9th Maths 6.1
NCERT Class 9th solution of Exercise 6.2
NCERT Class 9th solution of Exercise 6.3
Exercise 6.1
Q1. In the given figure lines `AB` and `CD` intersect at `O`, if `∠AOC + ∠BOE = 70°` and `∠BOD = 40°` then find the `∠BOE` and reflex `∠COE`.
`∠AOC=∠BOD`_______(1) [Vertically opp. angles]
`∠AOC=40°` ________(2) [ Given `∠BOD=40°`]
`∠AOC+∠BOE=70°`____(3) [Given]
`40°+∠BOE=70°` [ from eq. (2) ]
`∠BOE=70°-40°`
`∠BOE=30°`__________(4)
Now
`∠AOC+∠COE+∠BOE=180°` [ By Linear pair]
`40°+∠COE+30°=180°` [ from eq. (2) & (4) ]
`∠COE=180°-70°`
`∠COE=110°`
Reflex`∠COE=360°-∠COE`
Reflex`∠COE=360°-110°`
Reflex`∠COE=250°`
Answer :
The required `∠BOE=30°` and reflex`∠COE=250°`.
Q2. In the figure lines, `XY` and `MN` intersect at `O`. If `∠POY=90°` and `a:b=2:3`, find `c`.
`∠XOP+∠POY=180°` [ By linear pair ]
`(∠XOM+∠MOP)+90°=180°` [ Given `∠POY=90°`]
`(a+b)=180°-90°`
`a+b=90°`
Sum of the ratio
`a+b=2+3`
`a+b=5`
`a=frac(2)(5)times90°`
`a=36°`
`b=frac(3)(5)times90°`
`b=54°`_____________(1)
`c+b=180°` [ by linear pair ]
`c+54=180°` [ from eq. (1) ]
`c=180°-54°`
`c=126°`
Answer :
The required value of `c=126°`.
Q3. In the figure,`∠PQR=∠PRQ`, then prove that `∠PQS=∠PRT`.
`∠PQS+∠PQR=180°` [ By linear pair ]
`∠PQR=180°-∠PQS`________(1)
`∠PRQ+∠PRT=180°` [ By linear pair ]
`∠PRQ=180°-∠PRT`________(2)
`∠PQR=∠PRQ [ Given ]
`180°-∠PQS=∠PRT-180°` [ From eq. (1) & (2) ]
`∠PQS=∠PRT`
Proved.
Q4. In the figure, if `x+y=w+z`, then prove that: `AOB` is a line.
Sol.
`x+y+w+z=360°`____(1) [By angle sum property of a point]
`x+y=w+z` ____(2) [ Given ]
`x+y+x+y=360°`
`2x+2y=360°`
`2(x+y)=360°`
`x+y=frac(360°)(2)`
`x+y=180°`
`∠AOB=180°`
`AOB` is a line.
Proved.
Q5. In the figure, `POQ` is a straight line. Ray `OR` is perpendicular to line `PQ`. `OS` is another ray lying between ray `OP`. Prove that:
`∠ROS=frac(1)(2)(∠QOS-∠POS)`.
Sol.
`∠POR+∠ROQ=180°` [ By linear pair ]
`∠POR+90°=180°` [ Given `OR⊥PQ` ]
`∠POR=90°`
So
`∠POR=∠ROQ`_________(1)
`∠POR=∠POS+∠ROS`___(2)
`∠ROQ=∠QOS-∠ROS`___(3)
`∠POS+∠ROS=∠QOS-∠ROS`[ From eq. (1),(2) & (3) ]
`∠ROS+∠ROS=∠QOS-∠POS`
`2∠RPS=∠QOS-∠POS`
`∠ROS=frac(1)(2)(∠QOS-∠POS)`
Proved.
Q6. It is given that `∠XYZ=64°` and `XY` is produced to point `P`. Draw a figure from the given information. If ray `∠YO` bisects `∠ZYP`, find `∠XYQ` and reflex `∠QYP`.
Sol.
`∠ZYP+∠XYZ=180°` [ By linear pair ]
`∠ZYP+64°=180°` [ Given `∠XYZ=64°`]
`∠ZYP=180°-64°`
`∠ZYP=116°`
`∠ZYQ+∠QYP=116°`
`∠ZYQ+∠ZYQ=116°` [Given Ray `YQ` bisect `∠ZYP` ]
`2∠ZYQ=116°`
`∠ZYQ=frac(116°)(2)`
`∠ZYQ=58°=∠QYP`
`∠XYQ=∠XYZ+∠ZYQ`
`∠XYQ=64°+58°` [ `∠XYZ=64°`]
`∠XYQ=122°`
Reflex`∠QYP=360°-∠QYP`
Reflex`∠QYP=360°-58°`
Reflex`∠QYP=302°`
Answer :
The required `∠XYQ=122°` and Reflex `∠QYP=302°`.
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