9th Maths 6.2
NCERT Class 9th solution of Exercise 6.1
NCERT Class 9th solution of Exercise 6.3
Exercise 6.2
Q1. In the given figure, find the values of `x` and `y` and then show that `AB∥CD`.
According to the given figure
`x+50=180°` [ By linear pair ]
`x=180-50°`
`x=130°`____(1)
`y=130°`____(2) [ Vertically opp. angles ]
`x=y=130°` [ By eq. (1) & (2) ]
`AB∥CD`. [ Alternate angles are equal ]
Proved.
Sol. :
Q2. In the given figure, if `AB∥CD, CD∥EF` and `y:z=3:7`,find `x`.
Let ST is transversal line intersects `AB, CD,` & `EF` at `P, Q,` & `R` respectively,
According to the given figure
`∠DQR=∠PQC=y` [ Vertically opp. angles and given `∠PQC=y`]
`∠DQR=y`
`∠DQR+∠QRF=180°` [ linear pair and given `∠QRF=z` ]
`y+z=180°` [ Sum of interior angles on the same side ]
The Sum of the ratio of `y` & `z` is `3+7=10`
`y=frac(3)(10)times180°`
`y=54°`
`z=frac(7)(10)times180°`
`x=z=126°` [Alternate angles are equal]
Answer :
The required value of `y=54°` & `z=x=126°`.
Q3. In the given figure, if `AB∥CD`, `EF⊥CD` and `∠GED=126°` find `∠AGE, ∠GEF` and `∠FGE`.
Sol. :
According to the given figure
`GE` is transversal of `AB∥CD`
`∠AGE=∠GED` [ Alternate angle ]
`∠AGE=126°` [ Given `∠GED=126°` ]
`∠GEF+∠FED=∠GED`
`∠GEF+90°=126°` [ Given `EF⊥CD` ]
`∠GEF=126°-90°`
`∠GEF=36°`
`∠FGE+∠GED=180°`
`∠FGE+126°=180°`
`∠FGE=180°-126°`
`∠FGE=54°`
Answer :
The required values of `∠AGE=126°, ∠GEF=36°, and ∠FGE=54°`
Q4. In the given figure, if `PQ∥ST`, `∠PQR=110°` and `∠RST=130°`, find `∠QRS`.
Sol. :
According to the given figure `PQ∥ST`
Construction :
Draw `RD∥ST` through `R`.
Proof :
`ST∥RD`and `SR` is transversal.
`∠SRX+∠RST=180°` [ Sum of interior angles of the same side.]
`∠SRX+130°=180°` [ Given `RST=130°` ]
`∠SRX=180°-130°`
`∠SRX=50°`_________(1)
`PQ∥RX` and `QR` is transvesal.
`∠QRX=∠PQR` [ Alternate angles and given `∠PQR=110°` ]
`∠QRX=110°`
`∠QRS+SRX=110°`
`∠QRS+50°=110°` [ from eq. (1) ]
`∠QRS=110°-50°`
`∠QRS=60°`
Answer :
The required `∠QRS=60°`.
Q5. In the given figure, if `AB∥CD, ∠APQ=50°` and `∠PRD=127°`, find `x` and `y`.
Sol. :
According to the given figure
`AB∥CD` and `PQ` is transversal
`∠PQR=∠APQ` [ Alternate angles ]
`x=50°` [Given `PQR=x` and `APQ=50°`]
`∠PQR+∠QPR=∠PRD` [Exterior angle property ]
`50°+y=127°` [Given `∠PRD=127°` ]
`y=127°-50°`
`y=77°`
Answer :
The required values are `x=50° & `y=77°`.
Q6. In the given figure, `PQ` and `RS` are two mirrors placed parallel to each other. An incident ray `AB` strikes the mirror `PQ` at `B`. The reflected ray moves along the path `BC` and strikes the mirror `RS` at `C` and again reflected back along `CD`. Prove that `AB∥CD`.
Sol. :
`AB` is incident ray strikes at point `B` of mirror `PQ` and reflected along `BC`.
`∠PBA=∠QBC`_____(1) [ By the law of reflection ]
`BC` is incident ray strikes at point `C` of mirror `RS` and reflected along `CD`
`∠BCR=∠DCS`_____(2) [ By the law of reflection ]
`PQ∥RS` and transversal `BC` intersects, then at the point `B` and `C`
`∠QBC=∠BCR`_____(3) [ Alternate angles ]
`∠PBA+∠ABC+∠QBC=∠BCR+∠BCD+∠DCS=180°`___(4)
[Sum of the angles at a point of line ]
`∠PBA=∠QBC=∠BCR=∠DCS` [ from eq. (1), (2) & (3) ]
`∠PBA+∠QBC=∠BCD+∠DCS`_______(5)
`∠ABC=∠BCD` [ from eq. (4) - (5) ]
`∠ABC` and `∠BCD` are alternate angles.
`AB∥CD`
Proved.
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