Math, asked by mishrabijayakumar, 10 months ago

RD Sharma class 9 ex 9.2 question no. 4 iv

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Answered by pandeyvandana608
1

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

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