# NCERT Solutions for Class 9 Maths

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1.    We are given that ∠AOC + ∠BOE = 700 and ∠BOD = 400.

We need to find ∠BOE and reflex ∠COE

From the given figure, we can conclude that ∠COB and ∠COE form a linear pair. We know that sum of the angles of a linear pair is 1800.

∴ ∠COB + ∠COE = 1800

∴ ∠COB = ∠AOC + ∠BOE, or

∴ ∠AOC + ∠BOE + ∠COE = 1800

⇒ 700 + ∠COE = 1800

⇒ ∠COE = 1800 – 700

= 1100.

Reflex ∠COE = 360– ∠COE

= 3600 – 1100

= 2500.

∠AOC = ∠BOD                                    (Vertically opposite angles), or

∠BOD + ∠BOE = 700.

But, we are given that ∠BOD = 400.

400 + ∠BOE = 700 ∠BOE = 700 – 400

= 300.

Therefore, we can conclude that Reflex ∠COE = 2500 and ∠BOE = 300.

2.    We are given that ∠POY = 900 and a : b = 2 : 3 .

We need find the value of c in the given figure.

Let a be equal to 2x and b be equal to 3x.

∵ a + b = 900             ⇒              2x + 3x = 900            ⇒            5x = 900

⇒ x = 180

Therefore          b = 3 × 180 = 540

Now                 b + c = 1800                        [Linear pair]

⇒ 54+ c = 1800

⇒ c = 1800 – 540 = 1260

3.   We need to prove that ∠PQS = ∠PRT

We are given that ∠PQR = ∠PRQ

From the given figure, we can conclude that ∠PQS and ∠PQR, and ∠PRS and ∠PRT form a linear pair.

We know that sum of the angles of a linear pair is 1800.

∴ ∠PQS + ∠PQR = 1800, and                    (i)

∠PRQ + ∠PRT = 1800.                            (ii)

From equations (i) and (a), we can conclude that

∠PQS + ∠PQR = ∠PRQ + ∠PRT

But, ∠PQR = ∠PRQ.

∴ ∠PQS = ∠PRT

Therefore, the desired result is proved.

4.    We need to prove that AOB is a line.

We are given that x + y = w + z.

We know that the sum of all the angles around a fixed point is 3600. Thus, we can conclude that ∠A0C + ∠BOC + ∠AOD + ∠BOD = 3600 or y + x + z + w = 3600.

But, x + y = w + z                                          (Given).

2 (y + x) = 3600.

y + x = 1800.

From the given figure, we can conclude that y and x form a linear pair.

We know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is 1800.

y + x = 1800.

Therefore, we can conclude that AOB is a line.

We are given that OR is perpendicular to PQ, or

∠QOR = 900.

From the given figure, we can conclude that ∠POR and ∠QOR form a linear pair. We know that sum of the angles of a linear pair is 1800.

∴ ∠POR + ∠QOR = 1800, or

∠POR = 900.

From the figure, we can conclude that ∠POR = ∠POS + ∠ROS

⇒ ∠POS + ∠ROS = 900, or

∠ROS = 900 – ∠POS                               (i)

From the given figure, we can conclude that ∠QOS and ∠POS form a linear pair.

We know that sum of the angles of a linear pair is 1800.

∠QOS + ∠POS = 1800, or Therefore, the desired result is proved.

6.   We are given that ∠XYZ = 640, XY is produced to P and YQ bisects ∠ZYP

We can conclude the given below figure for the given situation: We need to find ∠XYQ and reflex ∠QYP

From the given figure, we can conclude that ∠XYZ and ∠ZYP form a linear pair. We know that sum of the angles of a linear pair is 1800.

∠XYZ + ∠ZYP = 1800.

But ∠XYZ = 640.

⇒ 64 + ∠ZYP =1800

⇒ ∠ZYP =1160.

Ray YQ bisects ∠ZYP , or

∠XYQ = ∠QYZ + ∠XYZ

= 580+ 640 = 1220.

Reflex ∠QYP = 3600 – ∠QYP

= 3600 – 580

= 3020.

Therefore, we can conclude that ∠XYQ = 1220 and Reflex ∠QYP = 3020.

7.   We need to find the value of x and y in the figure given below and then prove that

AB || CD. From the figure, we can conclude that

y = 130                                       (Vertically opposite angles), and

x and 500 form a pair of linear pair.

We know that the sum of linear pair of angles is 1800.

x + 500 = 1800

x = 1300.

x = y = 1300.

From the figure, we can conclude that x and y form a pair of alternate interior angles corresponding to the lines AB and CD. Therefore, we can conclude that x = 1300, y = 1300 and AB || CD.

8.   We are given that AB || CD , CD ⊥ EF and y : z = 3 : 7. We need to find the value of x in the figure given below.

We know that lines parallel to the same line are also parallel to each other.

We can conclude that AB || CD || EF.

Let y = 3a and z = 7 a.

We know that angles on same side of a transversal are supplementary.

x + y = 1800.

x = z                             (Alternate interior angles)

z + y = 1800, or

7 a + 3a = 1800               ⇒               10a =1800

a = 1800.

z = 7 a = 1260

y = 3a = 540.

Now                  x + 54 = 1800

x = 126.

Therefore, we can conclude that x = 126.

9.   We are given that AB || CD , EF ⊥ CD and ∠GED = 1260. We need to find the value of ∠AGS, ∠GEF and ∠FGE in the figure given below.

∠GED = 1260

∠GED = ∠FED + ∠GEF.

But, ∠FED = 900.

1260 = 900 + ∠GEF                     ⇒                     ∠GEF = 360.

∠AGE = ∠GED                                 (Alternate angles)

∠AGE =1260.

From the given figure, we can conclude that ∠FED and ∠FEC form a linear pair. We know that sum of the angles of a linear pair is 1800.

∠FED + ∠FEC = 1800                ⇒               90+ ∠FEC = 1800

∠FEC = 900

But ∠FEC = ∠GEF + ZGEC

900 = 360 + ∠GEC                    ⇒                 ∠GEC = 540.

∠GEC = ∠FGE = 540                          (Alternate interior angles).

Therefore, we can conclude that ∠AGE = 1260, ∠GEF = 360 and ∠FGE = 540.

10.  We are given that PQ || ST, ∠PQR = 1100 and ∠RST = 1300. We need to find the value of ∠QRS in the 1100 figure.

We need to draw a line RX that is parallel to the line ST, to get Thus, we have ST || RX.

We know that lines parallel to the same line are also parallel to each other. We can conclude that PQ || ST || RX .

∠PQR = ∠QRX                (Alternate interior angles), or

∠QRX =1100.

We know that angles on same side of a transversal are supplementary.

∠RST + ∠SRX = 1800                  ⇒                1300 + ∠SRX =1800

⇒ ∠SRX =1800 – 1300 = 500.

From the figure, we can conclude that

∠QRX = ∠SRX + ∠QRS               ⇒               1100 = 500 + ∠QRS

⇒ ∠QRS = 600.

Therefore, we can conclude that ∠QRS = 600.

11.  We are given that AB || CD, ∠APQ = 500 and ∠PRD = 1270. We need to find the value of x and y in the figure.

∠APQ = x = 500.                                     (Alternate interior angles)

∠PRD = ∠APR = 1270.                          (Alternate interior angles)

∠APR = ∠QPR + ∠APQ.

1270 = y + 500                                              y = 770.

Therefore, we can conclude that x = 500 and y = 770.

12.  We are given that PQ and RS are two mirrors that are parallel to each other. We need to prove that AB || CD in the figure.

Let us draw lines BX and CY that are parallel to each other, to get

We know that according to the laws of reflection

∠ABX = ∠CBX and ∠BCY = ∠DCY

∠BCY = ∠CBX                                   (Alternate interior angles)

We can conclude that ∠ABX = ∠CBX = ∠BCY = ∠DCY.

From the figure, we can conclude that

∠ABC = ∠ABX + ∠CBX, and

∠DCB = ∠BCY + ∠DCY .

Therefore, we can conclude that ∠ABC = ∠DCB.

From the figure, we can conclude that ∠ABC and ∠DCB form a pair of alternate interior angles corresponding to the lines AB and CD, and transversal BC.

Therefore, we can conclude that AB || CD.

13.  We are given that ∠SPR = 1350 and ∠PQT = 1100. We need to find the value of ∠PRQ in the figure given below. From the figure, we can conclude that

∠SPR and ∠RPQ, and ∠SPR and ∠RPQ form a linear pair.

We know that the sum of angles of a linear pair is 1800

∠SPR + ∠RPQ = 1800, and

∠PQT + ∠PQR = 1800.

135+ ∠RPQ = 1800, and

1100 + ∠PQR = 1800, or

∠RPQ = 450, and

∠PQR = 700.

From the figure, we can conclude that

∠PQR + ∠RPQ + ∠PRQ = 1800.                                    (Angle sum property)

⇒ 700 + 450 + ∠PRQ = 1800                    ⇒                   1150 + ∠PRQ = 1800

⇒ ∠PRQ = 650.

Therefore, we can conclude that ∠PRQ = 650.

14.  We are given that ∠X = 620, ∠XYZ = 540 and YO and ZO are bisectors of ∠XYZ and ∠XZY , respectively. We need to find ∠OZY and ∠YOZ in the figure. From the figure, we can conclude that in Δ XYZ

∠X + ∠XYZ + ∠XZY = 1800.  (Angle sum property)

⇒ 620 + 540 + ∠XZY = 1800            ⇒           1160 + ∠XZY = 1800

⇒ ∠XZY = 640.

We are given that OY and OZ are the bisectors of ∠XYZ and ∠XZY , respectively. From the figure, we can conclude that in Δ OYZ

∠OYZ + ∠OZY + ∠YOZ = 1800.                      (Angle sum property)

270 + 320 + ∠YOZ = 1800             ⇒              590 + ∠YOZ =1800

⇒ ∠YOZ = 1210.

Therefore, we can conclude that ∠YOZ = 1210 and ∠OZY = 320.

15.  We are given that AB || DE, ∠BAC = 350 and ∠CDE = 530. We need to find the value of ∠DCE in the figure given below.

From the figure, we can conclude that

∠BAC = 350 and ∠CDE = 530.

∠BAC = ∠CED = 350                       (Alternate interior angles)

From the figure, we can conclude that in Δ DCE

∠DCE + ∠CED + ∠CDE = 1800.       (Angle sum property)

∠DCE + 350 + 530 = 1800              ⇒            ∠DCE + 880 = 1800

⇒ ∠DCE = 920.

Therefore, we can conclude that ∠DCE = 920.

16.  We are given that ∠PRT = 400 ∠RPT = 950 and ∠TSQ = 750. We need to find the value of ∠SQT in the figure.

From the figure, we can conclude that in Δ RTP

∠PRT + ∠RTP + ∠RPT = 1800.                (Angle sum property)

400 + ∠RTP + 950 = 1800           ⇒            ∠RTP + 1350 = 1800

⇒ ∠RTP = 450.

From the figure, we can conclude that

∠RTP = ∠STQ = 450.                            (Vertically opposite angles)

From the figure, we can conclude that in Δ STQ

∠SQT + ∠STQ + ∠TSQ = 1800.           (Angle sum property)

∠SQT + 450 + 750 = 1800          ⇒        ∠SQT + 1200 = 1800

⇒ ∠SQT = 600.

Therefore, we can conclude that ∠SQT = 600.

17.  We are given that PQ ⊥ PS , PQ || SR, ∠SQR = 280 and ∠QRT = 650. We need to find the values of x and y in the figure.

We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”

From the figure, we can conclude that

∠SQR + ∠QSR = ∠QRT , or

280 + ∠QSR = 650             ⇒             ∠QSR = 370.

From the figure, we can conclude that

x = ∠QSR = 370                                                  (Alternate interior angles)

From the figure, we can conclude that Δ PQS

∠PQS + ∠QSP + ∠QPS = 1800.                         (Angle sum property)

∠QPS = 900                                (PQ ⊥ PS)

x + y + 900 = 1800        ⇒            x + 370+ 900 = 1800

x + 1270 = 1800             ⇒           x = 530.

Therefore, we can conclude that x = 530 and y = 370.

18.  We need to prove that in the figure given below. We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”

From the figure, we can conclude that in Δ QTR , ∠TRS is an exterior angle

∠QTR + ∠TQR = ∠TRS, or

∠QTR = ∠TRS – ∠TQR                        …(i)

From the figure, we can conclude that in Δ QTR , ∠TRS is an exterior angle

∠QPR + ∠PQR = ∠PRS.

We are given that QT and RT are angle bisectors of ∠PQR and ∠PRS.

∠QPR + 2∠TQR = 2∠TRS

∠QPR = 2 (∠TRS – ∠TQR).

We need to substitute equation (i) in the above equation, to get

∠QPR = 2∠QTR, or Therefore, we can conclude that the desired result is proved.

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