Std 9 Mathematics Part 2 · Chapter 1

1. Basic Concepts in Geometry

Solutions to Practice Sets 1.1 to 1.3 and Problem Set 1, with figures. Maharashtra State Board (Balbharati), Std 9 Mathematics Part 2 (Geometry).

Practice Set 1.1

1Find the distances with the help of the number line given below (Fig. 1.5).

Fig. 1.5 — co-ordinates: Q = −5, P = −4, K = −3, J = −2, H = −1, O = 0, A = 1, B = 2, C = 3, D = 4, E = 5

Subtract the smaller co-ordinate from the larger one.

d(B, E) = 5 − 2 = 3
d(J, A) = 1 − (−2) = 1 + 2 = 3
d(P, C) = 3 − (−4) = 3 + 4 = 7
d(J, H) = −1 − (−2) = −1 + 2 = 1
d(K, O) = 0 − (−3) = 3
d(O, E) = 5 − 0 = 5
d(P, J) = −2 − (−4) = −2 + 4 = 2
d(Q, B) = 2 − (−5) = 2 + 5 = 7

2If the co-ordinate of A is x and that of B is y, find d(A, B).

x = 1, y = 7 : 7 > 1, so d(A, B) = 7 − 1 = 6
x = 6, y = −2 : 6 > −2, so d(A, B) = 6 − (−2) = 6 + 2 = 8
x = −3, y = 7 : 7 > −3, so d(A, B) = 7 − (−3) = 7 + 3 = 10
x = −4, y = −5 : −4 > −5, so d(A, B) = −4 − (−5) = −4 + 5 = 1
x = −3, y = −6 : −3 > −6, so d(A, B) = −3 − (−6) = −3 + 6 = 3
x = 4, y = −8 : 4 > −8, so d(A, B) = 4 − (−8) = 4 + 8 = 12

3From the information given below, find which of the points is between the other two. If the points are not collinear, state so.

For three points, check whether the largest distance equals the sum of the other two. If it does, the point common to the two smaller distances lies between the other two. If it does not, the points are not collinear.

(i) d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, R) + d(Q, R) = 7 + 3 = 10 = d(P, Q)
∴ Point R is between P and Q (P − R − Q).

(ii) d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
The largest distance is d(R, S) = 8, but d(S, T) + d(R, T) = 6 + 4 = 10 ≠ 8
∴ Points R, S, T are not collinear.

(iii) d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(C, A) + d(B, C) = 9 + 7 = 16 = d(A, B)
∴ Point C is between A and B (A − C − B).

(iv) d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
The largest distance is d(M, N) = 12, but d(L, M) + d(N, L) = 11 + 8 = 19 ≠ 12
∴ Points L, M, N are not collinear.

(v) d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Z) + d(Y, Z) = 8 + 7 = 15 = d(X, Y)
∴ Point Z is between X and Y (X − Z − Y).

(vi) d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
The largest distance is d(E, F) = 8, but d(D, E) + d(D, F) = 5 + 6 = 11 ≠ 8
∴ Points D, E, F are not collinear.

4On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.

d(A, C) = 10 and d(C, B) = 8. Since A, B, C are on a number line, one of them lies between the other two.

Case 1 : A − C − B (C is between A and B)
d(A, B) = d(A, C) + d(C, B) = 10 + 8 = 18

Case 2 : A − B − C (B is between A and C)
d(A, B) + d(B, C) = d(A, C)
d(A, B) + 8 = 10 ∴ d(A, B) = 2

Case 3 : C − A − B (A is between C and B)
Then d(C, A) + d(A, B) = d(C, B), that is 10 + d(A, B) = 8, which gives d(A, B) = −2. A distance cannot be negative, so this case is not possible.

Case 1 : A − C − B

Case 2 : A − B − C

∴ d(A, B) = 18 or d(A, B) = 2

5Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8. Find d(X, Z).

d(X, Y) = 17 and d(Y, Z) = 8.

Case 1 : X − Y − Z
d(X, Z) = d(X, Y) + d(Y, Z) = 17 + 8 = 25

Case 2 : X − Z − Y
d(X, Z) + d(Z, Y) = d(X, Y)
d(X, Z) + 8 = 17 ∴ d(X, Z) = 9

Case 3 : Y − X − Z
Then d(Y, X) + d(X, Z) = d(Y, Z), that is 17 + d(X, Z) = 8, which is not possible because 17 > 8.

Case 1 : X − Y − Z

Case 2 : X − Z − Y

∴ d(X, Z) = 25 or d(X, Z) = 9

6Sketch proper figure and write the answers of the following questions.

(i) A − B − C, l(AC) = 11, l(BC) = 6.5

Since A − B − C, l(AB) + l(BC) = l(AC)
l(AB) + 6.5 = 11
l(AB) = 11 − 6.5 = 4.5

(ii) R − S − T, l(ST) = 3.7, l(RS) = 2.5

Since R − S − T, l(RT) = l(RS) + l(ST) = 2.5 + 3.7 = 6.2

(iii) X − Y − Z, l(XZ) = 3√7, l(XY) = √7

Since X − Y − Z, l(XY) + l(YZ) = l(XZ)
√7 + l(YZ) = 3√7
l(YZ) = 3√7 − √7 = 2√7

7Which figure is formed by three non-collinear points?

Three non-collinear points, when joined in pairs, give three segments which form a triangle.

Practice Set 1.2

1The following table shows points on a number line and their co-ordinates (A = −3, B = 5, C = 2, D = −7, E = 9). Decide whether the pairs of segments given below are congruent or not.

First find the length of each segment from the co-ordinates: A = −3, B = 5, C = 2, D = −7, E = 9.

(i) seg DE and seg AB
l(DE) = 9 − (−7) = 16; l(AB) = 5 − (−3) = 8
16 ≠ 8, so seg DE and seg AB are not congruent.

(ii) seg BC and seg AD
l(BC) = 5 − 2 = 3; l(AD) = −3 − (−7) = 4
3 ≠ 4, so seg BC and seg AD are not congruent.

(iii) seg BE and seg AD
l(BE) = 9 − 5 = 4; l(AD) = 4
l(BE) = l(AD), so seg BE ≅ seg AD

2Point M is the midpoint of seg AB. If AB = 8 then find the length of AM.

M is the midpoint of seg AB, so AM = MB = ½ AB.
AM = ½ × 8 = 4

3Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).

P is the midpoint of seg CD, so CP = PD = 2.5.
l(CD) = CP + PD = 2.5 + 2.5 = 5

4If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.

AB = 5 cm, AP = 3.4 cm and BP = 2 cm. Since 5 > 3.4 > 2,

seg AB > seg AP > seg BP (equivalently, seg BP < seg AP < seg AB).

Here AP + BP = 5.4 ≠ AB, so P is not a point of seg AB; the three points are not collinear. Segments are compared only by their lengths, so this does not affect the answer.

5Write the answers to the following questions with reference to figure 1.13.

Fig. 1.13 — the order of the points is T, S, R, P, Q

The opposite ray of ray RP is ray RS (the same ray can also be named ray RT).
Ray PQ lies entirely on ray RP, so ray PQ ∩ ray RP = ray PQ
seg PQ ∪ seg QR = seg PR
seg QR is a subset of ray RQ, ray SQ, ray TQ and ray QR. (Ray RQ is the same as ray RP; ray SQ is the same as ray SR or ray SP; ray QR is the same as ray QP, ray QS or ray QT.)
The pair of opposite rays with common end point R: ray RS and ray RP (or ray RT and ray RQ).
Two rays with common end point S: ray ST and ray SR (ray SP and ray SQ are also correct).
Ray SP and ray ST are opposite rays; their only common point is S. So ray SP ∩ ray ST = {S}, that is, point S.

6Answer the questions with the help of figure 1.14.

Fig. 1.14 — co-ordinates: R = −6, U = −5, Q = −4, L = −3, P = −2, A = 0, B = 2, C = 4, V = 5, D = 6

(i) Points equidistant from B (co-ordinate 2)
d(B, A) = 2 − 0 = 2 and d(B, C) = 4 − 2 = 2, so A and C are equidistant from B.
d(B, P) = 2 − (−2) = 4 and d(B, D) = 6 − 2 = 4, so P and D are also equidistant from B.

(ii) A pair of points equidistant from Q (co-ordinate −4)
d(Q, U) = −4 − (−5) = 1 and d(Q, L) = −3 − (−4) = 1, so U and L are equidistant from Q.
(Also, d(Q, R) = d(Q, P) = 2, so R and P form another such pair.)

(iii)

d(U, V) = 5 − (−5) = 5 + 5 = 10
d(P, C) = 4 − (−2) = 4 + 2 = 6
d(V, B) = 5 − 2 = 3
d(U, L) = −3 − (−5) = −3 + 5 = 2

Practice Set 1.3

1Write the following statements in ‘if-then’ form.

If a quadrilateral is a parallelogram, then its opposite angles are congruent.
If a quadrilateral is a rectangle, then its diagonals are congruent.
If a triangle is isosceles, then the segment joining its vertex and the midpoint of its base is perpendicular to the base.

2Write converses of the following statements.

The converse is obtained by interchanging the antecedent and the consequent.

(i) Conditional form: If two parallel lines are cut by a transversal, then the alternate angles formed are congruent.
Converse: If the alternate angles formed by two lines and their transversal are congruent, then the two lines are parallel.

(ii) Converse: If two lines are parallel, then the interior angles made by a transversal on the same side are supplementary.

(iii) Conditional form: If a quadrilateral is a rectangle, then its diagonals are congruent.
Converse: If the diagonals of a quadrilateral are congruent, then it is a rectangle.

The converses in (i) and (ii) are true. The converse in (iii) is not always true: an isosceles trapezium also has congruent diagonals but is not a rectangle. This shows that the converse of a true statement need not be true.

Problem Set 1

1Select the correct alternative from the answers of the questions given below.

(i) How many midpoints does a segment have? — (A) only one
Every segment has one and only one midpoint.
(ii) How many points are there in the intersection of two distinct lines? — (C) one
Two distinct lines can meet in at most one point. (If they are parallel they do not meet at all, but among the options, one is the answer for intersecting lines.)
(iii) How many lines are determined by three distinct points? — (C) one or three
One line if the points are collinear, three lines if they are non-collinear.
(iv) Find d(A, B), if co-ordinates of A and B are −2 and 5 respectively. — (C) 7
d(A, B) = 5 − (−2) = 7
(v) If P − Q − R and d(P, Q) = 2, d(P, R) = 10, then find d(Q, R). — (B) 8
d(Q, R) = d(P, R) − d(P, Q) = 10 − 2 = 8

2On 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.

Order on the number line: Q − P − R

Co-ordinates: P = 3, Q = −5, R = 6.
d(P, Q) = 3 − (−5) = 8; d(Q, R) = 6 − (−5) = 11; d(P, R) = 6 − 3 = 3

(i) d(P, Q) + d(Q, R) = d(P, R)
LHS = 8 + 11 = 19; RHS = 3. 19 ≠ 3, so the statement is False. (Q is not between P and R.)

(ii) d(P, R) + d(R, Q) = d(P, Q)
LHS = 3 + 11 = 14; RHS = 8. 14 ≠ 8, so the statement is False. (R is not between P and Q.)

(iii) d(R, P) + d(P, Q) = d(R, Q)
LHS = 3 + 8 = 11; RHS = 11. LHS = RHS, so the statement is True. (P is between R and Q.)

(iv) d(P, Q) − d(P, R) = d(Q, R)
LHS = 8 − 3 = 5; RHS = 11. 5 ≠ 11, so the statement is False.

3Co-ordinates of some pairs of points are given below. Hence find the distance between each pair.

3, 6 : 6 − 3 = 3
−9, −1 : −1 − (−9) = −1 + 9 = 8
−4, 5 : 5 − (−4) = 5 + 4 = 9
0, −2 : 0 − (−2) = 2
x + 3, x − 3 : x + 3 > x − 3, so the distance = (x + 3) − (x − 3) = x + 3 − x + 3 = 6
−25, −47 : −25 − (−47) = −25 + 47 = 22
80, −85 : 80 − (−85) = 80 + 85 = 165

4Co-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.

Let M and N be the points to the left and right of P respectively, each at a distance of 8 units.

Co-ordinate of M = −7 − 8 = −15
Co-ordinate of N = −7 + 8 = 1

∴ The required co-ordinates are −15 and 1.

Verification: d(P, M) = −7 − (−15) = 8 and d(P, N) = 1 − (−7) = 8.

5Answer the following questions.
(i) If A − B − C and d(A, C) = 17, d(B, C) = 6.5 then d(A, B) = ?
(ii) If P − Q − R and d(P, Q) = 3.4, d(Q, R) = 5.7 then d(P, R) = ?

(i) A − B − C, so d(A, B) + d(B, C) = d(A, C)
d(A, B) + 6.5 = 17
d(A, B) = 17 − 6.5 = 10.5

(ii) P − Q − R, so d(P, R) = d(P, Q) + d(Q, R)
d(P, R) = 3.4 + 5.7 = 9.1

6Co-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?

Let X and Y be the points to the left and right of A respectively, each 7 units away from A.

Co-ordinate of X = 1 − 7 = −6
Co-ordinate of Y = 1 + 7 = 8

∴ The required co-ordinates are −6 and 8.

7Write the following statements in conditional form.

If a quadrilateral is a rhombus, then it is a square.
If two angles are in a linear pair, then they are supplementary.
If a figure is a triangle, then it is formed by three segments.
If a number has only two divisors, then it is a prime number.

In (i) the statement itself is false: every square is a rhombus, but a rhombus is a square only when its angles are right angles. The question asks only to rewrite it in conditional form, which is done above.

8Write the converse of each of the following statements.

If a figure is a triangle, then the sum of the measures of its angles is 180°.
If two angles are complements of each other, then the sum of their measures is 90°.
If two lines are parallel, then the corresponding angles formed by a transversal of the lines are congruent.
If a number is divisible by 3, then the sum of its digits is divisible by 3.

9Write the antecedent (given part) and the consequent (part to be proved) in the following statements.

(i) If all sides of a triangle are congruent then its all angles are congruent.

Antecedent (given part): All sides of a triangle are congruent.
Consequent (part to be proved): All angles of the triangle are congruent.

(ii) Conditional form: If a quadrilateral is a parallelogram, then its diagonals bisect each other.

Antecedent (given part): The quadrilateral is a parallelogram.
Consequent (part to be proved): The diagonals of the quadrilateral bisect each other.

10*Draw a labelled figure showing information in each of the following statements and write the antecedent and the consequent.

(i) Two equilateral triangles are similar.

ΔABC and ΔPQR are equilateral triangles

Antecedent: ΔABC and ΔPQR are equilateral triangles.
Consequent: ΔABC ∼ ΔPQR

(ii) If angles in a linear pair are congruent then each of them is a right angle.

∠AOC and ∠BOC form a linear pair

Antecedent: ∠AOC and ∠BOC form a linear pair and ∠AOC ≅ ∠BOC.
Consequent: m∠AOC = m∠BOC = 90°, that is, each is a right angle.

(iii) If the altitudes drawn on two sides of a triangle are congruent then those two sides are congruent.

In ΔABC, seg BE ⊥ side AC and seg CF ⊥ side AB

Antecedent: In ΔABC, seg BE ⊥ side AC, seg CF ⊥ side AB and seg BE ≅ seg CF.
Consequent: side AC ≅ side AB

2. Parallel Lines →