RD Sharma class 9 ex 9.2 question no. 4 iv
Answers
Answer:
Question 4: Compute the value of x in each of the following figures:
(i)
RD Sharma Class 9 Maths chapter 9 ex 9.2 question 4 part 1
Solution:
∠BAC = 1800 – 1200 = 600 [Linear pair]
∠ACB = 1800 – 1120 = 680 [Linear pair]
Sum of all angles of a triangle = 1800
x = 1800 − ∠BAC − ∠ACB
= 1800 − 600 − 680 = 520
Answer: x = 520
(ii)
RD Sharma Class 9 Maths chapter 9 ex 9.2 question 4 part 2
Solution:
∠ABC = 1800 – 1200 = 600 [Linear pair]
∠ACB = 1800 – 1100 = 700 [Linear pair]
Sum of all angles of a triangle = 1800
x = ∠BAC = 1800 − ∠ABC − ∠ACB
= 1800 – 600 – 700 = 500
Answer: x = 500
(iii)
RD Sharma Class 9 Maths chapter 9 ex 9.2 question 4 part 3
Solution:
∠BAE = ∠EDC = 520 [Alternate angles]
Sum of all angles of a triangle = 1800
x = 1800 – 400 – 520 = 1800 − 920 = 880
Answer: x = 880
(iv)
RD Sharma Class 9 Maths chapter 9 ex 9.2 question 4 part 4
Solution:
CD is produced to meet AB at E.
RD Sharma Class 9 Maths chapter 9 ex 9.2 question 4 part 2 solution
∠BEC = 1800 – 450 – 500 = 850 [Sum of all angles of a triangle = 1800]
∠AEC = 1800 – 850 = 950 [Linear Pair]
Now, x = 950 + 350 = 1300 [Exterior angle Property]
Answer: x = 1300
Question 5: In figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.
RD Sharma Class 9 Maths chapter 9 ex 9.2 question 5
Solution:
Let ∠BAD = y, ∠BAC = 3y
∠BDA = ∠BAD = y (As AB=DB)
Now,
∠BAD + ∠BAC + 1080 = 1800 [Linear Pair]
y + 3y + 1080 = 1800
4y = 720
or y = 180
Now, In ΔADC
∠ADC + ∠ACD = 1080 [Exterior Angle Property]
x + 180 = 1800
x = 900
Answer: x = 900