Chapter 13: Geometrical Constructions (Exercise 13.1)

Chapter 13: Geometrical Constructions (Exercise 13.1)

Question 1:
Construct an angle of 90° at the initial point of a given ray AB and justify the construction.

Steps of Construction:
1. Draw a ray AB with initial point A.
2. With A as center and any convenient radius, draw a main arc intersecting AB at point P.
3. With P as center and the same radius, draw an arc intersecting the main arc at Q (60°).
4. With Q as center and the same radius, draw another arc intersecting the main arc at R (120°).
5. With Q and R as centers, draw two arcs intersecting each other at point C.
6. Join AC. The ray AC forms the required angle ∠CAB = 90°.

Question 2:
Construct an angle of 45° at the initial point of a given ray.

Steps of Construction:
1. Construct a perpendicular line at initial point A to get ∠CAB = 90°.
2. With the point of intersection of the 90° line on the main arc and point P (on ray AB) as centers, draw two arcs intersecting each other at point D.
3. Join AD. Ray AD is the angle bisector of 90°.
4. Thus, ∠DAB = 45°.

Question 3:
Construct a triangle ABC in which BC = 7 cm, ∠B = 75°, and AB + AC = 13 cm.

Solution:
1. Draw the base segment BC = 7 cm.
2. At point B, construct an angle of 75° using a compass.
3. Cut an arc of length BD = 13 cm (equal to AB + AC) on the ray forming 75°.
4. Join DC.
5. Draw the perpendicular bisector of line segment DC intersecting BD at point A.
6. Join AC to complete ΔABC.

Chapter 14: Probability (Exercise 14.1)

Question 1:
In a cricket match, a female batter hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Solution:
* Total number of balls played = 30
* Number of times boundary was hit = 6
* Number of times boundary was NOT hit = 30 - 6 = 24

Formula: P(Event) = (Number of favorable outcomes) / (Total number of trials)

P(Not hitting a boundary) = 24 / 30 = 4 / 5 (or 0.8)

Question 2:
1500 families with 2 children were selected randomly, and the following data were recorded:
* No. of girls in a family: 2 | 1 | 0
* No. of families: 475 | 814 | 211

Compute the probability of a family chosen at random having:
1. 2 girls
2. 1 girl
3. No girls

Solution:
Total number of families = 475 + 814 + 211 = 1500

1. Probability of family having 2 girls:
P(2 girls) = 475 / 1500 = 19 / 60

2. Probability of family having 1 girl:
P(1 girl) = 814 / 1500 = 407 / 750

3. Probability of family having no girls:
P(0 girls) = 211 / 1500

* Verification (Sum of probabilities):
(475 + 814 + 211) / 1500 = 1500 / 1500 = 1.

Question 3:
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
* Outcome: 3 heads | 2 heads | 1 head | No head
* Frequency: 23 | 72 | 77 | 28

If the three coins are simultaneously tossed again, compute the probability of getting 2 heads.

Solution:
* Total number of tosses = 200
* Frequency of getting 2 heads = 72

P(Getting 2 heads) = 72 / 200 = 9 / 25 (or 0.36)