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Chapter 1 Basic Concepts in Geometry Practice Set 1.2 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 1 Basic Concepts in Geometry Practice Set 1.2. Step-by-step solved exercises, numerical problems, and digest answers.

6 Solved Questions• 868 words

Practice Set 1.2 Geometry 9th Std Maths Part 2 Answers Chapter 1 Basic Concepts in Geometry

Question 1 Maharashtra Board Solution
The following table shows points on a number line and their co-ordinates. Decide whether the pair of segments given below the table are congruent or not. i. seg DE and seg AB ii. seg BC and seg AD iii. seg BE and seg AD
Solution & Step-by-Step Answer:
i. Co-ordinate of the point E is 9. Co-ordinate of the point D is -7. Since, 9 > -7 ∴ d(D, E) = 9 – (-7) = 9 + 7 = 16 ∴ l(DE) = 16 …(i) Co-ordinate of the point A is -3. Co-ordinate of the point B is 5. Since, 5 > -3 ∴ d(A, B) = 5 – (-3) = 5 + 3 = 8 ∴ l(AB) = 8 …(ii) ∴ l(DE) ≠ l(AB) …[From (i) and (ii)] ∴ seg DE and seg AB are not congruent.

ii. Co-ordinate of the point B is 5.
Co-ordinate of the point C is 2.
Since, 5 > 2
∴ d(B, C) = 5 – 2 = 3
∴ l(BC) = 3 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = -3 + 7 = 4
∴ l(AD) = 4...(ii)
∴ l(BC) ≠ l(AD) … [From (i) and (ii)]
∴ seg BC and seg AD are not congruent.

iii. Co-ordinate of the point E is 9.
Co-ordinate of the point B is 5.
Since, 9 > 5
∴ d(B, E) = 9 – 5 = 4
∴ l(BE) = 4 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = 4
∴ l(AD) = 4 …(ii)
∴ l(BE) =l(AD) …[From (i) and (ii)]
∴ seg BE and seg AD are congruent.
i.e, seg BE ≅ seg AD

Question 2 Maharashtra Board Solution
Point M is the midpoint of seg AB. If AB = 8, then find the length of AM.
Solution & Step-by-Step Answer:
Point M is the midpoint of seg AB and l(AB) = 8. …[Given]
Question 3 Maharashtra Board Solution
Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).
Solution & Step-by-Step Answer:
Point P is the midpoint of seg CD and l(CP) = 2.5 …[Given] ∴ l(CD) = 2.5 x 2 ∴ l(CD) = 5
Question 4 Maharashtra Board Solution
If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.
Solution & Step-by-Step Answer:
Given, l(AB) = 5 cm, l(BP) = 2 cm, l(AP) = 3.4 cm … [Given] r Since, 2 < 3.4 < 5 ∴ l(BP) < l(AP) < l(AB) i.e., seg BP < seg AP < seg AB
Question 5 Maharashtra Board Solution
Write the answers to the following questions with reference to the figure given below: i. Write the name of the opposite ray of ray RP ii. Write the intersection set of ray PQ and ray RP. iii. Write the union set of ray PQ and ray QR. iv. State the rays of which seg QR is a subset. v. Write the pair of opposite rays with common end point R. vi. Write any two rays with common end point S. vii. Write the intersection set of ray SP and ray ST.
Solution & Step-by-Step Answer:
i. Ray RS or ray RT ii. Ray PQ iii. Line QR iv. Ray QR, ray QS, ray QT, ray RQ, ray SQ, ray TQ v. Ray RP and ray RS, ray RQ and ray RT vi. Ray ST, ray SR vii. Point S
Question 6 Maharashtra Board Solution
Answer the questions with the help of figure given below. i. State the points which are equidistant from point B. ii. Write a pair of points equidistant from point iii. Find d(U,V), d(P,C), d(V,B), d(U, L).
Solution & Step-by-Step Answer:
i. Points equidistant from point B are a. A and C, because d(B, A) = d(B, C) = 2 b. D and P, because d(B, D) = d(B, P) = 4 ii. Points equidistant from point Q are a. L and U, because d(Q, L) = d(Q, U) = 1 b. P and R, because d(P, Q) = d(Q, R) = 2 iii. a. Co-ordinate of the point U is -5. Co-ordinate of the point V is 5. Since, 5 > -5 ∴ d(U, V) = 5 – (-5) = 5 + 5 ∴ d(U, V) = 10

b. Co-ordinate of the point P is -2.
Co-ordinate of the point C is 4.
Since, 4 > -2
∴ d(P, C) = 4 – (-2)
= 4 + 2
∴ d(P, C) = 6

c. Co-ordinate of the point V is 5.
Co-ordinate of the point B is 2.
Since, 5 > 2
∴ d(V, B) = 5 – 2
∴ d(V, B) = 3

d. Co-ordinate of the point U is -5.
Co-ordinate of the point L is -3.
Since, -3 > -5
∴ d(U, L) = -3 – (-5)
= -3 + 5
∴ d(U, L) = 2

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