Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 1 Basic Concepts in Geometry Set 1 Solutions

Question 1. Select the correct alternative answer for the questions given below.
 

(i) How many midpoints does a segment have ?
(A) only one
(B) two
(C) three
(D) many
Answer: (A) only one
In simple words: A line segment has exactly one midpoint, which divides it into two equal parts.

🎯 Exam Tip: Understanding basic geometric definitions is crucial for MCQs. Always remember the uniqueness of a segment's midpoint.

(ii) How many points are there in the intersection of two distinct lines ?
(A) infinite
(B) two
(C) one
(D) not a single
Answer: (C) one
In simple words: Two different straight lines can intersect at only one common point. If they don't intersect, they are parallel.

🎯 Exam Tip: This is a fundamental concept in Euclidean geometry- distinct lines either intersect at one point or are parallel (never intersect).

(iii) How many lines are determined by three distinct points?
(A) two
(B) three
(C) one or three
(D) six
Answer: (C) one or three
In simple words: If three points are on the same line (collinear), they define one line. If they are not on the same line (non-collinear), they define three distinct lines, forming a triangle.

🎯 Exam Tip: Remember the two cases: collinear points define one line, non-collinear points define three lines (one for each pair of points).

(iv) Find d(A, B), if co-ordinates of A and B are - 2 and 5 respectively.
(A) -2
(B) 5
(C) 7
(D) 3
Answer: Since, \( 5 > -2 \)
.. \( d(A, B) = 5 - (-2) = 5+2 = 7 \)
(C) 7
In simple words: The distance between two points on a number line is found by subtracting the smaller coordinate from the larger one.

🎯 Exam Tip: Always subtract the smaller coordinate from the larger one to get a positive distance. Be careful with negative numbers.

(v) If P - Q - R and d(P, Q) = 2, d(P, R) = 10, then find d(Q, R).
(A) 12
(B) 8
(C) √96
(D) 20
Answer:
\( d(P, R) = d(P, Q) + d(Q, R) \)
.. \( 10 = 2 + d(Q, R) \)
.. \( d(Q, R) = 8 \)
(B) 8
In simple words: When points P, Q, R are collinear in that order (P-Q-R), the distance from P to R is the sum of the distances from P to Q and Q to R.

🎯 Exam Tip: The segment addition postulate is key here. Make sure the order of points is correctly interpreted (P-Q-R means Q is between P and R).

Question 2. On a number line, co-ordinates of P, Q, R are 3,-5 and 6 respectively. State with reason whether the following statements are true or false.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक संख्या रेखा जिस पर तीन बिंदु Q, P और R अंकित हैं। बिंदु Q का निर्देशांक -5 है, बिंदु P का निर्देशांक 3 है, और बिंदु R का निर्देशांक 6 है। बिंदु P के नीचे 3 और Q के नीचे 0 भी दर्शाया गया है।
Answer:
Co-ordinate of the point P is 3.
Co-ordinate of the point Q is -5.
Since, \( 3 > -5 \)
\( d(P, Q) = 3 - (-5) = 3 + 5 \)
.. \( d(P,Q) = 8 \)
Co-ordinate of the point Q is -5.
Co-ordinate of the point R is 6.
Since, \( 6 > -5 \)
\( d(Q, R) = 6 - (-5) = 6 + 5 \)
.. \( d(Q, R) = 11 \)
Co-ordinate of the point P is 3.
Co-ordinate of the point R is 6.
Since, \( 6 > 3 \)
\( d(P, R) = 6 - 3 \)
.. \( d(P, R) = 3 \)

(i) d(P, Q) + d(Q, R) = 8 + 11
= 19 ...(i)
d(P, R) = 3 ...(ii)
.. \( d(P, Q) + d(Q, R) \neq d(P, R) \) ... [From (i) and (ii)]
.. The given statement is false.

(ii) d(P, R) + d(R, Q) = 3 + 11
d(P,Q) = 8 ...(ii)
.. d(P, R) + d(R, Q) + d(P, Q) ...[From (i) and (ii)]
.. The given statement is false.

(iii) d(R, P) + d(P, Q) = 3 + 8
= 11 ...(i)
d(R, Q) = 11 . -(ii)
.. \( d(R,P) + d(P,Q) = d(R,Q) \) ....[From (i) and (ii)]
.. The given statement is true.

(iv) d(P, Q) – d(P, R) = 8 – 3
= 5 ...(i)
d(Q,R) = 11 ..(h)
.. \( d(P, Q) - d(P, R) \neq d(Q, R) \) ... [From (i) and (ii)]
.. The given statement is false.
In simple words: To determine collinearity and relative positions of points on a number line, calculate the distances between all pairs of points. If the sum of two smaller distances equals the largest distance, the points are collinear, and the common point is between the other two.

🎯 Exam Tip: For problems involving distances on a number line, always draw a quick sketch to visualize the positions of points. Remember the segment addition postulate. Distances are always positive.

Question 3. Co-ordinates of some pairs of points are given below. Hence find the distance between each pair.
Answer:
(i) 3,6
Co-ordinate of first point is 3.
Co-ordinate of second point is 6.
Since, \( 6 > 3 \)
.. Distance between the points \( = 6 - 3 = 3 \)

(ii) -9, -1
Co-ordinate of first point is -9.
Co-ordinate of second point is -1.
Since, \( -1 > -9 \)
.. Distance between the points \( = -1 - (-9) = -1+9 = 8 \)

(iii) -4, 5
Co-ordinate of first point is -4.
Co-ordinate of second point is 5.
Since, \( 5 > -4 \)
.. Distance between the points \( = 5 - (-4) \)
\( = 5 + 4 = 9 \)

(iv) 0,-2
Co-ordinate of first point is 0.
Co-ordinate of second point is -2.
Since, \( 0 > -2 \)
.. Distance between the points \( = 0 - (-2) \)
\( = 0 + 2 \)
\( = 2 \)

(v) x + 3, x - 3
Co-ordinate of first point is x + 3.
Co-ordinate of second point is x - 3.
Since, \( x + 3 > x - 3 \)
.. Distance between the points \( = x + 3 - (x - 3) \)
\( = x + 3 - x + 3 = 3 + 3 \)
\( = 6 \)

(vi) -25, -47
Co-ordinate of first point is -25.
Co-ordinate of second point is -47.
Since, \( -25 > -47 \)
.. Distance between the points \( = -25 - (-47) \)
\( = -25 + 47 \)
\( = 22 \)

(vii) 80, -85
Co-ordinate of first point is 80.
Co-ordinate of second point is -85.
Since, \( 80 > -85 \)
.. Distance between the points \( = 80 - (-85) \)
\( = 80 + 85 \)
\( = 165 \)
In simple words: The distance between any two points on a number line is always a positive value, calculated by taking the absolute difference between their coordinates. It's usually easier to subtract the smaller coordinate from the larger one.

🎯 Exam Tip: Be careful with signs, especially when dealing with negative coordinates. Always ensure the distance is positive. For algebraic expressions, correctly expand and simplify terms.

Question 4. Co-ordinate of point P on a number line is - 7. Find the co-ordinates of points on the number line which are at a distance of 8 units from point P.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक संख्या रेखा जिस पर बिंदु P का निर्देशांक -7 है। बिंदु Q, P के बाईं ओर 8 इकाई की दूरी पर है, और बिंदु R, P के दाईं ओर 8 इकाई की दूरी पर है।
Answer:
Let point Q be at a distance of 8 units from P and on left side of P
Let point R be at a distance of 8 units from P and on right side of P.

(i) Let the co-ordinate of point Q be x.
Co-ordinate of point P is -7.
Since, point Q is to the left of point P.
.. \( -7 > x \)
.. \( d(P, Q) = -7 -x \)
.. \( 8 = -7 - x \)
.. \( x = -7 - 8 \)
.. \( x = -15 \)

(ii) Let the co-ordinate of point R be y.
Co-ordinate of point P is -7.
Since, point R is to the right of point P.
.. \( y > -7 \)
.. \( d(P, R) = y - (-7) \)
.. \( 8 = y + 7 \)
.. \( 8 - 7 = y \)
∴ \( y = 1 \)
.. The co-ordinates of the points at a distance of 8 units from P are -15 and 1.
In simple words: To find points at a certain distance from a given point on a number line, consider two possibilities: one point to the left (subtract the distance) and one point to the right (add the distance).

🎯 Exam Tip: Always consider both directions (left and right) on the number line when finding points at a specified distance. Be precise with sign conventions for coordinates.

Question 5. Answer the following questions.
Answer:
(i) If A – B – C and d(A, C) = 17, d(B, C) = 6.5, then d (A, B) = ?
Given, \( d(A, C) = 17 \), \( d(B, C) = 6.5 \)
\( d(A, C) = d(A, B) + d(B, C) \) ...[A – B – C]
.. \( 17 = d(A, B) + 6.5 \)
.. \( d(A,B)= 17 - 6.5 \)
.. \( d(A, B) = 10.5 \)

(ii) If P – Q – R and d(P, Q) = 3.4, d(Q, R) = 5.7, then d(P, R) = ?
Given, \( d(P, Q) = 3.4 \), \( d(Q, R) = 5.7 \)
\( d(P,R) = d(P,Q) + d(Q,R) \) ...[P – Q – R]
\( = 3.4 + 5.7 \)
.. \( d(P, R) = 9.1 \)
In simple words: For three collinear points, if the middle point is known, the total distance between the two outer points is the sum of the distances from the middle point to each outer point.

🎯 Exam Tip: The segment addition postulate is fundamental. Ensure the order of points is correctly understood for applying the sum or difference of distances.

Question 6. Co-ordinate of point A on a number line is 1. What are the co-ordinates of points on the number line which are at a distance of 7 units from A ?
ℹ️ चित्र व्याख्या (Diagram Explanation): एक संख्या रेखा जिस पर बिंदु A का निर्देशांक 1 है। बिंदु C, A के बाईं ओर 7 इकाई की दूरी पर है और बिंदु B, A के दाईं ओर 7 इकाई की दूरी पर है।
Answer:
Let point C be at a distance of 7 units from A and on left side of A
Let point B be at a distance of 7 units from A and on right side of A.

(i) Let the co-ordinate of point C be x.
Co-ordinate of point A is 1.
Since, point C is to the left of point A.
. \( 1 > x \)
.. \( d(A, C) = 1-x \)
.. \( 7 = 1 -x \)
.. \( x = 1-7 \)
.. \( x = - 6 \)

(ii) Let the co-ordinate of point B be y.
Co-ordinate of point A is 1.
Since, point B is to the right of point A.
\( y > 1 \)
.. \( d(A, B) = y-1 \)
.. \( 7 = y - 1 \)
.. \( 7 + 1 = y \)
.. \( y = 8 \)
.. The co-ordinates of the points at a distance of 7 units from A are -6 and 8.
In simple words: Similar to finding coordinates at a distance, you add the distance for points to the right and subtract for points to the left of the given coordinate.

🎯 Exam Tip: Always state the assumption of the point being to the left or right of the given point clearly to avoid confusion in calculations. Remember that distances are non-negative.

Question 7. Write the following statements in conditional form.
Answer:
(i) Every rhombus is a square.
If a quadrilateral is a rhombus, then it is a square.

(ii) Annies in a linear pair are supplementary.
If two angles are in a linear pair, then they are supplementary.

(iii) A triangle is a figure formed by three segments
If a figure is a triangle, then it is formed by three segments.

(iv) A number having only two divisors is called a prime number.
If a number has only two divisors, then it is a prime number.
In simple words: A conditional statement is an "If-then" statement, where the "If" part is the hypothesis (antecedent) and the "then" part is the conclusion (consequent).

🎯 Exam Tip: Practice identifying the subject and predicate of a statement to correctly phrase it into the "If-then" conditional form.

Question 8. Write the converse of each of the following statements.
Answer:
(i) If the sum of measures of angles in a figure is 180°, then the figure is a triangle.
Converse: If a figure is a triangle, then the sum of the measures of its angles is 180°.

(ii) If the sum of measures of two angles is 90°, then they are complement of each other.
Converse: If two angles are complement of each other, then sum of their measures is 90°.

(iii) If the corresponding angles formed by a transversal of two lines are congruent, then the two lines are parallel.
Converse: If two lines are parallel, then the corresponding angles formed by a transversal of two lines are congruent.

(iv) If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
Converse: If a number is divisible by 3, then the sum of its digits is also divisible by 3.
In simple words: The converse of an "If P, then Q" statement is "If Q, then P." You swap the hypothesis (P) and the conclusion (Q).

🎯 Exam Tip: When forming the converse, ensure you accurately swap the entire hypothesis and conclusion. The truth value of the converse is not necessarily the same as the original statement.

Question 9. Write the antecedent (given part) and the consequent (part to be proved) in the following statements.
Answer:
(i) If all sides of a triangle are congruent, then its all angles are congruent.
Antecedent (Given): All the sides of the triangle are congruent.
Consequent (To prove): All the angles are congruent.

(ii) The diagonals of a parallelogram bisect each other.
Conditional statement: "If a quadrilateral is a parallelogram then its diagonals bisect each other.
Antecedent (Given): Quadrilateral is a parallelogram.
Consequent (To prove): Its diagonals bisect each other.
In simple words: In a conditional statement ("If P, then Q"), the antecedent is 'P' (what is given), and the consequent is 'Q' (what is to be proved or concluded).

🎯 Exam Tip: Clearly separate the "if" clause (antecedent) and the "then" clause (consequent) in any given conditional statement.

Question 10. Draw a labelled figure showing information in each of the following statements and write the antecedent and the consequent.
Answer:
(i) Two equilateral triangles are similar.
Conditional statement: "If two triangles are equilateral, then they are similar.
Antecedent (Given): Two triangles are equilateral.
i.e. ΔΑΒC and ΔPQR are equilateral triangles.
Consequent (To prove): Triangles are similar
i.e. \( \Delta ABC \sim \Delta PQR \)
ℹ️ चित्र व्याख्या (Diagram Explanation): दो समबाहु त्रिभुज, ΔABC और ΔPQR, दर्शाए गए हैं। त्रिभुज ΔABC में सभी भुजाएँ बराबर हैं और त्रिभुज ΔPQR में भी सभी भुजाएँ बराबर हैं। दोनों त्रिभुजों को समान रूप से लेबल किया गया है।

(ii) If angles in a linear pair are congruent, then each of them is a right angle.
Antecedent (Given): Angles in a linear pair are congruent.
\( \angle ABC \) and \( \angle ABD \) are angles in a linear pair i.e. \( \angle ABC = \angle ABD \)
Consequent (To prove): Each angle is a right angle.
i.e. \( \angle ABC = \angle ABD = 90^\circ \)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक सीधी रेखा पर बिंदु D, B, C स्थित हैं। एक किरण BA बिंदु B से ऊपर की ओर निकल रही है, जिससे \( \angle ABC \) और \( \angle ABD \) बन रहे हैं। ये कोण एक रेखीय युग्म बनाते हैं।

(iii) If the altitudes drawn on two sides of a triangle are congruent, then these two sides are congruent.
Antecedent (Given): Altitude drawn on two sides of triangle are congruent.
In \( \Delta ABC \), AD \( \perp \) BC. and BE \( \perp \) AC. seg AD \( \cong \) seg BE
Consequent (To prove): Two sides are congruent.
side BC \( \cong \) side AC
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज ABC दर्शाया गया है। बिंदु B से भुजा AC पर एक लंब BE खींचा गया है, और बिंदु A से भुजा BC पर एक लंब AD खींचा गया है। ये लंब (ऊंचाई) त्रिभुज की भुजाओं पर स्थित हैं।
In simple words: This question requires breaking down a geometric statement into its 'given' and 'to prove' parts, and illustrating the scenario with a clear, labeled diagram.

🎯 Exam Tip: For geometry problems, a well-labeled diagram is often crucial. It helps in visualizing the given information (antecedent) and what needs to be proved (consequent).

Maharashtra Board Class 9 Maths Chapter 1 Basic Concepts In Geometry Problem Set 1 Intext Questions And Activities

Question 1. Points A, B, C are given below. Check, with a stretched thread, whether the three points are collinear or not. If they are collinear, write which one of them is between the other two. (Textbook pg. no. 4)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक सीधी रेखा पर तीन बिंदु A, B और C दर्शाए गए हैं, जो संरेखीय हैं। बिंदु B, बिंदु A और C के बीच में स्थित है।
Answer: Point B is between the points A and C.
In simple words: Three points are collinear if they lie on the same straight line. You can test this by trying to draw a single straight line through all of them.

🎯 Exam Tip: Collinearity can be visually checked with a straightedge or, mathematically, by verifying if the sum of two smaller distances between pairs of points equals the largest distance.

Question 2. Given below are four points P, Q, R, and S. Check which three of them are collinear and which three are non collinear. In the case of three collinear points, state which of them is between the other two. (Textbook pg. no. 4)
ℹ️ चित्र व्याख्या (Diagram Explanation): चार बिंदु P, Q, R और S दर्शाए गए हैं। बिंदु P, R और S एक सीधी रेखा पर स्थित हैं (संरेखीय हैं), जबकि बिंदु Q इन बिंदुओं से अलग है।
Answer: Points P, R and S are collinear.
Point R is between the points P and S.
In simple words: To identify collinear points among several, look for any three points that form a straight line. If they do, then one point will lie between the other two.

🎯 Exam Tip: For multiple points, test combinations of three points to find collinear sets. Visual inspection combined with understanding of segment relationships helps determine which point is between the others.

Question 3. Students are asked to stand in a line for mass drill. How will you check whether the students standing are in a line or not ? (Textbook pg. no. 4)
Answer: If one stands in front of the line and observes only the first student standing in the line, then all the students standing in that line are collinear i.e., standing in the same line. We can use this property of collinearity to check whether the students are standing in the same line or not.
In simple words: To check if students are in a straight line, an observer can stand at one end and see if the first student blocks the view of all subsequent students. If so, they are collinear.

🎯 Exam Tip: This question tests practical application of the collinearity concept. Relate it to real-world scenarios for better comprehension.

Question 4. How had you verified that light rays travel in a straight line? Recall an experiment in science which you have done in a previous standard. (Textbook pg. no. 4)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक प्रयोग दिखाया गया है जिसमें प्रकाश स्रोत के रूप में एक मोमबत्ती है। तीन कार्डबोर्ड शीटों में छेद किए गए हैं और उन्हें एक सीधी रेखा में व्यवस्थित किया गया है। मोमबत्ती की लौ तभी दिखाई देती है जब सभी छेद एक सीधी रेखा में हों, यह दर्शाता है कि प्रकाश सीधी रेखा में चलता है।
Answer: The flame of the candle can be seen only when the pin holes in all cardboards are in the same straight line. We can use the set up shown in the figure above to verify that light rays travels in a straight line.
In simple words: The pinhole experiment demonstrates that light travels in straight lines because the light source is only visible when the pinholes are perfectly aligned, forming a straight path.

🎯 Exam Tip: This question connects geometry to basic physics. Understanding that light travels in straight lines (rectilinear propagation) is a fundamental principle illustrated by collinearity.