CBSE 9 Class Exams Questions and Answers

NCERT

Calculate the molar mass of the following substances. (c) baking soda

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A Amancha Dathu

84

NCERT

Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

2. Plot the points (x, y) given in the following table on the plane, choosing suitable unitsof distance on the axes.

x	-2	-1	0	1	3
y	8	7	-1.25	2	-1

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A Ankit

So, the graph of the given points is

NCERT

In which quadrant or on which axis do each of the points (– 2, 4), (3, – 1), (– 1, 0),(1, 2) and (– 3, – 5) lie

1. In which quadrant or on which axis do each of the points (– 2, 4), (3, – 1), (– 1, 0),(1, 2) and (– 3, – 5) lie? Verify your answer by locating them on the Cartesian plane.

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A Ankit

(-2,4) lies in second quadrant (3,-1) lies in fourth quadrant (-1,0) lies in second quadrant (1,2) lies in first quadrant (-3,-5) lies in third quadrant

NCERT

Find the remainder when x power 3 plus 3x power 2 plus 3x plus 1 is divided by 5 plus 2x

1.(v) Find the remainder when $x^ 3 + 3x^ 2 + 3x + 1$ is divided by
(v) $5+2x$

NCERT

How will you construct a 15° angle?

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D Divya Prakash Singh

First, we make and then its bisector. The steps of constructions are: 1. Draw a ray OA. 2. Taking O as the center and any radius of your own choice, draw an arc cutting OA at B. 3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. 4. Draw the ray OD passing through the C. Thus, Now, we draw bisector of 5. Taking C and...

NCERT

In Fig. 6.44, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1 / 2 ∠ QPR.

6. In Fig. 6.44, the side QR of $\Delta$ PQR is produced to a point S. If the bisectors of $\angle$ PQR and $\angle$ PRS meet at point T, then prove that $\angle$ QTR = $\frac{1}{2}$ $\angle$ QPR.

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M manish painkra

We have, PQR is produced to a point S and bisectors of PQR and PRS meet at point T, By exterior angle sum property, PRS = P + PQR Now, ................(i) Since QT and QR are the bisectors of PQR and PRS respectively. Now, in QRT, ..............(ii) From eq (i) and eq (ii), we get Hence proved

NCERT

In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.

5. In Fig. 6.43, if PQ $\bot$ PS, PQ $||$ SR, $\angle$ SQR = 28° and $\angle$ QRT = 65°, then find the values of x and y.

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M manish painkra

We have, PQ PS, PQ || SR, SQR = 28° and QRT = 65° Now, In QRS, the side SR produced to T and PQ || RS therefore, QRT = = So, Also, QRT = RSQ + SQR (By exterior angle property of a triangle) Therefore, RSQ = QRT - SQR Now, in PQS, P + PQS + PSQ =

NCERT

In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°, find ∠ SQT.

4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that $\angle$ PRT = 40°, $\angle$ RPT = 95° and $\angle$ TSQ = 75°, find $\angle$ SQT.

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M manish painkra

We have, lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75° In PRT, by using angle sum property PRT + PTR + TPR = So, PTR = Since lines, PQ and RS intersect at point T therefore, PTR = QTS (Vertically opposite angles) QTS = Now, in QTS, By using angle sum property TSQ + STQ + SQT = So, SQT =

NCERT

In Fig. 6.41, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.

3. In Fig. 6.41, if AB $||$ DE, $\angle$ BAC = 35° and $\angle$ CDE = 53°, find $\angle$ DCE.

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M manish painkra

We have, AB || DE, BAC = 35° and CDE = 53° AE is a transversal so, BAC = AED = Now, In CDE, CDE + DEC + ECD = (By angle sum property) Therefore, ECD =

NCERT

In Fig. 6.40, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ, find ∠ OZY and ∠ YOZ.

2. In Fig. 6.40, $\angle$ X = 62°, $\angle$ XYZ = 54°. If YO and ZO are the bisectors of $\angle$ XYZ and $\angle$ XZY respectively of $\Delta$ XYZ, find $\angle$ OZY and $\angle$ YOZ.

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M manish painkra

We have X = , XYZ = YO and ZO bisects the XYZ and XZY Now, In XYZ, by using angle sum property XYZ + YZX + ZXY = So, YZX = YZX = and, OYZ = also, OZY = Now, in OYZ Y + O + Z = [Y = and Z = ] So, YOZ =

NCERT

In Fig. 6.39, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠ SPR = 135° and ∠ PQT = 110°, find ∠ PRQ.

1. In Fig. 6.39, sides QP and RQ of $\Delta$ PQR are produced to points S and T respectively. If $\angle$ SPR = 135° and $\angle$ PQT = 110°, find $\angle$ PRQ.

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M manish painkra

Given, PQR is a triangle, SPR =, PQT = Now, TQP + PQR = (Linear pair) So, PQR = Since the side of QP of the triangle, PQR is produced to S So, PQR + PRQ = (Exterior angle property of triangle)

NCERT

In Fig. 6.33, 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 reflects back along CD. Prove that AB || CD.

6. In Fig. 6.33, 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 reflects back along CD. Prove that AB $||$ CD.

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M manish painkra

Draw a ray BL PQ and CM RS Since PQ || RS (Given) So, BL || CM and BC is a transversal LBC = MCB (Alternate interior angles).............(i) It is known that, angle of incidence = angle of reflection So, ABL = LBC and MCB = MCD ..................(ii) Adding eq (i) and eq (ii), we get ABC = DCB Both the interior angles are equal Hence AB || CD

NCERT

In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.

5. In Fig. 6.32, if AB $||$ CD, $\angle$ APQ = 50° and $\angle$ PRD = 127°, find x and y.

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M manish painkra

Given, AB || CD, APQ = and PRD = PQ is a transversal. So, APQ = PQR= (alternate interior angles) Again, PR is a transversal. So, y + = (Alternate interior angles)

NCERT

In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS. [Hint : Draw a line parallel to ST through point R.]

4. In Fig. 6.31, if PQ $||$ ST, $\angle$ PQR = 110° and $\angle$ RST = 130°, find $\angle$ QRS.
[Hint : Draw a line parallel to ST through point R.]

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M manish painkra

Draw a line EF parallel to the ST through R. Since PQ || ST and ST || EF EF || PQ PQR = QRF = (Alternate interior angles) QRF = QRS + SRF .............(i) Again, RST + SRF = (Interior angles of two parallels ST and RF) (RST = , given) Thus, QRS =

NCERT

In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.

3. In Fig. 6.30, if AB $||$ CD, EF $\bot$ CD and $\angle$ GED = 126°, find $\angle$ AGE, $\angle$ GEF and $\angle$ FGE.

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M manish painkra

Given AB || CD, EFCD and GED = In the above figure, GE is transversal. So, that AGE = GED = [Alternate interior angles] Also, GEF = GED - FED = Since AB is a straight line Therefore, AGE + FGE = So, FGE =

NCERT

In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

2. In Fig. 6.29, if AB $||$ CD, CD $||$ EF and y : z = 3 : 7, find x.

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M manish painkra

Given AB || CD and CD || EF and therefore, AB || EF and (alternate interior angles)..............(i) Again, CD || AB .............(ii) Put the value of in equation (ii), we get Then By equation (i), we get the value of

NCERT

In Fig. 6.28, find the values of x and y and then show that AB || CD.

1. In Fig. 6.28, find the values of x and y and then show that AB $||$ CD.

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M manish painkra

Given that, In the figure, CD and PQ intersect at F Therefore, (vertically opposite angles) PQ is a straight line. So, Hence AB || CD (since and are alternate interior angles)

NCERT

It is given that anlge XYZ = 64 degrees and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects anlge ZYP, find anlge XYQ and reflex anlge QYP.

6. It is given that $\angle$ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects $\angle$ ZYP, find $\angle$ XYQ and reflex $\angle$ QYP.

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M manish painkra

Given that, XYZ = and XY produced to point P and Ray YQ bisects ZYP Now, XYP is a straight line So, XYZ + ZYQ + QYP = Thus reflex of QYP = Since XYQ = XYZ + ZYQ [ =

NCERT

In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that angle ROS = 1 / 2 ( angle QOS - angle POS).

5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})$

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M manish painkra

Given that, POQ is a line, OR PQ and ROQ is a right angle. Now, POS + ROS + ROQ = [since POQ is a straight line] .............(i) and, ROS + ROQ = QOS ..............(ii) Add the eq (i ) and eq (ii), we get hence proved

NCERT

In Fig. 6.16, if x + y + w = z, then prove that AOB is a line.

4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

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M manish painkra

Given that, ..............(i) It is known that, the sum of all the angles at a point = ..............(ii) From eq (i) and eq (ii), we get Hence proved AOB is a line.

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