CBSE Class 10 Mathematics Probability Worksheet Set 03

 Probability

Question. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow ?
Answer: P(E) = 0.85
P(E) = 1 – P(E) = 1 – P(E) = 1 – .085 = 0.15

Question. A die is thrown once. Find the probability of getting multiple of 2 or 3.
Answer: (1) S = {1,2,3,4,5,6} n(S) = 6
E = {2,3,4,6} n(E) = 4 P(E) = 4/6 = 2/3

Question. Write a sample space when two coins are tossed simultaneously?
Answer: S = {HH,HT,TH,TT}

Question. Savita and Hamida are friends what is the probability that both will have
i) the same birthday?
ii) different birthday?
Answer: (1) 1/365 (2) 364/365

Question. What is the probability of
a) A sure event
b) Impossible event
Answer: (a) P(S) = 1 (b) P(2) = 0

Question. All the three face cards of spade are removed from a well shuffled pack of 52 cards & card is drawn from the remaining pack. Find the probability of getting
a) a black face card
b) a queen of diamond
c) a spade
d) a black card
Answer: (1) 9/49 (2) 1/49 (3) 10/49 (4) 23/49

Question. One card is drawn from well-shuffled pack of 52 cards. Find the probability of getting
a) the jack of hearts
b) a face card.
Answer: (1) 1/52 (2) 12/52 = 3/13 (Jack, King & Queen)

Question. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday) each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
i) the same day?
ii) consecutive day?
iii) Different day?
Answer: S= {Tue, Wed, Thur, Fri, Sat} n(S) = 25
(1) 5/25 = 1/5 (2) 8/25 (3) 20/25 = 4/5

Question. Find the probability of getting 53 Fridays in a leap year.
Answer: A leap year consist of 52 weeks and 2 extra days.
n(S) = 7 , n(E) = 2 , P(E) = 2/7

Question. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuvi will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to him. What is the probability that
i) She will buy it
ii) She will not buy it
Answer: 20 Defected & 124 Non-Defected
(1) 124/144 = 31/36 (2) 20/144 = 5/36

Question. A letter is chosen at random from the letters of the word “ASSASSI NATION” and the probabilities that the letter chosen is a
i) vowel
ii) consonant
Answer: A,A,I,A,I,O = 6’s Vowels S,S,S,S,N,T,N = 7’S Constants
(1) 6/13 (2) 7/13

More Question-

CONSTRUCTIONS

Key Points

1. Division of a line segment in the given ratio.

2. Construction of triangles:-

a. When three sides are given.

b. When two sides and included angle given.

c. When two angles and one side given.

d. Construction of a right angled triangle.

3. Construction of triangle similar to a given triangle as per given scale factor.

4. Construction of tangents to a circle.

EXPECTED LEARNING OUTCOMES

1. Correct use of Mathematical instruments.

2. Drawing a line segment and an angle as per the given data.

3. To divide the given line segment in the given ratio accurately.

4. Neatness and accuracy in drawing.

5. The concept of similar triangles.

6. To Construct a triangle as per the conditions given.

7. To construct similar triangle to a given triangle as per the given ratio.

8. To know that when the ratio is a proper fraction then the similar triangle lies inside the given

Triangle and when improper then the similar triangle lies outside the given triangle.

9. To construct tangents to a circle from an external point given.

CONCEPT MAP

CONSTRUCTIONS

DIVISION OF A LINE SEGMENT

CONSTRUCTION OF A TANGENT TO A CIRCLE

CONSTRUCT SIMILAR TRIANGLES AS PER GIVEN RATIO

KNOWLEDGE OF BASIC PROPORTIONALITY THEOREM

When given ratio is proper fraction the similar triangle lies inside the given triangle

Two triangles are similar if their corresponding sides are proportion

LEVEL – I

1. Draw a line segment AB=8cm and divide it in the ratio 4:3.

2. Divide a line segment of 7cm internally in the ratio 2:3.

3. Draw a circle of radius 4 cm. Take a point P on it. Draw tangent to the given circle at P.

4. Construct an isosceles triangle whose base is 7.5 cm and altitude is 4.2 cm.

5. Draw a line segment of length 9 cm. and divide it in seven equal parts.

LEVEL –II

1. Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. (CBSE 2013)

2. Construct a triangle similar to a given ΔABC such that each of its sides is 2/3rd of the corresponding sides of ΔABC. It is given that AB=5cm BC=6cm and AC=7cm. Also write the steps of construction.

3. Draw a pair of tangents to a circle of radius 4cm, which are inclined to each other at an angle of 600. (CBSE 2013)

4. Draw a circle of radius 5cm. From a point 8cm away from its centre construct the pair of tangents to the circle and measure their lengths.

5. Construct a triangle PQR in which QR=6cm, Q=600 and R=450. Construct another triangle similar to ΔPQR such that its sides are 5/6 of the corresponding sides of ΔPQR.

6. Draw a line segment AB= 7.5cm and locate a point P on AB such that AP= 3/7 AB. Give justification of the construction.

LEVEL-III

1. Draw a circle with centre O and radius 3.5cm. Take a horizontal diameter. Extend it to both sides to point P and Q such that OP=OQ=7cm. Draw tangents PA and QB, one above the diameter and the other below the diameter. Is PA||BQ.

2. Construct a Δ ABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct another ΔAB’C’ similar to ΔABC with base AB’ = 8 cm. (CBSE 2015)

3. Draw a right triangle ABC in which B=900, AB=5cm, BC=4cm, then construct another triangle A’BC’ whose sides are 5/3 times the corresponding sides of ΔABC. Is the new triangle also a right triangle?

4. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

5. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP/AB = 3/5. (CBSE 2011)

6. Construct an isosceles triangle whose base is 8 cm. and altitude 4 cm. and then construct another triangle whose sides are ¾ times the corresponding sides of the isosceles triangle. (CBSE 2011)

7. ABC is a right triangle in which AB=5.4 cm, BC= 7 cm and <B = 900.Draw BD perpendicular on AC and a circle through B, C, D. Construct a pair of tangents from A to this circle.

8. Construct a triangle ABC in which AB=5cm,<B=600and altitude CD=3 cm. Construct a triangle PQR similar to ΔABC such that each side of ΔPQR is 1.5 times that of the corresponding sides of ΔABC.

9. Construct a tangent to a circle of radius 3.5 from a point on the concentric circle of radius 6.5 cm and measure its length. Also, verify the measurement by actual calculation.

Self-Evaluation

1.Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.

2. Draw an isosceles triangle ABC in which AB=AC=6 cm and BC=5 cm. Construct a triangle PQR similar to ΔABC in which PQ=8 cm. Also justify the construction.

3.Two line segments AB and AC include an angle of 600where AB=5 cm and AC=7 CM. Locate points P and Q on AB and AC respectively such that AP=3/4 AB and AQ=1/4 AC. Join P and Q and measure the length PQ.

4. Draw a triangle ABC in which AB=4 cm, BC=6 cm and AC=9 cm. Construct a triangle similar to ΔABC with scale factor 3/2. Justify your construction.

5.Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 450.

6.Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre another circle of radius 2.5 cm. Construct tangents to each circle from the centre of the other circle.

Value Based Question

(1) Two trees are to be planted at two positions A and B in the middle of a park and the third tree is to be planted at a position C in such a way that AC: BC= 3:4. How it can be done? What value is indicated from the above action?

(2) Draw a circle of radius 5 cm. Draw tangents from the end points of its diameter. What do you observe?

Key Points

Core Concepts

• Division of a Line Segment: Dividing a given line segment internally in a specified ratio (e.g., \( m:n \)).
• Construction of Triangles:
  • (a) When all three side lengths are specified.
  • (b) When two sides and the angle between them (included angle) are specified.
  • (c) When two angles and the side between them are specified.
  • (d) Construction of a right-angled triangle given its hypotenuse and one side.
• Construction of Similar Triangles: Constructing a triangle similar to a given triangle such that its sides are in a given ratio (scale factor) relative to the original triangle.
  • If the scale factor is a proper fraction (less than 1), the newly constructed similar triangle will lie entirely inside the original triangle.
  • If the scale factor is an improper fraction (greater than 1), the similar triangle will expand and lie outside the original triangle.
• Construction of Tangents to a Circle: Drawing tangents to a circle from a point lying on its circumference or from an external point.

Expected Learning Outcomes

• Accurate and precise use of geometric instruments (ruler, compasses, protractor).
• Constructing line segments and angles as per given quantitative data.
• Dividing line segments in a given ratio using proportional division techniques.
• Applying the concepts of similar triangles and Thales' Theorem (Basic Proportionality Theorem) to practical geometric tasks.
• Constructing tangents to a circle from external points using perpendicular bisectors.

LEVEL – I

Question 1. Draw a line segment AB=8cm and divide it in the ratio 4:3.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw a horizontal line segment \( AB = 8\text{ cm} \) using a ruler.
2. From point \( A \), draw an acute angle ray \( AX \) pointing downwards (e.g., at \( 30^\circ \) or \( 40^\circ \)).
3. Since the ratio is \( 4:3 \), the total number of divisions on the ray is \( 4 + 3 = 7 \).
4. Using a compass with a fixed radius, mark 7 equidistant points on ray \( AX \) starting from \( A \). Label them as \( A_1, A_2, A_3, A_4, A_5, A_6, A_7 \).
5. Connect the final point \( A_7 \) to the endpoint \( B \) with a straight line.
6. From point \( A_4 \), construct a line parallel to \( A_7B \). To do this, construct an angle at \( A_4 \) equal to \( \angle AA_7B \). Let this parallel line intersect \( AB \) at point \( P \).
7. The point \( P \) divides the line segment \( AB \) internally in the ratio \( 4:3 \), such that \( AP:PB = 4:3 \).

In simple words: We draw a line of length 8 cm and a downward ray. We mark 7 equal spaces on the ray, connect the last point to B, and draw a parallel line from the 4th mark to split the main line into a 4:3 ratio.

Exam Tip: Always make sure to count the sum of the ratio parts (\( m + n \)) to find the correct number of marks needed on the acute ray.

Question 2. Divide a line segment of 7cm internally in the ratio 2:3.
Answer: Let's carry out the construction step-by-step:

Steps of Construction:
1. Draw a straight line segment \( AB = 7\text{ cm} \) using a metric ruler.
2. Draw a ray \( AX \) making an acute angle with \( AB \) pointing downwards.
3. Add the terms of the ratio: \( 2 + 3 = 5 \).
4. Using a compass set to any convenient radius, mark 5 equidistant points on ray \( AX \). Label them \( A_1, A_2, A_3, A_4, A_5 \).
5. Connect the last point \( A_5 \) to the end of the line segment, \( B \).
6. At point \( A_2 \), construct a line parallel to \( A_5B \) by replicating the angle \( \angle AA_5B \) at \( A_2 \).
7. Let this parallel line intersect \( AB \) at point \( P \). Point \( P \) divides the line segment \( AB \) internally in the ratio \( 2:3 \).

In simple words: We draw a 7 cm line segment and an acute angle ray. We mark 5 equal steps along the ray, join the 5th mark to B, and construct a parallel line from the 2nd mark to get the desired \( 2:3 \) split.

Exam Tip: Double check your measurements with a ruler after completing the construction; the lengths \( AP \) and \( PB \) should be exactly 2.8 cm and 4.2 cm respectively.

Question 3. Draw a circle of radius 4 cm. Take a point P on it. Draw tangent to the given circle at P.
Answer: Let's construct the circle and its tangent:

Steps of Construction:
1. Mark a point \( O \) to serve as the center of the circle.
2. Using a compass set to a radius of 4 cm, draw a circle with its center at \( O \).
3. Mark any point \( P \) on the circumference of the circle.
4. Draw the radial segment \( OP \).
5. At point \( P \), construct a perpendicular line to \( OP \) (representing a \( 90^\circ \) angle). To do this, draw an arc intersecting \( OP \), use those intersections to mark \( 60^\circ \) and \( 120^\circ \) arcs, and bisect the space between them.
6. Extend this perpendicular line in both directions to form the line \( TPT' \).
7. The line \( TPT' \) is the required tangent to the circle at point \( P \).

In simple words: Draw a circle of radius 4 cm with center O, mark a point P on its edge, and draw a radius line OP. Constructing a vertical line at right angles to OP at point P gives the tangent.

Exam Tip: A tangent to a circle is always perpendicular to the radius at the point of contact; explicitly constructing a \( 90^\circ \) angle is the most critical part of this task.

Question 4. Construct an isosceles triangle whose base is 7.5 cm and altitude is 4.2 cm.
Answer: Let's follow these steps of construction:

Steps of Construction:
1. Draw a horizontal line segment \( BC = 7.5\text{ cm} \).
2. Construct the perpendicular bisector of \( BC \). To do this, place the compass needle at \( B \), open it to more than half the length of \( BC \), and draw arcs above and below the line. Repeat from \( C \). Join the intersection points of these arcs with a dashed line. Let this bisector intersect \( BC \) at its midpoint \( D \).
3. Using a ruler and compass, measure a radius of 4.2 cm. Place the compass needle at \( D \) and draw an arc cutting the perpendicular bisector. Label this intersection point as \( A \).
4. Connect point \( A \) to \( B \) and point \( A \) to \( C \) with straight lines.
5. The triangle \( ABC \) is the required isosceles triangle with base \( BC = 7.5\text{ cm} \) and altitude \( AD = 4.2\text{ cm} \).

In simple words: Draw a base line BC of 7.5 cm, then construct its vertical midpoint line. Measure 4.2 cm straight up along this middle line to locate point A, and join it to B and C.

Exam Tip: Since an isosceles triangle is symmetric, its top apex must lie exactly on the perpendicular bisector of its base.

Question 5. Draw a line segment of length 9 cm. and divide it in seven equal parts.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw a line segment \( AB = 9\text{ cm} \) using a ruler.
2. Draw a ray \( AX \) making an acute angle with \( AB \) pointing downwards.
3. Using a compass set to a small, fixed radius, mark 7 equidistant points on ray \( AX \) starting from \( A \). Label them \( A_1, A_2, A_3, A_4, A_5, A_6, A_7 \).
4. Connect the last division point \( A_7 \) to point \( B \).
5. Using a compass and ruler, draw lines parallel to \( A_7B \) passing through each of the points \( A_1, A_2, A_3, A_4, A_5, A_6 \). Let these lines intersect \( AB \) at points \( P_1, P_2, P_3, P_4, P_5, P_6 \) respectively.
6. These points divide the line segment \( AB \) into 7 equal parts.

In simple words: Draw a 9 cm line segment and an acute ray. Mark 7 equal units along the ray, join the last mark to B, and draw parallel lines from each of the other marks to divide the main line into 7 equal parts.

Exam Tip: Keep the radius of your compass completely locked when marking the division points along the ray to ensure equal segments.

LEVEL – II

Question 1. Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw the base of the triangle \( BC = 6\text{ cm} \).
2. With \( B \) as center and a radius of 5 cm, draw an arc. With \( C \) as center and a radius of 4 cm, draw another arc intersecting the first arc at point \( A \).
3. Join \( AB \) and \( AC \) to obtain the original triangle \( ABC \).
4. Draw a ray \( BX \) pointing downwards and making an acute angle with \( BC \).
5. Mark 3 equidistant points \( B_1, B_2, B_3 \) on ray \( BX \) because the denominator of the scale factor \( \frac{2}{3} \) is 3.
6. Connect the last point \( B_3 \) to \( C \).
7. From point \( B_2 \), construct a line parallel to \( B_3C \) to intersect \( BC \) at point \( C' \).
8. From point \( C' \), construct a line parallel to \( CA \) to intersect \( AB \) at point \( A' \).
9. The triangle \( A'BC' \) is the required similar triangle with sides \( \frac{2}{3} \) of the corresponding sides of \( \Delta ABC \).

In simple words: First construct the main triangle ABC with the given side lengths. Draw a downward ray from B with 3 equal steps, join the 3rd mark to C, and project parallel lines from the 2nd mark to scale down the triangle to \( 2/3 \) of its size.

Exam Tip: Ensure that your parallel lines look strictly parallel to the corresponding original lines; examiners look closely at these parallel properties to grade your drawing.

Question 2. Construct a triangle similar to a given ∆ABC such that each of its sides is 2/3rd of the corresponding sides of ∆ABC. It is given that AB=5cm BC=6cm and AC=7cm. Also write the steps of construction.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw the base of the triangle \( BC = 6\text{ cm} \) using a ruler.
2. With \( B \) as center and a radius of 5 cm, draw an arc. With \( C \) as center and a radius of 7 cm, draw another arc intersecting the first arc at point \( A \).
3. Join \( AB \) and \( AC \) to obtain the original triangle \( ABC \).
4. Draw a ray \( BX \) pointing downwards and making an acute angle with \( BC \).
5. Mark 3 equidistant points \( B_1, B_2, B_3 \) on ray \( BX \) because the denominator of the scale factor \( \frac{2}{3} \) is 3.
6. Connect the last point \( B_3 \) to \( C \).
7. From point \( B_2 \), construct a line parallel to \( B_3C \) to intersect \( BC \) at point \( C' \).
8. From point \( C' \), construct a line parallel to \( CA \) to intersect \( AB \) at point \( A' \).
9. The triangle \( A'BC' \) is the required similar triangle with sides \( \frac{2}{3} \) of the corresponding sides of \( \Delta ABC \).
In simple words: Construct the original triangle ABC first. Then draw a downward ray from B with 3 divisions, join the 3rd point to C, and draw parallel lines starting from the 2nd division to create the smaller similar triangle.

Exam Tip: When writing the "steps of construction", use clear numbered points and name each geometric tool (ruler, compasses) where applicable.

Question 3. Draw a pair of tangents to a circle of radius 4cm, which are inclined to each other at an angle of 600.
Answer: Let's carry out the construction step-by-step:

Steps of Construction:
1. Mark a point \( O \) as the center and draw a circle of radius 4 cm.
2. The angle between the two tangents is \( 60^\circ \). The angle between the radii at the center must be supplementary to it:
\[ \text{Angle at center} = 180^\circ - 60^\circ = 120^\circ \]
3. Draw any radius \( OA \).
4. Using a protractor or compass, construct an angle of \( 120^\circ \) at center \( O \) from \( OA \), and draw another radius \( OB \) such that \( \angle AOB = 120^\circ \).
5. At point \( A \), construct a perpendicular line to \( OA \) (\( 90^\circ \) angle).
6. At point \( B \), construct a perpendicular line to \( OB \) (\( 90^\circ \) angle).
7. Let these two perpendicular lines intersect each other at point \( P \).
8. The lines \( PA \) and \( PB \) are the required tangents to the circle, and they are inclined to each other at an angle of \( 60^\circ \).

In simple words: Draw a circle of radius 4 cm. Since the tangents meet at 60 degrees, the two radius lines must meet at 120 degrees at the center. Drawing perpendiculars at the ends of these radii gives the tangents.

Exam Tip: The sum of the angles in quadrilateral \( OAPB \) is \( 360^\circ \). Since two angles are right angles, the remaining two angles (\( \angle AOB \) and \( \angle APB \)) must add up to \( 180^\circ \).

Question 4. Draw a circle of radius 5cm. From a point 8cm away from its centre construct the pair of tangents to the circle and measure their lengths.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Mark a point \( O \) as the center and draw a circle of radius 5 cm.
2. Draw a line segment \( OP = 8\text{ cm} \) from the center \( O \) to an external point \( P \).
3. Construct the perpendicular bisector of the segment \( OP \) to locate its midpoint \( M \).
4. With \( M \) as the center and radius \( MO = MP \), draw another circle that intersects the original circle at two points, \( A \) and \( B \).
5. Join \( PA \) and \( PB \).
6. The segments \( PA \) and \( PB \) are the required pair of tangents from point \( P \).
7. Measuring their lengths with a ruler, we find \( PA = PB \approx 6.24\text{ cm} \) (mathematically, \( \sqrt{8^2 - 5^2} = \sqrt{39} \approx 6.24\text{ cm} \)).

In simple words: Draw the circle and the 8 cm line OP. Bisect OP to find its center M, draw a second dashed circle centered at M through O, and connect P to the points where the two circles cross.

Exam Tip: You can always verify your drawn tangent length by calculating \( \sqrt{OP^2 - r^2} \) using Pythagoras' theorem in right-angled \( \Delta OAP \).

Question 5. Construct a triangle PQR in which QR=6cm, Q=600 and R=450. Construct another triangle similar to ∆PQR such that its sides are 5/6 of the corresponding sides of ∆PQR.
Answer: Let's follow these steps of construction:

Steps of Construction:
1. Draw base \( QR = 6\text{ cm} \).
2. At point \( Q \), construct an angle of \( 60^\circ \) using a compass or protractor.
3. At point \( R \), construct an angle of \( 45^\circ \) using a compass or protractor.
4. Let the two rays intersect at point \( P \), completing the triangle \( PQR \).
5. Draw a ray \( QX \) pointing downwards and making an acute angle with \( QR \).
6. Mark 6 equidistant points \( Q_1, Q_2, Q_3, Q_4, Q_5, Q_6 \) on the ray \( QX \).
7. Connect the last point \( Q_6 \) to \( R \).
8. From point \( Q_5 \), construct a line parallel to \( Q_6R \) to intersect \( QR \) at point \( R' \).
9. From point \( R' \), construct a line parallel to \( RP \) to intersect \( PQ \) at point \( P' \).
10. The triangle \( P'QR' \) is the required similar triangle with sides \( \frac{5}{6} \) of the corresponding sides of \( \Delta PQR \).
In simple words: Construct the original triangle PQR first. Draw an acute ray from Q with 6 equal divisions, join the 6th point to R, and construct parallel lines from the 5th point to scale down the triangle.

Exam Tip: Since the scale factor is \( \frac{5}{6} < 1 \), the new similar triangle \( P'QR' \) lies completely inside the original triangle \( PQR \).

Question 6. Draw a line segment AB= 7.5cm and locate a point P on AB such that AP= 3/7 AB. Give justification of the construction.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw a line segment \( AB = 7.5\text{ cm} \) using a ruler.
2. Draw an acute angle ray \( AX \) pointing downwards.
3. Since we want \( AP = \frac{3}{7}AB \), point \( P \) must divide the segment \( AB \) in the ratio \( 3:4 \) (since \( 7 - 3 = 4 \)).
4. Mark 7 equidistant points \( A_1, A_2, A_3, A_4, A_5, A_6, A_7 \) on ray \( AX \).
5. Join the final point \( A_7 \) to point \( B \).
6. From point \( A_3 \), draw a line parallel to \( A_7B \) intersecting \( AB \) at point \( P \).
7. Point \( P \) is the required point on \( AB \) such that \( AP = \frac{3}{7}AB \).

Justification:
By construction, the line \( A_3P \) is parallel to \( A_7B \). In \( \Delta AA_7B \), applying the Basic Proportionality Theorem (B.P.T.):
\[ \frac{AP}{PB} = \frac{AA_3}{A_3A_7} = \frac{3}{4} \]
Adding 1 to the reciprocal:
\[ \frac{PB}{AP} + 1 = \frac{4}{3} + 1 \implies \frac{AP + PB}{AP} = \frac{7}{3} \]
\[ \implies \frac{AB}{AP} = \frac{7}{3} \implies AP = \frac{3}{7}AB \]
Hence, justified.
In simple words: To get \( AP = \frac{3}{7}AB \), we divide the segment AB in the ratio of \( 3:4 \). We use an acute ray with 7 equal steps, join the 7th step to B, and construct a parallel line from the 3rd step.

Exam Tip: Be careful with the ratio choice: a fraction of \( \frac{3}{7} \) means dividing the line in the ratio \( 3:(7-3) = 3:4 \).

LEVEL-III

Question 1. Draw a circle with centre O and radius 3.5cm. Take a horizontal diameter. Extend it to both sides to point P and Q such that OP=OQ=7cm. Draw tangents PA and QB, one above the diameter and the other below the diameter. Is PA||BQ.
Answer: Let's follow these steps to carry out the construction:

Steps of Construction:
1. Draw a circle of radius 3.5 cm with center \( O \).
2. Draw its horizontal diameter and extend it on both sides to points \( P \) and \( Q \) such that \( OP = OQ = 7\text{ cm} \).
3. Bisect the segment \( OP \) to find its midpoint \( M_1 \). Draw a circle with center \( M_1 \) and radius \( M_1P \) intersecting the original circle at point \( A \) (above the diameter). Join \( PA \).
4. Bisect the segment \( OQ \) to find its midpoint \( M_2 \). Draw a circle with center \( M_2 \) and radius \( M_2Q \) intersecting the original circle at point \( B \) (below the diameter). Join \( QB \).

Is \( PA \parallel BQ \)?
Yes, \( PA \) is parallel to \( BQ \).
In right-angled triangles \( \Delta OAP \) and \( \Delta OBQ \):
\( \sin(\angle APO) = \frac{OA}{OP} = \frac{3.5}{7} = \frac{1}{2} \implies \angle APO = 30^\circ \).
Similarly, \( \angle BQO = 30^\circ \).
Since \( \angle APO = \angle BQO = 30^\circ \), and they represent alternate interior angles relative to the transversal line \( PQ \), the tangent lines \( PA \) and \( BQ \) are parallel to each other (\( PA \parallel BQ \)).
In simple words: We draw the circle and extend the diameter to P and Q. After constructing the tangents PA (above) and QB (below), we find they both make a \( 30^\circ \) angle with the central line, proving they are parallel.

Exam Tip: Calculate the angles mathematically using trigonometric ratios to verify and justify your geometric parallel conclusions.

Question 2. Construct a ∆ ABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm.
Answer: Let's follow these steps to construct the similar triangle:

Steps of Construction:
1. Draw a line segment \( AB = 6\text{ cm} \).
2. Construct an angle of \( 30^\circ \) at point \( A \) and \( 60^\circ \) at point \( B \). Let these rays intersect at point \( C \), forming the triangle \( ABC \).
3. Extend the line segment \( AB \) to point \( B' \) such that \( AB' = 8\text{ cm} \).
4. At point \( B' \), construct a line parallel to \( BC \) (by replicating \( \angle B = 60^\circ \)).
5. Extend the ray \( AC \) to intersect this parallel line at point \( C' \).
6. The triangle \( AB'C' \) is the required similar triangle with base \( AB' = 8\text{ cm} \).
In simple words: Draw the original triangle ABC. Extend the base line AB to 8 cm to get point B', and construct a line parallel to BC from B' to meet the extended line AC at C'.

Exam Tip: Since the scale factor is \( \frac{8}{6} = \frac{4}{3} > 1 \), the new similar triangle \( AB'C' \) is larger and lies outside the original triangle \( ABC \).

Question 3. Draw a right triangle ABC in which B=900, AB=5cm, BC=4cm, then construct another triangle A’BC’ whose sides are 5/3 times the corresponding sides of ∆ABC. Is the new triangle also a right triangle?
Answer: Let's carry out the construction:

Steps of Construction:
1. Draw base \( BC = 4\text{ cm} \).
2. At point \( B \), construct a perpendicular line segment \( BA = 5\text{ cm} \) making \( \angle B = 90^\circ \).
3. Join \( AC \) to obtain the right-angled triangle \( ABC \).
4. Draw a ray \( BX \) pointing downwards and making an acute angle with \( BC \).
5. Mark 5 equidistant points \( B_1, B_2, B_3, B_4, B_5 \) on ray \( BX \) since the numerator of the scale factor \( \frac{5}{3} \) is 5.
6. Connect the 3rd point \( B_3 \) to \( C \).
7. From point \( B_5 \), construct a line parallel to \( B_3C \) intersecting the extended line \( BC \) at point \( C' \).
8. From point \( C' \), construct a line parallel to \( CA \) intersecting the extended line \( BA \) at point \( A' \).
9. The triangle \( A'BC' \) is the required similar triangle.

Is the new triangle also a right triangle?
Yes, the new triangle \( A'BC' \) is also a right-angled triangle because \( \angle A'BC' = \angle ABC = 90^\circ \) (as they share the same angle \( B \)).
In simple words: Draw right triangle ABC. Extend the sides, and scale them up by a factor of \( 5/3 \) using parallel lines from our division ray. Since they share the right angle at B, the new triangle is also a right triangle.

Exam Tip: Similar triangles always preserve angle measures, so any triangle similar to a right triangle must also be a right triangle.

Question 4. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Answer: Let's follow these steps to construct the tangents:

Steps of Construction:
1. Draw a line segment \( AB = 8\text{ cm} \).
2. With \( A \) as center, draw a circle of radius 4 cm.
3. With \( B \) as center, draw a circle of radius 3 cm.
4. Construct the perpendicular bisector of \( AB \) to locate its midpoint \( M \).
5. With \( M \) as center and radius \( MA = MB = 4\text{ cm} \), draw a circle.
6. Let this new circle intersect the circle centered at \( A \) at points \( P \) and \( Q \). Join \( BP \) and \( BQ \). These are the tangents from \( B \) to the circle centered at \( A \).
7. Let the same circle centered at \( M \) intersect the circle centered at \( B \) at points \( R \) and \( S \). Join \( AR \) and \( AS \). These are the tangents from \( A \) to the circle centered at \( B \).

In simple words: Draw the 8 cm line AB and the two circles at its ends. Find the midpoint M, draw a large helper circle from M passing through A and B, and connect each center to the intersection points on the opposite circle.

Exam Tip: A single helper circle centered at the midpoint \( M \) of line \( AB \) is sufficient to construct all four required tangents.

Question 5. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP/AB = 3/5.
Answer: Let's carry out the construction:

Steps of Construction:
1. Draw a line segment \( AB = 7\text{ cm} \) using a ruler.
2. Draw an acute angle ray \( AX \) pointing downwards.
3. The ratio is \( \frac{AP}{AB} = \frac{3}{5} \), which means point \( P \) divides \( AB \) in the ratio \( 3:2 \) (since \( 5 - 3 = 2 \)).
4. Mark 5 equidistant points \( A_1, A_2, A_3, A_4, A_5 \) on ray \( AX \).
5. Connect the last point \( A_5 \) to point \( B \).
6. From point \( A_3 \), construct a line parallel to \( A_5B \) intersecting \( AB \) at point \( P \).
7. The point \( P \) divides \( AB \) such that \( \frac{AP}{AB} = \frac{3}{5} \).
In simple words: Draw a 7 cm segment AB and an acute ray. Mark 5 equal steps along the ray, join the last step to B, and draw a parallel line from the 3rd step to divide the line in the desired \( 3:2 \) ratio.

Exam Tip: Be careful with the ratio: a fraction of \( \frac{3}{5} \) means dividing the segment into parts in the ratio \( 3:(5-3) = 3:2 \).

Question 6. Construct an isosceles triangle whose base is 8 cm. and altitude 4 cm. and then construct another triangle whose sides are ¾ times the corresponding sides of the isosceles triangle.
Answer: Let's follow these steps to construct the triangles:

Steps of Construction:
1. Draw base \( BC = 8\text{ cm} \).
2. Construct the perpendicular bisector of \( BC \) intersecting \( BC \) at its midpoint \( D \).
3. Mark point \( A \) on the bisector such that \( AD = 4\text{ cm} \). Join \( AB \) and \( AC \) to obtain the isosceles triangle \( ABC \).
4. Draw an acute angle ray \( BX \) pointing downwards.
5. Mark 4 equidistant points \( B_1, B_2, B_3, B_4 \) on ray \( BX \) since the denominator of the scale factor \( \frac{3}{4} \) is 4.
6. Connect \( B_4 \) to \( C \).
7. From point \( B_3 \), construct a line parallel to \( B_4C \) intersecting \( BC \) at point \( C' \).
8. From point \( C' \), construct a line parallel to \( CA \) intersecting \( AB \) at point \( A' \).
9. The triangle \( A'BC' \) is the required similar triangle.
In simple words: Draw the original isosceles triangle ABC with base 8 cm and height 4 cm. Use a 4-step division ray from B to scale down the triangle to \( 3/4 \) of its original size using parallel lines.

Exam Tip: Since the scale factor \( \frac{3}{4} \) is less than 1, the new similar triangle \( A'BC' \) lies completely inside the original triangle \( ABC \).

Question 7. ABC is a right triangle in which AB=5.4 cm, BC= 7 cm and
Answer: Let's carry out this construction step-by-step:

Steps of Construction:
1. Draw the right-angled triangle \( ABC \) with \( BC = 7\text{ cm} \), \( \angle B = 90^\circ \), and \( AB = 5.4\text{ cm} \).
2. From vertex \( B \), draw a perpendicular line \( BD \) to the hypotenuse \( AC \).
3. Since \( \angle BDC = 90^\circ \), a circle passing through \( B, C, \) and \( D \) will have \( BC \) as its diameter (as the angle in a semicircle is a right angle).
4. Find the midpoint of \( BC \), say \( O \). Draw a circle with center \( O \) and radius \( OB = OC \). This circle passes through \( B, C, \) and \( D \).
5. Join \( AO \) (where \( A \) is the external point).
6. Bisect the segment \( AO \) to locate its midpoint \( M \).
7. With \( M \) as center and radius \( MA \), draw a circle intersecting our first circle at point \( B \) and another point, say \( P \).
8. Join \( AP \).
9. The lines \( AB \) and \( AP \) are the required pair of tangents to the circle from point \( A \) (note that \( AB \) is already a tangent since \( \angle ABO = 90^\circ \)).
In simple words: Construct the right-angled triangle and draw the altitude BD. Since the circle through B, C, D has BC as its diameter, we locate its center at the midpoint of BC. Constructing the second circle from the midpoint of AO gives the tangent point P.

Exam Tip: Since \( AB \) is perpendicular to the diameter \( BC \), \( AB \) is automatically one of the tangents; you only need to construct the second tangent \( AP \).

Question 8. Construct a triangle ABC in which AB=5cm,
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw a line segment \( AB = 5\text{ cm} \).
2. Construct a parallel line to \( AB \) at a distance of 3 cm above it (to represent the altitude \( CD = 3\text{ cm} \)).
3. At point \( B \), construct an angle of \( 60^\circ \) and extend the ray to intersect the parallel line at point \( C \).
4. Join \( AC \) to complete the triangle \( ABC \).
5. Extend \( AB \) to \( Q \) such that \( AQ = 1.5 \times AB = 7.5\text{ cm} \) (or use an acute ray division method with a \( 3:2 \) scale factor).
6. From point \( Q \), construct a line parallel to \( BC \) intersecting the extended line \( AC \) at point \( R \).
7. The triangle \( APQ \) (or \( PQR \)) is the required similar triangle whose sides are 1.5 times the sides of \( \Delta ABC \).
In simple words: We draw the base and construct a parallel helper line 3 cm above it. Drawing a 60-degree angle from B locates the top point C. We then scale up this triangle by a factor of 1.5 using parallel lines.

Exam Tip: An altitude of 3 cm means the vertex \( C \) must lie on a line parallel to the base \( AB \) at a perpendicular distance of exactly 3 cm.

Question 9. Construct a tangent to a circle of radius 3.5 from a point on the concentric circle of radius 6.5 cm and measure its length. Also, verify the measurement by actual calculation.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Mark a center \( O \) and draw two concentric circles of radii 3.5 cm and 6.5 cm.
2. Mark any point \( P \) on the outer circle of radius 6.5 cm. Join \( OP \).
3. Construct the perpendicular bisector of \( OP \) to locate its midpoint \( M \).
4. With \( M \) as center and radius \( MO = MP \), draw a circle (or arc) intersecting the inner circle of radius 3.5 cm at point \( T \).
5. Join \( PT \).
6. The line \( PT \) is the required tangent.
7. Measuring \( PT \) with a ruler, we find \( PT \approx 5.48\text{ cm} \).

Actual Calculation Verification:
In right-angled \( \Delta OTP \) (since the radius is perpendicular to the tangent at the point of contact):
\[ OP^2 = OT^2 + PT^2 \]
\[ (6.5)^2 = (3.5)^2 + PT^2 \]
\[ 42.25 = 12.25 + PT^2 \]
\[ PT^2 = 42.25 - 12.25 = 30 \]
\[ PT = \sqrt{30} \approx 5.48\text{ cm} \]
The measured value matches the calculated value.
In simple words: Draw two concentric circles. From a point P on the outer circle, we use a helper circle to find where the tangent touches the inner circle at T. Measuring the tangent gives about 5.48 cm, which matches our Pythagoras calculation.

Exam Tip: Concentric circles have the same center; always write down the Pythagoras calculation explicitly to show the verification step.

Self-Evaluation

Question 1. Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.
Answer: Let's follow these steps to divide the segment:

Steps of Construction:
1. Draw a line segment \( AB = 7\text{ cm} \).
2. Draw an acute angle ray \( AX \) pointing downwards.
3. Since the ratio is \( 3:5 \), the total number of division points is \( 3 + 5 = 8 \).
4. Mark 8 equidistant points \( A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8 \) on ray \( AX \) with a compass.
5. Join the final point \( A_8 \) to point \( B \).
6. From point \( A_3 \), draw a line parallel to \( A_8B \) intersecting \( AB \) at point \( P \).
7. Point \( P \) divides the line segment \( AB \) internally in the ratio \( 3:5 \).
In simple words: Draw a 7 cm segment AB. Using an acute ray with 8 equal divisions, we connect the last mark to B, and construct a parallel line from the 3rd mark to split AB into a 3:5 ratio.

Exam Tip: Ensure that your dividing parallel line is drawn carefully using equal alternate interior angles to guarantee accuracy.

Question 2. Draw an isosceles triangle ABC in which AB=AC=6 cm and BC=5 cm. Construct a triangle PQR similar to ∆ABC in which PQ=8 cm. Also justify the construction.
Answer: Let's carry out the construction:

Steps of Construction:
1. Draw base \( BC = 5\text{ cm} \).
2. With \( B \) and \( C \) as centers and a radius of 6 cm, draw arcs intersecting at point \( A \). Join \( AB \) and \( AC \) to complete \( \Delta ABC \).
3. Since \( AB = AC = 6\text{ cm} \) and the new similar triangle \( PQR \) has \( PQ = 8\text{ cm} \) corresponding to side \( AB \), the scale factor is:
\[ \text{Scale Factor} = \frac{8}{6} = \frac{4}{3} \]
4. Draw a ray \( BX \) making an acute angle with \( BC \) pointing downwards.
5. Mark 4 equidistant points \( B_1, B_2, B_3, B_4 \) on ray \( BX \).
6. Connect \( B_3 \) to \( C \).
7. From point \( B_4 \), construct a line parallel to \( B_3C \) intersecting the extended line \( BC \) at point \( R \).
8. From point \( R \), construct a line parallel to \( CA \) intersecting the extended line \( BA \) at point \( P \).
9. The triangle \( PQR \) (which can also be written as \( A'BC' \)) is the required similar triangle.

Justification:
By construction, \( PR \parallel AC \). Therefore:
\[ \Delta PQR \sim \Delta ABC \]
Also, \( \frac{PQ}{AB} = \frac{QR}{BC} = \frac{PR}{AC} = \frac{BB_4}{BB_3} = \frac{4}{3} \).
Since \( AB = 6\text{ cm} \), \( PQ = \frac{4}{3} \times 6 = 8\text{ cm} \).
Hence, justified.
In simple words: Draw the original triangle ABC. Since we want a similar triangle with side 8 cm instead of 6 cm, we scale it up by a factor of \( 4/3 \). Using parallel lines from our 4-division ray gives the enlarged triangle.

Exam Tip: Since the scale factor \( \frac{4}{3} \) is greater than 1, the new similar triangle will lie outside the original triangle.

Question 3. Two line segments AB and AC include an angle of 600where AB=5 cm and AC=7 CM. Locate points P and Q on AB and AC respectively such that AP=3/4 AB and AQ=1/4 AC. Join P and Q and measure the length PQ.
Answer: Let's follow these steps to locate the points and measure \( PQ \):

Steps of Construction:
1. Draw a line segment \( AC = 7\text{ cm} \).
2. At point \( A \), construct an angle of \( 60^\circ \) using a protractor or compass and draw a ray.
3. Mark point \( B \) on this ray such that \( AB = 5\text{ cm} \).
4. **To locate \( P \) on \( AB \):**
We can calculate \( AP = \frac{3}{4} \times 5 = 3.75\text{ cm} \). Using a ruler, mark point \( P \) on \( AB \) at a distance of 3.75 cm from \( A \).
5. **To locate \( Q \) on \( AC \):**
We can calculate \( AQ = \frac{1}{4} \times 7 = 1.75\text{ cm} \). Mark point \( Q \) on \( AC \) at a distance of 1.75 cm from \( A \).
6. Join \( P \) and \( Q \) with a straight line.
7. Measuring the length \( PQ \) with a ruler, we find \( PQ \approx 3.3\text{ cm} \).
In simple words: Draw two segments AB (5 cm) and AC (7 cm) meeting at 60 degrees. Mark point P at 3.75 cm along AB, and point Q at 1.75 cm along AC. Joining them gives a segment PQ of approximately 3.3 cm.

Exam Tip: If the scaling factors are simple, you can calculate the exact distances (\( 3.75\text{ cm} \) and \( 1.75\text{ cm} \)) directly with a ruler to save construction time.

Question 4. Draw a triangle ABC in which AB=4 cm, BC=6 cm and AC=9 cm. Construct a triangle similar to ∆ABC with scale factor 3/2. Justify your construction.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw base \( BC = 6\text{ cm} \).
2. With \( B \) as center and a radius of 4 cm, draw an arc. With \( C \) as center and a radius of 9 cm, draw another arc intersecting the first arc at point \( A \).
3. Join \( AB \) and \( AC \) to obtain the original triangle \( ABC \).
4. Draw a ray \( BX \) pointing downwards and making an acute angle with \( BC \).
5. Mark 3 equidistant points \( B_1, B_2, B_3 \) on ray \( BX \) since the scale factor is \( \frac{3}{2} \).
6. Connect \( B_2 \) to \( C \).
7. From point \( B_3 \), construct a line parallel to \( B_2C \) intersecting the extended line \( BC \) at point \( C' \).
8. From point \( C' \), construct a line parallel to \( CA \) intersecting the extended line \( BA \) at point \( A' \).
9. The triangle \( A'BC' \) is the required similar triangle.

Justification:
By construction, \( A'C' \parallel AC \). Therefore, \( \Delta A'BC' \sim \Delta ABC \).
The ratio of their corresponding sides is:
\[ \frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{BB_3}{BB_2} = \frac{3}{2} \]
Hence, justified.
In simple words: Draw the original triangle ABC. Since the scale factor \( 3/2 \) is greater than 1, we draw a 3-step division ray from B, join the 2nd point to C, and project parallel lines from the 3rd point to construct the enlarged similar triangle.

Exam Tip: Since the scale factor \( \frac{3}{2} \) is greater than 1, the newly constructed similar triangle will lie outside the original triangle.

Question 5. Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 450.
Answer: Let's follow these steps to construct the tangents:

Steps of Construction:
1. Draw a circle of radius 4.5 cm with center \( O \).
2. The angle between the two tangents is \( 45^\circ \). Therefore, the angle between the radii at the center must be:
\[ \text{Angle at center} = 180^\circ - 45^\circ = 135^\circ \]
3. Draw any radius \( OA \).
4. At point \( O \), construct an angle of \( 135^\circ \) from \( OA \) and draw another radius \( OB \) such that \( \angle AOB = 135^\circ \).
5. At point \( A \), construct a perpendicular line to \( OA \).
6. At point \( B \), construct a perpendicular line to \( OB \).
7. Let these two perpendicular lines intersect each other at point \( P \).
8. The lines \( PA \) and \( PB \) are the required tangents inclined to each other at an angle of \( 45^\circ \).
In simple words: Draw a circle of radius 4.5 cm. Draw two radius lines meeting at \( 135^\circ \) at the center. Drawing perpendicular lines at the ends of these radii gives the tangents, which will meet at \( 45^\circ \).

Exam Tip: Use a protractor to measure the angle between the tangents at \( P \) to verify that it is exactly \( 45^\circ \) after completing your drawing.

Question 6. Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre another circle of radius 2.5 cm. Construct tangents to each circle from the centre of the other circle.
Answer: Let's follow these steps to perform the construction:

Steps of Construction:
1. Draw a line segment \( AB = 7\text{ cm} \).
2. With \( A \) as center, draw a circle of radius 3 cm.
3. With \( B \) as center, draw a circle of radius 2.5 cm.
4. Construct the perpendicular bisector of \( AB \) to locate its midpoint \( M \).
5. With \( M \) as center and radius \( MA = MB = 3.5\text{ cm} \), draw a circle.
6. Let this new circle intersect the circle centered at \( A \) at points \( P \) and \( Q \). Join \( BP \) and \( BQ \). These are the tangents from \( B \) to the circle of radius 3 cm.
7. Let the same circle centered at \( M \) intersect the circle centered at \( B \) at points \( R \) and \( S \). Join \( AR \) and \( AS \). These are the tangents from \( A \) to the circle of radius 2.5 cm.
In simple words: Draw a 7 cm segment AB and circles at its ends. Find the midpoint M, draw a large helper circle from M passing through A and B, and connect each center to the intersection points on the opposite circle.

Exam Tip: Using a single helper circle centered at the midpoint \( M \) of line \( AB \) is sufficient to construct all four required tangents.

Value Based Question

Question (1) Two trees are to be planted at two positions A and B in the middle of a park and the third tree is to be planted at a position C in such a way that AC: BC= 3:4. How it can be done? What value is indicated from the above action?
Answer: Let's follow these steps to perform the division:

Method of division:
1. Join the two tree positions \( A \) and \( B \) with a straight line segment \( AB \).
2. Draw an acute angle ray \( AX \) pointing downwards.
3. Since the ratio of planting is \( AC:BC = 3:4 \), mark \( 3 + 4 = 7 \) equidistant points on ray \( AX \).
4. Join the 7th point \( A_7 \) to point \( B \).
5. From the 3rd point \( A_3 \), draw a line parallel to \( A_7B \) intersecting \( AB \) at point \( C \).
6. The third tree should be planted at this point \( C \) on the line segment \( AB \).

Value Indicated:
The action indicates environmental responsibility, care for nature, green landscaping planning, and planting trees to preserve ecological balance.
In simple words: We join A and B with a line and divide it in the ratio of \( 3:4 \) using our division ray method to find point C. Planting trees shows our love and care for the environment.

Exam Tip: In value-based questions, make sure to write down both the geometric division steps and the conceptual values clearly to secure full marks.

Question (2) Draw a circle of radius 5 cm. Draw tangents from the end points of its diameter. What do you observe?
Answer: Let's carry out the construction:

Steps of Construction:
1. Draw a circle of radius 5 cm with center \( O \).
2. Draw any diameter \( AB \) passing through center \( O \).
3. At point \( A \), construct a perpendicular line to the diameter \( AB \).
4. At point \( B \), construct a perpendicular line to the diameter \( AB \).

Observation:
We observe that the two tangents drawn at the endpoints of the diameter are **parallel** to each other.
This happens because the angles made by the tangents with the diameter are both \( 90^\circ \). Since the sum of the co-interior angles is \( 90^\circ + 90^\circ = 180^\circ \), the lines must be parallel.
In simple words: Draw a circle and a straight diameter line through its center. Constructing perpendicular tangents at both ends of this diameter line shows that they run perfectly parallel to each other.

Exam Tip: The theorem "tangents drawn at the ends of a diameter of a circle are parallel" is a fundamental property; make sure to write the angle sum justification in your answer.