Lines And Angles – Maharashtra Board Class 9 Solutions for Geometry
AlgebraGeometryScience and TechnologyHindi
Exercise – 1.1
Solution 1:
Solution 2:
i.
Six lines can be drawn through four given points such that no three points are collinear.
The six lines are line PQ, line QR, line RS, line SP, line PR, and line QS.
ii.
Let Q, R and S be the three collinear points.
Four lines can be drawn through given four points such that three points are collinear.
The four lines are line PQ, line PS, line PR, and line QS.
Solution 3:
The sets of collinear points are:
1. P, F, R, B
2. P, S, T, Q
3. A, R, E, Q
4. A, F, S, D
5. B, E, T, D
Solution 4:
i. Line PQ, line SR, and line DC are parallel to line AB.
ii. Yes, line AD and point R lie in the same plane.
[There is exactly one plane passing through a line and a point, not on the line (axiom)].
iii. Yes, points A, S, R, and B are coplanar (since these points lie in the same plane ASRB).
iv. Plane APSD, plane APQB and plane ABCD pass through point A.
v. Points A, S, R, B, and V.
Exercise – 1.2
Solution 1:
i. Co-ordinates of points C, S, Q, and D are -3, 4, 2 and -4 respectively.
ii. The points whose co-ordinates are 4, 5, 0, and -2 are S, T, O, and B respectively.
iii. d(Q, T)
Co-ordinate of Q is 2 and co-ordinate of T is 5, 2 < 5
∴d(Q, T) = 5 – 2 = 3
d(E, B)
Co-ordinate of E is -5 and co-ordinate of B is -2, -5 < -2
∴d(E, B) = -2 – (-5) = -2 + 5 = 3
d(O, C)
Co-ordinate of O is 0 and co-ordinate of C is -3, -3 < O
∴d(O, C) = 0 – (-3) = 3
d(O, R)
Co-ordinate of O is 0 and co-ordinate of R is 3, 0 < 3
∴d(O, R) = 3 – 0 = 3
iv. There are two cases :
a. The point can be towards the positive side i.e. point S (∵ d(O, S) = 4 – 0 = 4)
b. The point can be towards the negative side i.e. point D (∵d(O, D) = 0 – (-4) = 4)
Solution 2:
i. Co-ordinate of point P is x = 7
Co-ordinate of point Q is y = 10
10 > 7
∴d(P, Q) = 10 – 7 = 3
∴d(P, Q) = 3
ii. Co-ordinate of point P is x = -2
Co-ordinate of point Q is y = 11
11 > -2
∴d(P, Q) = 11 – (-2) = 11 + 2 = 13
∴d(P, Q) = 13
iii. Co-ordinate of point P is x = -8
Co-ordinate of point Q is y = -3
-3 > -8
∴d(P, Q) = -3 -(-8 ) = -3 + 8 = 5
∴d(P, Q) = 5
iv. Co-ordinate of point P is x = 5
Co-ordinate of point Q is y = -9
5 > -9
∴d(P, Q) = 5-(-9) = 5 + 9 = 14
∴d(P, Q) = 14
Solution 3:
If points P, Q and R are three distinct collinear points and if d(P, Q) + d(Q, R) = d(P, R), then the point Q is said to be between the points P and R. When point Q is between the points P and R, we write P – Q – R to represent the betweenness among P, Q and R.
i. d(A, B) + d(B, D) = 5 + 8 = 13
d(A, D) = 11
∴d(A, B) + d(B, D) ≠d(A,D)
∴There is no betweenness among the points A, B and D
ii. d(B, D) + d(A, D) = 6 + 5 = 11
d(A, B) = 11
∴d(B, D) + d(A, D) = d(A, B)
∴There exists a betweeness among the points A, B, and D.
The point D lies between A and B.
Hence we write, A – D – B.
iii. d(A, B) + d(B, D) = 2 + 15 = 17
d(A, D) = 17
∴d(A, B) + d(B, D) = d(A, D)
∴There exists a betweenness among the points A, B and D.
The point B lies between A and D.
Hence we write, A – B – D.
Solution 4:
Solution 5:
l(PL) + l(LN) = l(PN) (P-L-N)
∴ l(PL) + 5 = 11
∴ l(PL) =11 – 5
∴ l(PL) = 6 units
l(MN) + l(NR) = l(MR) (M-N-R)
∴7 + l(NR) = 13
∴ l(NR) = 13 – 7
∴ l(NR) = 6 units
l(LM) + l(MQ) = l(LQ) (L-M-Q)
∴6 + 2 = l(LQ)
∴ l(LQ) = 8 units
Solution 6:
l(AB) + l(BC) = l(AC) (A-B-C)
∴ l(AB) + 5 = 8
∴ l(AB) = 8 – 5
∴ l(AB) = 3 units … (i)
seg AC ≅ seg BD (given)
l(BD) = 8
l(BC) + l(CD) = l(BD) (B-C-D)
∴5 + l(CD) = 8
∴ l(CD) = 8 – 5
∴ l(CD) = 3 units … (ii)
seg BD ≅ seg CE (given)
l(CE) = 8
l(CD) + l(DE) = l(CE) (C-D-E)
∴3 + l(DE) = 8
∴ l(DE) = 8 – 3
∴ l(DE) = 5 units … (iii)
l(BC) = l(DE) = 5 units [from (iii) and given that l(BC) = 5]
∴seg BC ≅ seg DE
l(AB) = l(CD) = 3 units [from(i) and (ii)]
∴seg AB ≅ seg CD
Solution 7:
Co-ordinate of point P is -3
Co-ordinate of point Q is 5
5 > -3
∴d(P, Q) = 5 – (-3 ) = 5 + 3 = 8 units
∴ l(PQ) = 8 units
Co-ordinate of point P is -3
Co-ordinate of point R is 2
2 > -3
∴d(P, R) = 2 – (-3 ) = 2 + 3 = 5 units
∴ l(PR) = 5 units
Co-ordinate of point P is -3
Co-ordinate of point S is -7
-3 > -7
∴d(P, S) = -3 – (-7 ) = -3 + 7 = 4 units
∴ l(PS) = 4 units
Co-ordinate of point P is -3
Co-ordinate of point T is 9
9 > -3
∴d(P, T) = 9 – (-3 ) = 9 + 3 = 12 units
∴ l(PT) = 12 units
Co-ordinate of point Q is 5
Co-ordinate of point R is 2
5 > 2
∴d(Q, R) = 5 – 2= 3 units
∴ l(QR) = 3 units
Co-ordinate of point Q is 5
Co-ordinate of point S is -7
5 > -7
∴d(Q, S) = 5 – (-7 ) = 5 + 7 = 12 units
∴ l(QS) = 12 units
Co-ordinate of point Q is 5
Co-ordinate of point T is 9
9 > 5
∴d(Q, T) = 9 – (5 ) = 4 units
∴ l(QT) = 4 units
Co-ordinate of point R is 2
Co-ordinate of point S is -7
2 > -7
∴d(R, S) = 2 – (-7 ) = 2 + 7 = 9 units
∴ l(RS) = 9 units
Co-ordinate of point R is 2
Co-ordinate of point T is 9
9 > 2
∴d(RT) = 9 – 2 = 7 units
∴ l(RT) = 7 units
Co-ordinate of point S is -7
Co-ordinate of point T is 9
9 > -7
∴d(S, T) = 9 – (-7 ) = 9 + 7 = 16 units
∴ l(ST) = 16 units
Solution 8:
Solution 9:
Q is the midpoint of CD
∴ l(CD) = 2 l(CQ)
∴ l(CD) = 2×4.5
∴ l(CD) = 9 cm
Solution 10:
7 > 5.4 > 4
∴ l(AB) > l(AP) > l(BP)
∴seg(AB) > seg(AP) > seg(BP)
∴ AB > AP >BP
Solution 11:
l(AB) = l(AC) = 5 cm
∴seg AB ≅ seg AC
l(BC) = l(DE) = 5.5 cm
∴seg BC ≅ seg DE
l(CD) = l(CE) = 4 cm
∴seg CD ≅ seg CE
Exercise – 1.3
Solution 1:
- Yes, two acute angles measuring 30° and 60° have their sum 90°.
- No, because the sum of the measures of two obtuse angles cannot be 90°.
- No, because the sum of the measures of two right angles is 180°
- No, because the sum of the measures of two acute angles is always less than 180°
- No, because the sum of the measures of two obtuse angles is always greater than 180°.
- Yes, because the sum of the measures of two right angles is 180°
- Yes, because a linear pair of angles is adjacent as well as supplementary.
- Yes, if the sum of the measures of adjacent angles adds to 90°
- Yes, AOB and COB are obtuse and adjacent angles.
Solution 2(i):
Measure of the given angle = 60°
Measure of its supplementary angle = 180° – 60° = 120°
Solution 2(ii):
Measure of the given angle = 138°
Measure of its supplementary angle = 180° – 138° = 42°
Solution 2(iii):
Solution 2(iv):
Measure of the given angle = (180 – r)°
Measure of its supplementary angle = 180°- (180 – r)°= r°
Solution 2(v):
Measure of the given angle = (90 + r)°
Measure of its supplementary angle = 180° – (90 + r)°
= 180° – 90° – r°
= (90 – r)°
Solution 2(vi):
Measure of the given angle = 87°
Measure of its supplementary angle = 180° – 87° = 93°
Solution 2(vii):
Measure of the given angle = 124°
Measure of its supplementary angle = 180° – 124° = 56°
Solution 2(viii):
Measure of the given angle = 108°
Measure of its supplementary angle = 180° – 108° = 72°
Solution 3(i):
Measure of the given angle = 58°
Measure of its complementary angle = 90° – 58° = 32°
Solution 3(ii):
Measure of the given angle = 16°
Measure of its complementary angle = 90° – 16° = 74°
Solution 3(iii):
Solution 3(iv):
Measure of the given angle = (a + b)°
Measure of its complementary angle = 90° – (a + b)°= (90 – a – b)°
Solution 3(v):
Measure of the given angle = (90 – r )°
Measure of its complementary angle = 90° – (90 – r)°
= 90° – 90° + r°
= r°
Measure of the given angle = 78°
Measure of its complementary angle = 90° – 78° = 12°
Solution 3(vii):
Measure of the given angle = 68°
Measure of its complementary angle = 90° – 68° = 22°
Solution 3(viii):
Measure of the given angle = 56°
Measure of its complementary angle = 90° – 56° = 34°
Solution 4:
Solution 5:
Exercise – 1.4
Solution 1:
- Alternate angles are congruent. (Converse of alternates angle test)
- All three lines are parallel to each other. (If two lines are parallel to the same line then they are parallel to each other.)
- Both angles are congruent.
- One and only one such line can be drawn.
- Let the interior angles formed be 2x and 7x.
The converse of interior angles test:
If two lines are parallel then the interior angles formed by a transversal are supplementary.
∴2x + 7x = 180°
∴9x = 180°
∴x = 20°
The measure of the greater angle = 7x
= 7 × 20°
= 140°
Solution 2:
Solution 3:
Solution 4:
