructions of Triangles

Case 1: Let AB > AC that is AB AC is given

Draw the base BC and at point B make an angle say XBC equal to the given angle.

Cut the line segment BD equal to AB and AC from ray BX.

Join DC and draw the perpendicular bisector, say PQ of DC.

Let it intersect BX at a point A. Join AC.

Case 2: Let AB < AC that is AC AB is given

Draw the base BC and at point B make an angle say XBC equal to the given angle

Cut the line segment BD equal to AC AB from the line BX extended on opposite side of line segmentBC.

Join DC and draw the perpendicular bisector, sayPQ of DC.

Let PQ intersect BX at A. Join AC

3. Construct a scalene triangle, given its perimeter and its two base angles.

Constructions of Triangles

Draw a line segment, say XY equal to BC + CA + AB.

Make angles LYX equal to B and MYX equal to C.

Bisect LYX and MYX. Let these bisectors intersect at a point A

Draw perpendicular bisectors PQ of AX and RS of AY.

Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC

Then ABC is the required triangle

Solved Examples for You

Question 1: For constructing a triangle whose perimeter and both base angles are given, the first step is to:

Draw base of any length

Draw base of any length = Perimeter

To draw the base angles from the random line

Draw base of length = 1/3 × perimeter

Solution: B is the correct option. For constructing a triangle whose perimeter and both base angles are given, the first step is to draw the base of any length = Perimeter.

Question 2: For constructing a triangle when the base, one base angle and the difference between lengths of the other two sides are given, the base length is equal to:

The difference between the length of the other two sides.

Given base length

The largest side

None of these

Solution: B is the correct option. Here the base length is equal to the given base.

C

onstruction Related to Line Segments

You must have seen many times that a point divides a line segment in a particular ratio. In this module, we will learn how to divide a given line segment in the desired ratio x:y without using a ruler.

Dividing a given line segment in a particular ratio

Let’s say we have been given a line segment AB which we want to divide it by a ratio of 3:2. Here, x = 3 and y = 2. To construct such a line segment, follow the steps as given below:

Steps of Construction

1. Draw any ray AZ making an acute angle with AB.
2. Locate 5 points (x + y = 3 + 2 = 5), say, A₁, A₂, A₃, A₄, and A₅ using a compass on AZ such that these points are equally distanced from each other.
   1. Open the compass of a specific width and make an arc on the ray AZ keeping the needle on the point A. The point where the arc intersects with AZ is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the steps 3 more times making AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
3. Join A₅ with B making a line segment A₅B. (See Figure 1)
4. Since we need to divide the line segment in 3:2, draw a line through the point A₃ which is parallel to A₅B by making ∠AA₃C = ∠AA₅B such that A₃C intersects AB at the point C.
   1. To construct a parallel line using Angle Copy Method, adjust the width of the compass to roughly half of the length of A₄A₅. Keeping the needle at point A₅, make an arc cutting the ray AZ and the line segment A₅B. This is arc 1.
   2. Make a similar arc from point A₃ using the same compass width. This is arc 2.
   3. Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AZ, the pencil should be on the point where arc 1 meets the line segment A₅B.
   4. Keeping the same width, cut the arc 2 making an intersecting point G.
   5. Make a line segment from A₃ through G meeting the line segment AB at C. The line segment A₃C || A₅B.
5. Therefore, AC : CB = 3 : 2. (See Figure 1)

 

Figure 1: AC : CB = 3 : 2

Rationale and Proof

Since A₃C || A₅B

∴ AA3A3A5 = ACCB [Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally]

AA3A3A5 = \( \frac{3}{2} \)          [By construction]

∴ ACCB = 32

Alternate Method

Following is the alternative method to divide a given line segment AB into a said ratio (same as above, that is, 3:2).

Steps of Construction

1. Draw any ray AG making an acute angle with AB.
2. Draw a ray BH parallel to AG by making ∠ABH equal to ∠BAG using the same method as described above in steps from 4(a) to 4(e).
3. Locate the pointsA₁, A₂, and A₃ (x = 3) on AG and B1, B2 (y = 2) on BH such that AA₁ = A₁A₂ = A₂A₃ = BB₁ = B₁B₂ using the same step as described in 2(a) above.
4. Join A₃B₂. Let it intersect AB at a point C (See Figure 2).
5. Then AC : CB = 3 : 2

Figure 2: AC : CB = 3 : 2 (Alternate method)

Rationale and Proof

Here,

 ∠CAA₃ = ∠CBB₂                    [By construction]

∠ACA₃ = ∠BCB₂                     [Vertically opposite angles are always equal]

∠AA₃C = ∠BB₂C                     [Alternate interior angles within two parallel lines are always equal]

∴ ∆ACA₃ ~ ∆BCB₂                 [By AAA criteria of triangle similarity]

⇒ ACBC = CA3CB2 = AA3BB2

Now, using construction AA3BB2 = 32

∴ ACBC = 32

Construction Related to Triangles

We have studied about similar triangles and their properties in previous classes. Have you ever wondered how these similar triangles are designed? Let’s take a look how.

Constructing a triangle similar to a given triangle

There are a few ways by which we can draw a triangle similar to the given one. Here, we shall study how to construct the same using the Scale Factor method without the help of a ruler.

Scale Factor is the ratio of two corresponding sides of the similar triangles. For example, let’s say that ∆ABC is similar to ∆DEF (∆ABC ~ ∆DEF) wherein AB = 3, BC = 4, and AC = 5; whereas DE = 6, EF = 8 and DF = 10. Here, the Scale Factor of two corresponding sides is

ABDE = BCEF = ACDF = 12

So, let’s see how we can construct a triangle similar to the given triangle (∆PQR) with a scale factor of 34, that is, to construct a smaller triangle (∆P′QR′) compared to ∆PQR wherein the length of the sides of ∆P′QR′ will be 3/4^th to that of ∆PQR.

Steps of Construction using Scale Factor

1. Draw any ray QX making an acute angle with QR on the side opposite to the vertex P.
2. Locate 4 points Q₁, Q₂, Q₃, and Q₄ on QX so that QQ₁ = Q₁Q₂= Q₂Q₃ = Q₃Q₄ using the same method as described above in step 2(a) (Dividing a line segment in a given ratio). Note: We located 4 points because 4 is the greater value in the Scale Factor ratio of 34
3. Considering the given triangle, ∆PQR, join a line Q₄R.
4. Draw a line through Q3 parallel to Q₄R using the same method as described above in steps from 4(a)  to 4(e) (Dividing a line segment in a given ratio) intersecting the line QR at R′. This gives us Q₃R′ || Q₄R and ∠QQ₃R′ = ∠QQ₄R (See Figure 3).
5. Similarly, draw a line through R′ parallel to the line PR intersecting the line PQ at a point, P′. This gives us P′R′ || PR and ∠P′R′Q = ∠PRQ (See Figure 3).
6. Subsequently, ∆P′QR′ is the required triangle similar to ∆PQR with Scale Factor of 34.

Figure 3: ∆ P′QR′ ~ ∆PQR

Rationale and Proof

Since Q₃R′ || Q₄R and QQ₃ is proportionate to QQ₄ in the ratio of 3:4 (by construction)

∴ QQ3QQ4 = QR′QR = 34 [By Triangle Proportionality Theorem]

Here,

∠P′QR′ =  ∠PQR         [Common angle]

∠P′R′Q  = ∠PRQ         [By Construction]

∠R′P′Q = ∠RPQ          [If two corresponding angles of two triangles are equal, the third angle will always be equal]

∴ ∆P′QR′ ~ ∆PQR       [By AAA criteria of triangle similarity]

Since ∆P′QR′ ~ ∆PQR and QR′QR = 34

∴ ∆P′QR′ ~ ∆PQR with Scale Factor of  34.

Solved Examples for You

Question 1: Given a line segment AB. Find a point, describing the steps, which divides the line segment in 2:1 without using a ruler. 

Answer : Steps of Construction

1. Draw any ray AX, making an acute angle with AB.
2. Locate 3 points (x + y = 2 + 1 = 3), say, A₁, A₂, and A₃ using a compass on AX such that these points are equally distanced from each other.
   1. Open the compass of a specific width and make an arc on the ray AX keeping the needle on the point A. The point where the arc intersects with AX is now A₁. Repeat the step keeping the same width from A₁ making a point A₂. Repeat the step one more times making AA₁ = A₁A₂ = A₂A₃.
3. Join A₃ with B making a line segment A₃B. (See Figure A)
4. Since we need to divide the line segment in 2:1, draw a line through the point A₂ which is parallel to A₃ by making ∠AA₂B = ∠AA₃B intersecting AB at the point C.
   1. To construct a parallel line, adjust the width of the compass to roughly half of the length of A₂A₃. Keeping the needle at point A₃, make an arc cutting the ray AX and the line segment A₃B. This is arc 1.
   2. Make a similar arc from point A₂ using the same compass width. This is arc 2.
   3. Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AX, the pencil should be on the point where arc 1 meets the line segment A₃B.
   4. Keeping the same width, cut the arc 2 making an intersecting point O.
   5. Make a line segment from A₂ through O meeting the line segment AB at C. The line segment A₂C || A₃B.
5. This point C divides AB as AC : CB = 2 : 1. (See Figure A)

Figure A. AC : CB = 2 : 1

Question 2: What do you mean by segment in maths?

Answer: Segment refers to the part of a line that connects two points. It happens to the shortest distance that exists between the two points. Furthermore, segment has a length. The word “segment” is important because a line can go on continuously without end. In contrast, a line segment has definite endpoints.

Question 3: Explain what is meant by segments of circle?

Answer: A chord of a circle divides a circle into two parts. These parts are known as the segments of circle.

Question 4: How can one measure a line segment?

Answer: A line segment is nothing more than a part of a line. In order to measure the length of a line segment, put the segment’s endpoint on the ruler’s zero mark. They must see where it ends. In inches, one must round to the nearest ¼, while in centimeters; one must measure to the nearest centimeter.

Question 5: How can one find out the midpoint of a segment?

Answer: One can find out the midpoint of a segment by dividing the segment’s length by 2 and counting that value from any endpoint.

T

angents

A common tangent can be any line, a ray or a segment that can be a tangent to more than one circle at the same time. There can be one to four number of common tangents to two circles. The point to be noted is that a tangent of a circle touches the circle but never enters it. Also, multiple tangents to the two circles can be an outer tangent or an inner one to each other’s circles.

Figure 1: Circle with a centre O and tangent as PQ. X is the point of tangency

Note: There can only be two tangents drawn from one point outside the circle. In this module, we’ll learn the art of constructing two tangents to a given circle from a point outside the circle without the help of a scale.

Browse more Topics under Constructions

• Basic Constructions
• Constructions Related to Segment and Triangles
• Construction of Triangles

Constructing Tangent of a Circle

We can come across two cases while constructing tangent of a circle. These are described as follows:

Case I: When the centre of a circle is known

Let’s say that we are given a circle with centre O and a point P outside it, and we have to construct two tangents from the point P to the circle without the help of a ruler. Follow the steps of construction as below:

1. Join OP and bisect it.  To bisect OP, take a compass, and open it slightly more than half of the length of the line segment.
2. From point P, mark a minimal arc above and below the line segment. Repeat the similar step from point O keeping the opening of the compass as same as it was from point P. Two points will be created where the two arcs, produced from point O and point P, meet.
3. Joint these two points with a line segment using a scale. This line segment bisects the OP. Let’s consider H as the mid-point of PO.
4. Taking the point H as a centre and HO as a radius, draw a circle. Let it intersect the given circle at the points, T and T’.
5. Join PT and PT’. Then PT and PT’ are the required two tangents.

Figure 2: Drawing tangents PT and PT’ (marked in red) to the circle

Proof and Rationale

To check if the construction is correct:

• Join OT. Then ∠PTO is an angle in the semicircle and, therefore, ∠PTO = 90° [ angle in a semicircle is always 90°]
• Check with the 180° protractor ensuring that ∠PTO = 90°.

Since OT is the radius of the circle with centre O and ∠PTO = 90°, therefore, PT has to be a tangent to the circle. This is because of the property of the circle which states that the radius from the centre of the circle to the point of tangency is perpendicular to the tangent line. Similarly, PT’ also becomes a tangent to the circle.

Case II: When the centre of the given circle is not known

In that case:

1. Draw any two non-parallel chords in the given circle
2. Draw perpendicular bisectors to both of the chords (as described in steps 2 and 3 above in Case I) (See Figure 3)
3. Find the point of intersection of two perpendicular bisectors which will be the centre of the given circle.
4. Follow the same steps from 1 to 5 as described above in Case I.

Figure 3: Intersection of two perpendicular bisectors is the centre point O

Solved Examples for You

Question 1: Given a circle without a centre point and a point P outside the circle. Find the centre point of the circle and draw two tangents from the point P to the circle. Describe the steps.

Answer : Steps of Construction:-

1. Draw any two non-parallel chords (CD and EF) in the given circle
2. Draw perpendicular bisectors to both of the chords.
   1. To bisect CD, take a compass, and open it slightly more than half of the length of the line segment.
   2. From point D, mark a minimal arc above and below the line segment. Repeat the similar step from point C keeping the opening of the compass as same as it was from point D. Two points will be created where the two arcs, produced from point C and point D, meet.
   3. Joint these two points with a line segment using a protractor. This line segment bisects the CD.
   4. Repeat steps from 2 a to 2 c for drawing a perpendicular bisector for EF.
3. Find the point of intersection of two perpendicular bisectors which will be the centre of the given circle, say point O. (See Figure A)
4. Join the centre point O to the given point P outside the circle.
5. Bisect OP by following steps from 2 a to 2 c. Taking the point G as a centre and OG as a radius, draw a circle. Let it intersect the given circle at the points, T and T’.
6. Join PT and PT’. Then PT and PT’ are the required two tangents.

Question 2: What does it mean if a line is tangent to a circle?

Answer: A tangent line refers to a line that is intersecting a circle at one point. We call such a line the tangent to that circle. Moreover, the point at which the circle and the line intersect is referred to as the point of tangency. In other words, for any tangent line, there will be a perpendicular radius

Question 3: How many tangents can a circle have?

Answer: A circle has 0 tangents. We can draw exactly 0 tangents through a circle. A tangent will intersect a circle in only one point so that it does not enter the area of the circle. Moreover, we can draw an infinite number of tangents to a certain circle as there won’t be any limits to the points for a tangent to intersect.

Question 4: Is a tangent line perpendicular?

Answer: A tangent to a circle is a line that intersects the circle at precisely one point. It is the point of tangency or tangency point. A significant result is that the radius from the centre of the circle to the point of tangency will be perpendicular to the tangent line

Question 5: What does tangent mean?

Answer: In geometry, the tangent line or sometimes simply called the tangent, is a plane curve at a certain point is the straight line which “just touches” the curve at that point. Leibniz stated that it is the line through a pair of infinitely close points on the curve. Moreover, the word “tangent” is derived from the Latin ‘tangere’ which means ‘to touch’.

Figure A: Tangents PT and PT’ from the given point P outside the given circle (highlighted)