GSEB Solutions for Class 9 Mathematics – Some Primary Concepts in Geometry : 1

GSEB Solutions for Class 9 Mathematics – Some Primary Concepts in Geometry : 1 (English Medium)

Exercise 7:

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5(1):

Solution 5(2):

Solution 5(3):

Solution 5(4):

Solution 5(5):

Solution 6:

Solution 7:

Solution 8(1):

Solution 8(2):

Solution 8(3):

Solution 8(4):

Solution 8(5):

Solution 8(6):

Solution 8(7):

Solution 8(8):

Solution 8(9):

b. 2
Every line has at least 2 distinct points.

Exercise 7.1:

Solution 1:

Solution 2:

Exercise 7.2:

Solution 1:

Solution 2:

Solution 3:

Exercise 7.3:

Solution 1:

1. Distance postulate: With each pair of points, there corresponds one and only one non-negative real number called the distance between these points.
2. X-Y-Z symbolises that the distinct points X, Y and Z are collinear and point Y lies between the points X and Z.
   Also if x, y and z are the real numbers corresponding to X, Y and Z respectively, then x > y > z or x
   < y < z.
   Also, X-Y-Z gives XY + YZ = XZ.
3. Ruler Postulate:
   1. Corresponding to each point on a line, there is one and only one real number.
   2. Corresponding to each real number, there is one and only one point on the line.
   3. There is a one-to-one correspondence between points on a line and real numbers such that for every pair of distinct points on the line, the positive difference of corresponding real numbers is the distance between them.
4. Condition of ‘betweeness‘: If p, q, and r are real numbers corresponding to the points P, Q and R respectively, and p > q > r or p < q < r then Q is between P and R. So, we can write it as P-Q-R or R-Q-P.

Solution 2:

Let X ↔  x, Y ↔ y and Z ↔ z.
Then, x = 6, y = -3 and z = -1.
XY = |x – y| = |6 – (-3)| = |9| = 9
YZ = |y – z| = |(-3) – (-1) | = |-2| = 2
ZX = |z – x| = |(-1) – 6 | = |-7| = 7

Solution 3:

Let P ↔ p, Q ↔ q and R ↔ r.
Then, p = 7, q = -3 and r = 3.
Now, we know, -3 < 3 < 7.
∴ q < r < p
Hence, point R lies between the points P and Q,
i.e. Q-R-P or P-R-Q.

OR

Le P ↔ p, Q ↔ q and R ↔ r.
Then, p = 7, q = -3 and r = 3.
PQ = |p – q| = |7 – (-3)| = |10| = 10
QR = |q – r| = |(-3) – 3| = |-6| = 6
RP = |r – p| = |3 – 7 | = |-4| = 4
Thus, we have PQ = QR + RP
Hence, point R lies between the points P and Q,
i.e. Q-R-P or P-R-Q.

Solution 4:

Let A ↔ a and B ↔ b.
Then, a = -3 and AB = 5
AB = |a – b|
∴ 5 = |-3 – b|
∴ -3 – b = 5 or -3 – b = -5
∴ -b = 5 + 3 or -b = -5 + 3
∴ -b = 8 or -b = -2
∴ b = -8 or b = 2
Thus, the number corresponding to B is -8 or 2.

Solution 5:

Given: A-B-C, BC = 3 and AC = 9
∴ AB + BC = AC
∴ AB + 3 = 9
∴ AB = 9 – 3
∴ AB = 6

Solution 6:

Solution 7:

Exercise 7.4:

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5: