Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: With Center as B and radius 3cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
Solution: Steps to construct: Step 1: Draw a line segment AB = 4cm. Step 2: At point B, draw a ray BX making an angle of 90o and cut off BC = 6cm. Step 3: Join AC. Step 4: Draw the perpendicular...
Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: With centers B and C, draw two arcs of radius 4cm which intersects each other at point A. Step 3: Join AB and AC. Step 4:...
Solution: Steps to construct: Step 1: Draw a line segment AB = 8cm. Step 2: With center as A and radius 4cm, with center as B and radius 3cm, draw circles. Step 3: Draw the third circle AB as...
(a) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate (i) ∠QBC (ii) ∠BCP Solution: (i) ABCD is a cyclic quadrilateral ∠A + ∠C = 180o 30o + p =...
(b) In the figure given below, O is the center of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC. Solution: (a) In the given figure, ABCD is a cyclic quadrilateral ∠ADC =...
(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find: (i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008) Solution: (a) ∠DBC = 58° BD is diameter ∠DCB = 90° (Angle in...
Solution: (a) Given, ∠AOC = 150° and AD = CD We know that an angle subtends by an arc of a circle at the center is twice the angle subtended by the same arc at any point on the remaining part of the...
Solution: From the figure (i) ABCD is a cyclic quadrilateral Ext. ∠DCE = ∠BAD ∠BAD = xo Now arc BD subtends ∠BOD at the center And ∠BAD at the remaining part of the circle. ∠BOD = 2 ∠BAD = 2 x 2 x =...
(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate (i)∠CAB (ii)∠OAC Solution: Given that (a) Arc AB subtends ∠APB at the center...
Solution (a) (i) ∠PRB = ∠BAP (Angles in the same segment of the circle) ∴ ∠PRB = 35° (∵ ∠BAP = 35° given)
Solution: In ∆APB, ∠APB = 90° (Angle in a semi-circle) But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle) ∠A + 90° + 42°= 180° ∠A + 132° = 180° ⇒ ∠A = 180° – 132° = 48° But ∠A = ∠PQB (Angles in the...
Solution (a) ∠NYB = 50°, ∠YNB = 20°. In ∆YNB, ∠NYB + ∠YNB + ∠YBN = 180o 50o + 20o + ∠YBN = 180o ∠YBN + 70o = 180o ∠YBN = 180o – 70o = 110o But ∠MAN = ∠YBN (Angles in the same segment) ∠MAN = 110o...
(a) ABCD is cyclic Quadrilateral ∠B + ∠D = 1800 Y + 400 + 45o = 180o (y + 85o = 180o) Y = 180o – 85o = 95o ∠ACB = ∠ADB xo = 40 (a) Arc ADC Subtends ∠AOC at the centre and ∠ ABC at the remaining part...
Solution: (a) Construction: Join AB ∠A = ∠C = 350 (Alt Angles) ∠ABC = 35o (b) ∠AOC + reflex ∠AOC = 360o 130o + Reflex ∠AOC = 360o Reflex ∠AOC = 360o – 130o = 230o Now arc BC Subtends reflex ∠AOC at...
Solution: (i) ∠ACB = ∠ADB (Angles in the same segment of a circle) But ∠ADB = x° ∠ABC = xo Now in ∆ABC ∠CAB + ∠ABC + ∠ACB = 180o 40o + 900 + xo = 180o (AC is the diameter) 130o + xo = 180o xo =...
Solution: (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° = 42o + x + 50o = 180o = 92o + x = 180o x = 180o – 92o x = 88o (ii) In the given figure we...
Given, (i) (sin 35o cos 55o + cos 35o sin 55 o)/ (cosec2 10o – tan2 80 o) (ii) sin2 34o + sin2 56o + 2 tan18o tan 72o – cot2 30o = sin2 34o + sin2 (90o – 34o) + 2 tan18o tan (90o – 18o) – cot2 30o =...
Given, (i) (sin 65o/ cos 25o) + (cos 32o/sin 58o) – sin 28o sec 62o + cosec2 30o = (sin 65o/ cos (90o – 65o)) + (cos 32o/sin (90o – 32o)) – sin 28o sec (90o – 28o) + 22 = (sin 65o/sin 65o) + (cos...
Given, (i) cos2 26o + cos 64o sin 26o + (tan 36o/ cot 54o) = cos2 26o + cos (90o – 16o) sin 26o + [tan 36o/ cot (90o – 54o)] = [cos2 26o + sin2 26o] + (tan 36o/ tan 36o) = 1 + 1 = 2 (ii) (sec 17o/...
Given, 12 cosec θ = 13 ⇒ cosec θ = 13/12 In right ∆ ABC, ∠A = θ So, cosec θ = AC/BC = 13/12 AC = 13 and BC = 12 By Pythagoras theorem, AB = √(AC2 – BC2) = √[(13)2 – (12)2] = √(169 – 144) = √25 = 5...
In right ∆ ABC, tan A = BC/AB = 1/√3 So, BC = 1 and AB = √3 By Pythagoras theorem, AC = √(AB2 + BC2) = √[(√3)2 + (1)2] = √(3 + 1) = √4 = 2 Hence, sin A = BC/AC = ½ cos A = AB/AC = √3/2 cot A = 1/tan...
We know that, sin2 A + cos2 A = 1 So, cos A = √(1 – sin2 A) tan A = sin A/cos A = sin A/ √(1 – sin2 A) sec A = 1/cos A = 1/ (√1 – sin2 A)
sec A = 17/8 and A is an acute angle So, in ∆ ABC we have ∠B = 90o And, AC = 17 and AB = 8 By Pythagoras theorem, BC = √(AC2 – AB2) = √(172 – 82) = √(289 – 64) = √225 = 15 Now, sin A = BC/AC = 15/17...
Solution:- From the question it is given that, a model of a ship is made to a scale of 1 : 250 (i) Given, the length of the model is 1.6 m Then, length of the ship = (1.6 × 250)/1 = 400 m (ii)...
Solution:- From the question it is given that, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O. (i) Now consider the ∆AOD and ∆EDC, ∠AOD = ∠EOC … [because Vertically...
Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC. If 2AB = 3DC. So, 2AB = 3DC AB/DC = 3/2 Now, consider ∆AOB and ∆COD ∠AOB = ∠COD … [because vertically opposite...
Solution:- From the given figure, The diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E. In the figure four triangles have equal area. So, area of ∆OAB = ¼ area...
Solution:- From the question it is given that, ∠ABD = ∠CAD AB = 5 cm, AC = 3 cm and AD = 4 cm Now, consider the ∆ABC and ∆ACD ∠C = ∠C … [common angle for both triangles] ∠ABC = ∠CAD … [from the...
Solution:- From the question it is given that, the areas of two similar triangles are 360 cm² and 250 cm². one side of the first triangle is 8 cm So, PQR and XYZ are two similar triangles, So, let...
Solution:- From the question it is given that, In a ∆ABC, D and E are points on the sides AB and AC respectively. DE || BC AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm Consider the ∆ABC, Given,...
Solution:- From the question it is given that, 2 AD = BD, EC || BH (i) Given, E is mid-point of BD 2DE = BD … [equation (i)] 2AD = BD … [equation (ii)] From equation (i) and equation (ii) we get,...
Solution:- From the question it is given that, CD = ½ AC BF || AG (i) We have to prove that, CE || AG Consider, CD = ½ AC AC = 2BC … [because from the figure B is mid-point of AC] So, CD = ½ (2BC)...
Solution:- From the figure it is given that, ∠AED = ∠ABC Consider the ∆ABC and ∆ADE ∠AED = ∠ABC … [from the figure] ∠A = ∠A … [common angle for both triangles] Therefore, ∆ABC ~ ∆ADE … [by AA axiom]...
Solution:- From the question it is given that, ∠1 = ∠2 and ∠3 = ∠4 We have to prove that, PT x QR = PR x ST Given, ∠1 = ∠2 Adding ∠6 to both LHS and RHS we get, ∠1 + ∠6 = ∠2 + ∠6 ∠SPT = ∠QPR...
Solution:- From the question it is given that, a model of a ship is made to a scale of 1 : 200 (i) Given, the length of the model is 4 m Then, length of the ship = (4 × 200)/1 = 800 m (ii) Given,...
Solution:- From the question it is given that, The model of a building is constructed with the scale factor 1 : 30 So, Height of the model/Height of actual building = 1/30 (i) Given, the height of...
Solution:- From the question it is given that, Map drawn to a scale of 1: 25000 AB = 12 cm, BG = 16 cm Consider the ∆ABC, From the Pythagoras theorem, AC2 = AB2 + BC2 AC = √(AB2 + BC2) = √((12)2 +...
Solution:- From the question it is given that, Map drawn to a scale of 1: 250000 AB = 3 cm, BC = 4 cm and ∠ABC = 90o (i) We have to find the actual length of AB in km. Let us assume scale factor K =...
Solution:- Consider the two isosceles triangle PQR and XYZ, ∠P = ∠X … [from the question] So, ∠Q + ∠R = ∠Y + ∠Z ∠Q = ∠R and ∠Y = ∠Z [because opposite angles of equal sides] Therefore, ∠Q = ∠Y and ∠R...
Solution:- From the question it is given that, ∠ABC = 90° AB and DE is perpendicular to AC (i) Consider the ∆ADE and ∆ACB, ∠A = ∠A … [common angle for both triangle] ∠B = ∠E … [both angles are equal...
Solution:- From the question it is given that, ∠ABC = ∠DAC AB = 8 cm, AC = 4 cm, AD = 5 cm (i) Now, consider ∆ACD and ∆BCA ∠C = ∠C … [common angle for both triangles] ∠ABC = ∠CAD … [from the...
Solution:- From the question it is given that, DC is parallel to AB AB = 9 cm, DC = 6 cm and BB = 12 cm (i) Consider the ∆APB and ∆CPD ∠APB = ∠CPD … [because vertically opposite angles are equal]...
Solution:- From the question it is given that, AB || DC AB = 2 DC, AD = 3 cm, BC = 4 cm Now consider ∆EAB, EA/DA = EB/CB = AB/DC = 2DC/DC = 2/1 (i) EA = 2, DA = 2 × 3 = 6 cm Then, ED = EA – DA = 6 –...
Solution:- From the question it is given that, DE || BC The ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5 Now, consider the ∆ABC and ∆ADE ∠A = ∠A … [common angle for both triangles] ∠D = ∠B...
Solution:- From the question it is given that, ABCD is a parallelogram. BP: PC = 1: 2 area of ∆CPQ = 20 cm² Construction: draw QN perpendicular CB and Join BN. Then, area of ∆BPQ/area of ∆CPQ =...
Solution:- From the question it is given that, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O. AE : EB = 2 : 3 (i) We have to find EF : AD So, AB/BE = AD/EF EF/AD =...
Solution:- From the question it is given that, ABCD is a parallelogram, AM ⊥ DC and AN ⊥ CB AM = 6 cm AN = 10 cm The area of parallelogram ABCD is 45 cm² Then, area of parallelogram ABCD = DC × AM =...
Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC and AB = 2 CD, Then, ∠OAB = ∠OCD … [because alternate angles are equal] ∠OBA = ∠ODC Then, ∆AOB ~ ∆COD So, area of...
Solution:- From the question it is given that, PB = 2: 3 PO is parallel to BC and is extended to Q so that CQ is parallel to BA. (i) we have to find the area ∆APO: area ∆ABC, Then, ∠A = ∠A … [common...
Solution:- (i) We have to find the ratios AD/AB, DE/BC, From the question it is given that, AD/DB = 3/2 Then, DB/AD = 2/3 Now add 1 for both LHS and RHS we get, (DB/AD) + 1 = (2/3) + 1 (DB + AD)/AD...
Solution:- (i) Consider the ∆ABC and ∆DEC, ∠ABC = ∠DEC … [both angles are equal to 90o] ∠C = ∠C … [common angle for both triangles] Therefore, ∆ABC ~ ∆DEC … [by AA axiom] (ii) AC/CD = AB/DE...
Solution:- (i) From the question it is given that, DE || BC We have to prove that, ∆ADE and ∆ABC are similar ∠A = ∠A … [common angle for both triangles] ∠ADE = ∠ABC … [because corresponding angles...
Solution:- From the question it is given that, DE || BC and AD : DB = 1 : 2, ∠D = ∠B, ∠E = ∠C … [corresponding angles are equal] Consider the ∆ADE and ∆ABC, ∠A = ∠A … [common angles for both...
Solution:- From the question it is given that, DE || BC, DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm From the fig, ∠D = ∠B and ∠E = ∠C … [corresponding angles are equal] Now consider the ∆ADE...
Solution:- From the question it is given that, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm (i) We have to prove that, ∆OAB ~ ∆OCD So, consider the ∆OAB and ∆OCD ∠AOB = ∠COD …...
Solution:- From the question it is given that, PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm² From the figure, Consider the ∆AOQ and ∆BOP, ∠OAQ = ∠OBP … [both angles are equal to 90o] ∠AOQ =...
Solution:- From the question it is given that, The area of two similar triangles are 36 cm² and 25 cm². Let us assume ∆PQR ~ ∆XYZ, PM and XN are their altitudes. So, area of ∆PQR = 36 cm2 Area of...
Solution:- From the question it is given that, ∆ABC ~ ∆DEF BC = 3 cm, EF = 4 cm Area of ∆ABC = 54 sq. cm. We know that, Area of ∆ABC/ area of ∆DEF = BC2/EF2 54/area of ∆DEF = 32/42 54/area of ∆DEF =...
Solution:- From the question it is given that, ∆ABC ~ DEF Area of ∆ABC = 9 sq. cm Area of ∆DEF =16 sq. cm We know that, area of ∆ABC/area of ∆DEF = BC2/EF2 area of ∆ABC/area of ∆DEF = BC2/EF2 9/16 =...
Solution:- From the question it is given that, AD is bisector of ∠BAC AB = 6 cm, AC = 4 cm and BD = 3cm Construction, from C draw a straight line CE parallel to DA and join AE ∠1 = ∠2 … [equation...
Solution:- From the question it is given that, AB || CR and LM || QR (i) We have to prove that, BM/MC = AL/LQ Consider the ∆ARQ LM || QR … [from the question] So, AM/MR = AL/LQ … [equation (i)] Now,...
Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O Now consider the ∆OAB and ∆OCD, ∠AOB = ∠COD [because vertically...
Solution:- Consider the ∆POQ AB || PQ … [given] So, OA/AP = OB/BQ … [equation (i)] Then, consider the ∆OPR AC || PR OA/AP = OC/CR … [equation (ii)] Now by comparing both equation (i) and equation...
Solution:- From the given figure, ∠D = ∠E and AD/BD = AE/EC, We have to prove that, BAC is an isosceles triangle So, consider the ∆ADE ∠D = ∠E … [from the question] AD = AE … [sides opposite to...
Solution:- From the given figure, CD || LA and DE || AC, Consider the ∆BCA, BE/BC = BD/BA By using the corollary of basic proportionality theorem, BE/(BE + EC) = BD/AB 4/(4 + 2) = BD/AB … [equation...
Solution:- From the figure it is given that, AB || DE and BD || EF. We have to prove that, DC² = CF x AC Consider the ∆ABC, DC/CA = CE/CB … [equation (i)] Now, consider ∆CDE CF/CD = CE/CB …...
Solution:- From the question it is given that, DE || BC and BD = CE So, we have to prove that ABC is an isosceles triangle. Consider the triangle ABC, AD/DB = AE/EC Given, DB = EC … [equation (i)]...
Solution:- From the dimensions given in the question, Consider the ∆PQR So, PQ/PE = 1.28/0.18 = 128/18 = 64/9 Then, PR/PF = 2.56/0.36 = 256/36 = 64/9 By comparing both the results, 64/9 = 64/9...
Solution:- From the given dimensions, Consider the ∆PQR So, PE/EQ = 3.9/3 = 39/30 = 13/10 Then, PF/FR = 8/9 By comparing both the results, 13/10 ≠ 8/9 Therefore, PE/EQ ≠ PF/FR So, EF is not parallel...
Solution:- (i) From the figure, it is given that, Consider the ∆ABC, AD/DB = AE/EC x/(x – 2) = (x + 2)/(x – 1) By cross multiplication we get, X(x – 1) = (x – 2) (x + 2) x2 – x = x2 – 4 -x = -4 x =...
Solution:- From the figure, XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, So, PX/QX = PY/YR 1/3 = PY/4.5 By cross multiplication we get, (4.5 × 1)/3 = PY PY = 45/30 PY = 1.5 Then, ∠X =...
Solution:- From the figure, DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm (i) AE: EC So, AD/BD = AE/EC AE/EC = AD/BD AE/EC = ¾ AE: EC = 3: 4 (ii) consider ∆ADE and ∆ABC ∠D = ∠B ∠E = ∠C Therefore,...
Solution:- From the question it is given that, Height of pole (PQ) = 6m Height of a woman (MN) = 1.5m So, shadow NR = 3m Therefore, pole and woman are standing in the same line PM ||MR ∆PRQ ~ ∆MNR...
Solution:- From the question it is given that, Height of a tower PQ = 15m It’s shadow QR = 24 m Let us assume the height of a telephone pole MN = x It’s shadow NO = 16 m Given, at the same time,...
Solution:- From the figure, consider ∆ABC, So, ∠A = 90o And AD ⊥ BC ∠BAC = 90o Then, ∠BAD + ∠DAC = 90o … [equation (i)] Now, consider ∆ADC ∠ADC = 90o So, ∠DCA + ∠DAC = 90o … [equation (ii)] From...
Solution:- From the figure, AF, BE and CD are parallel lines. Consider the ∆AEF and ∆CED ∠AEF and ∠CED [because vertically opposite angles are equal] ∠F = ∠C [alternate angles are equal] Therefore,...
Solution:- From the figure it is given that, AB, EF and CD are parallel lines. (i) Consider the ∆EFG and ∆CGD ∠EGF = ∠CGD [Because vertically opposite angles are equal] ∠FEG = ∠GCD [alternate angles...
Solution:- From the figure it is given that, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. (i) We have to prove that, EF = FC From the figure, D is the midpoint of...
Solution:- Consider the ∆ABC, Where, the altitude BN and CM of ∆ABC meet at H. and construction: join MN (i) We have to prove that, CN × HM = BM × HN In ∆BHM and ∆CHN ∠BHM = ∠CHN [because vertically...
Solution:- From the question it is give that, Consider the ∆RLQ and ∆PLN, ∠RLQ = ∠NLP [vertically opposite angles are equal] ∠RQL = ∠LNP [alternate angle are equal] Therefore, ∆RLQ ~ ∆PLN So, QR/PN...
Solution:- From the figure, ABCD is a parallelogram, Then, E is a point on AD and produced and BE intersects CD at F. We have to prove that ∆ABE ~ ∆CFB Consider ∆ABE and ∆CFB ∠A = ∠C [opposite...
Solution:- Consider the ∆ABC, So, we have to prove that, AB × AE = AC × AD Now, consider the ∆ADB and ∆AEC, ∠A = ∠A [common angle for both triangles] ∠ADB = ∠AEC [both angles are equal to 90o] ∆ADB...
Solution:- From the given figure, ABCD is a trapezium in which AB || DC, The diagonals AC and BD intersect at O. So we have to prove that, AO/OC = BO/OD Consider the ∆AOB and ∆COD, ∠AOB = ∠COD …...
Solution:- Consider the two triangles, ∆MNO and ∆XYZ From the question it is given that, two triangles are similar triangles So, ∆MNO ~ ∆XYZ If two triangles are similar, the corresponding angles...
Solution:- From the figure, ∠P = ∠P (common angle for both triangles) ∠PQR = ∠PRS [from the question] So, ∆PQR ~ ∆PRS Then, PQ/PR = PR/PS = QR/SR Consider PQ/PR = PR/PS PQ/8 = 8/4 PQ = (8 × 8)/4 PQ...
Solution:- From the given figure, (i) ∠A = ∠A (common angle for both triangles) ∠ACB = ∠ADE [given] Therefore, ∆ABC ~ ∆AED (ii) from (i) proved that, ∆ABC ~ ∆AED So, BC/DE = AB/AE = AC/AD AD = AB –...
Solution:- From the given figure, ∠P = ∠RTS So we have to prove that ∆RPQ ~ ∆RTS In ∆RPQ and ∆RTS ∠R = ∠R (common angle for both triangle) ∠P = ∠RTS (from the question) ∆RPQ ~ ∆RTS (b) In the figure...
Solution:- (i) We have to prove that, ∆ACO ~ ∆BDO. So, from the figure Consider ∆ACO and ∆BDO Then, ∠AOC = ∠BOD [from vertically opposite angles] ∠A = ∠B Therefore, ∆ACO = ∆BDO Given, BD = 2.4 cm,...
Solution:- From the question it is given that, AB||DE AC = 3 cm CE = 7.5 cm BD = 14 cm From the figure, ∠ACB = ∠DCE [because vertically opposite angles] ∠BAC = ∠CED [alternate angles] Then, ∆ABC ~...
Solution:- Let us assume that, ∆ABC ~ ∆DEF ∆ABC is BC = 6cm ∆ABC ~ ∆DEF So, AB/DE = BC/EF = AC/DF Consider AB/DE = BC/EF AB/8 = 6/4 AB = (6 × 8)/4 AB = 48/4 AB = 12 Now, consider BC/EF = AC/DF 6/4 =...
Solution:- From the question it is given that, ∆ABC ~ ∆PQR Perimeter of ∆ABC = 32 cm Perimeter of ∆PQR = 48 cm So, AB/PQ = AC/PR = BC/QR Then, perimeter of ∆ABC/perimeter of ∆PQR = AC/PR 32/48 =...
Solution; Now, we have to find out the perimeter of ΔABC Let ΔABC ~ ΔDEF So, AB/DE = AC/DF = BC/EF Consider, AB/DE = AC/DE 4/6 = AC/12 By cross multiplication we get, AC = (4 × 12)/6 AC = 48/6 AC =...
From the question it is given that, ΔDEF ~ ΔLMN So, AB/ED = AC/EF = BC/DF Consider AB/ED = AC/EF 5/12 = 7/EF By cross multiplication, EF = (7 × 12)/5 EF = 16.8 cm Now, consider AB/ED = BC/DF 5/12 =...
Solution:- From the figure, two line segments are intersecting each other at P. In ΔBCP and ΔDPE 5/10 = 6/12 Dividing LHS and RHS by 2 we get, ½ = ½ Therefore, ΔBCD ~ ΔDEP
Solution:- From the question is given that, ∆DEF ~ ∆RPQ ∠D = ∠R and ∠F = ∠Q not ∠P No, ∠F ≠ ∠P
Solution:- (i) From the ΔABC and ΔPQR AB/PQ = 3.2/4 = 32/40 Divide both numerator and denominator by 8 we get, = 4/5 AC/PR = 3.6/4.5 = 36/45 Divide both numerator and denominator by 9 we get, = 4/5...
(iii)Let E be an event of getting a sum divisible by 5. Favourable outcomes = {(1,4),(2,3), (3,2), (4,1),(4,6), (5,5), (6,4)} Number of favourable outcomes = 7 P(E) = 7/36 Probability of getting a...
Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...
Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...
(iii)Let E be an event of getting no tails. Favourable outcomes = HH Number of favourable outcomes = 1 P(E) = 1/4 Probability of getting no tails is 1/4 . (iv)Let E be an event of getting atmost one...
Solution: When 2 coins are tossed, the possible outcomes are HH. HT, TH, TT. Number of possible outcomes = 4 (i)Let E be an event of getting 2 tails. Favourable outcomes = TT Number of favourable...
Solution: When 2 coins are tossed, the possible outcomes are HH. HT, TH, TT. Number of possible outcomes = 4 (i)Let E be an event of getting 2 heads. Favourable outcomes = HH Number of favourable...
Solution: Total number of cards = 52-4 = 48 [∵4 cards fell down] So number of possible outcomes = 48 (i) Let E be the event of getting black card. There will be 23 black cards since a black jack and...
(v) Let E be the event of getting a spade. There will be 10 spades. Number of favourable outcomes = 10 P(E) = 10/49 Hence the probability of getting a spade is 10/49. (vi) Let E be the event of...
(iii) Let E be the event of getting a black card. There will be 23 black cards remaining since 3 spades are removed. Number of favourable outcomes = 23 P(E) = 23/49 Hence the probability of getting...
Solution: Total number of cards = 52-3 = 49. [since 3 face cards of spade are removed] So number of possible outcomes = 49. (i) Let E be the event of getting black face card. There will be 3 black...
(xi) Let E be the event of getting a neither a spade nor a jack. There are 13 spades and 3 other jacks. So remaining cards = 52-13-3 = 36 There will be 36 cards which are neither a spade nor a jack....
(ix) Let E be the event of getting a non ace card. There will be 48 non ace cards. Number of favourable outcomes = 48 P(E) = 48/52 = 24/26 = 12/13 Hence the probability of getting a non ace card is...
(vii) Let E be the event of getting a black face card. There will be 6 black face cards. Number of favourable outcomes = 6 P(E) = 6/52 = 3/26 Hence the probability of getting a black face card is...
(iii) Let E be the event of getting a king of red colour. There will be 2 cards of king of red colour. Number of favourable outcomes = 2 P(E) = 2/52 = 1/26 Hence the probability of getting a king of...
Solution: Total number of cards = 52. So number of possible outcomes = 52. (i) Let E be the event of getting ‘2’ of spades. There will be only one card of ‘2’ spades. Number of favourable outcomes =...
Solution: Total number of balls = 24 Number of red balls = x. Number of white balls = 2x. Number of blue balls = 3x. x+2x+3x = 24 6x = 24 x = 24/6 = 4 Number of red balls = x = 4 Number of white...
Solution: Number of red balls = 6 Let number of blue balls be x. Total number of balls = 6+x Probability of drawing a red ball = 6/(6+x) Probability of drawing a blue ball = x/(6+x) Given the...
Solution: Total number of balls in the bag = 15. Let the number of white balls be x. Then number of red balls = 15-x. The probability of drawing a white ball = x/15. Probability of drawing a red...
(iii) Let E be the event of getting the number on the card is a perfect square. Outcomes favourable to E are {4,9,16,25,36,49,64,81,100} Number of favourable outcomes = 9 P(E) = 9/100 Hence the...
Solution: The possible outcomes are {2,3,…101} Number of possible outcomes = 100 (i) Let E be the event of getting the number on the card is an even number. Outcomes favourable to E are...
Solution: The possible outcomes are {1,2,3,…90} Number of possible outcomes = 90 (i) Let E be the event of getting the number on the disc is a two-digit number. Outcomes favourable to E are...
Solution: The possible outcomes are {3,5,7,9..…29} Number of possible outcomes = 14 (i) Let E be the event of getting the number on the ticket is a prime number. Outcomes favourable to E are...
Solution: The possible outcomes are {13,14,15,…60} Number of possible outcomes = 48 (i) Let E be the event of getting the number on the card is divisible by 5. Outcomes favourable to E are...
(iii) Let E be the event of getting the number on the ball is neither divisible by 5 nor by 10. Outcomes favourable to E are {1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19} Number of favourable outcomes =...
Solution: The possible outcomes are {1,2,3,4…19} Number of possible outcomes = 19 (i) Let E be the event of getting the number on the ball is a prime number. Outcomes favourable to E are...
(v) Let E be the event of getting the number on the card is divisible by 3 or 2 Outcomes favourable to E are {2,3,4,6,8,9,10,12,14,15} Number of favourable outcomes = 10 P(E) = 10/15 = 2/3 Hence the...
(iii) Let E be the event of getting the number on the card is divisible by 3. Outcomes favourable to E are {3,6,9,12,15} Number of favourable outcomes = 5 P(E) = 5/15 = 1/3 Hence the probability of...
Solution: The possible outcomes are {1,2,3,4…15} Number of possible outcomes = 15 (i) Let E be the event of getting the number on the card is odd. Outcomes favourable to E are {1,3,5,7,9,11,13,15}...
Solution: The possible outcomes are {1,2,3,4 ….25} Number of possible outcomes = 25 (i) Let E be the event of getting the number on the card is an even number. Outcomes favourable to E are...
Solution: The possible outcomes are {1,2,3,….20} Number of possible outcomes = 20 (i) Let E be the event of getting the number on the card is a prime number. Outcomes favourable to E are...
Solution: Number of integers between 0 and 100 = 99 Number of possible outcomes = 99 (i) Let E be the event of getting an integer divisible by 7. Outcomes favourable to E are...
Solution: The possible outcomes are {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p} Number of possible outcomes = 16 (i) Let E be the event of getting a vowel. Outcomes favourable to E are { a,e,i,o} Number of...
Solution: There are 7 ways in which the month of February can occur, each starting with a different day of the week. (i)Only 1 way is possible for 5 Wednesdays to occur in February with 29 days....
Solution: For a leap year there are 366 days. Number of days in January = 31 Total number of January month types = 7 Number of January months with 5 Mondays = 3 (i)Probability that the month of...
(iii) Let E be the event of arrow pointing a number greater than 2. Outcomes favourable to E are {3,4,5,6,7,8} Number of favourable outcomes = 6 P(E) = 6/8 = 3/4 Hence the probability of arrow...
Solution: The possible outcomes of the game are {1,2,3,4,5,6,7,8} Number of possible outcomes = 8 (i) Let E be the event of arrow pointing 8. Outcomes favourable to E is 8. Number of favourable...
Solution: When a die is thrown, the possible outcomes are {1,2,3,-1,-2,-3} Number of possible outcomes = 6 (i) Let E be the event of getting a positive integer. Outcomes favourable to E are {1,2,3}...
(vii) Let E be the event of getting a number between 3 and 6. Outcomes favourable to E is 4,5. Number of favourable outcomes = 2 P(E) = 2/6 = 1/3 Hence the probability of getting a number between 3...
(v) Let E be the event of getting a number less than 8. Outcomes favourable to E is 1,2,3,4,5,6. Number of favourable outcomes = 6 P(E) = 6/6 = 1 Hence the probability of getting a number less than...
(iii)Let E be the event of getting a number greater than 5. Outcomes favourable to E is 6. Number of favourable outcomes = 1 P(E) = 1/6 Hence the probability of getting a number greater than 5 is...
Solution: When a die is thrown, the possible outcomes are 1,2,3,4,5,6. Number of possible outcomes = 6 (i) Let E be the event of getting an odd number. Outcomes favourable to E are 1,3,5. Number of...
Solution: When a die is thrown, the possible outcomes are 1,2,3,4,5,6. So Sample space = { 1,2,3,4,5,6} Number of possible outcomes = 6 Even numbers are (2,4,6). Number of favourable outcomes = 3...
Solution: Total number of shirts = 100 Number of good shirts = 88 Number of shirts with minor defects = 8 Number of shirts with major defects = 4 Peter accepts only good shirts. So number of shirts...
Solution: Number of 50 paisa coins = 100 Number of 1 rupee coins = 50 Number of 2 rupee coins = 20 Number of 5 rupee coins = 10 Total number of coins = 100+50+20+10 = 180 (i) Probability of getting...
(iii)Probability of not green = Probability of getting red, white and black = (6+8+3)/22 = 17/22 Hence the probability of not green is 17/22. (iv) Probability of neither white nor black =...
Solution: Number of red balls = 6 Number of white balls = 8 Number of green balls = 5 Number of black balls = 3 Total number of marbles = 6+8+5+3 = 22 (i)Probability of white balls, = 8/22 = 4/11...
(iii) Probability of not black = Probability of white and blue = (7+8)/20 = 15/20 = 3/4 Hence the probability of not black is 3/4. (iv) Since there are no green marbles in the box, the probability...
Solution: Number of blue marbles = 7 Number of white marbles = 8 Number of black marbles = 5 Total number of marbles = 7+8+5 = 20 (i) Probability of getting black , = 5/20 = 1/4 Hence the...
Solution: Number of black balls = 5 Number of red balls = 7 Number of white balls = 3 Total number of balls = 5+7+3 = 15 (i)Probability that the ball drawn is red, = 7/15 (ii) Probability of black...
Solution: Total number of alphabets = 26 Number of vowels = 5 Total number of consonants = 26-5 = 21 Probability that the letter chosen is a consonant , = 21/26 Hence the required probability is...
Solution: Number of vowels in the word ‘TRIANGLE’ = 3 Total number of letters = 8 Probability that the letter chosen is a vowel , P(E) = 3/8 Hence the probability that the letter chosen is a vowel...
Solution: Total number of students = 40 Number of girls = 25 Number of boys = 40-25 = 15 (i) Probability of getting a girl, P(E) = 25/40 = 5/8 Hence the probability of getting a girl is 5/8. (ii)...
Solution: (i) Number of red balls = 3 Number of black balls = 5 Total number of balls = 3+5 = 8 Probability that the ball drawn is red , P(E) = 3/8 Hence the probability that the ball drawn is red...
Solution: Probability of Sania winning the match, P(E) = 0.69 Probability of Sonali winning = Probability of Sania losing, = 1-0.69 = 0.31 Hence the probability of Sonali winning is...
Solution: Given probability of winning the game, P(E) = 5/11 We know that, Probability of losing game, = 1-5/11 = (11-5)/11 = 6/11 Hence the probability of losing game is 6/11.
Solution: Number of defective pens = 12 Number of good pens = 132. Total number of pens = 132+12 = 144 Probability of getting a good pen, P(E) P(E) = 132/144 = 11/12 Hence the required probability...
Solution: Number of prized tickets = 5 Number of blank tickets = 995 Total number of tickets = 5+995 = 1000 The probability of winning a prize, P(E) P(E) = 5/1000 = 1/200 Hence the required...
Solution: Total number of screws = 600 Number of possible outcomes = 600 Number of rusted screws = one tenth of 600 = (1/10)×600 = 60 Number of remaining good screws = 600-60 = 540 Number of...
Solution: Anjali takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them. Let Y be the event ‘the ball taken out is yellow’, B be the event...
(i)Given radius of the cylinder, r = 3.5 cm Diameter of the sphere = height of the cylinder = 3.5×2 = 7 cm So radius of sphere, r = 7/2 = 3.5 cm Height of cylinder, h = 7 cm Total surface area of...
Solution: Given speed of water flow = 15 km/h Diameter of pipe = 14 cm So radius of pipe, r = 14/2 = 7 cm = 0.07 m Dimensions of cuboidal pond = 50 m × 44 m Height of water in pond = 21 cm = 0.21 m...
Solution: (i)Given surface area of the solid metallic sphere = 1256 cm2 4R2 = 1256 4×3.14×R2 = 1256 R2 = 1256/4×3.14 R2 = 100 R = 10 Hence the radius of solid sphere is 10 cm. (ii)Volume of the...
Solution: Given surface area of the sphere = 616 cm2 4R2 = 616 4×(22/7)R2 = 616 R2 = 616×7/4×22 R2 = 49 R = 7 Volume of the solid metallic sphere = (4/3)R3 = (4/3)×73 = (1372/3) cm3 Diameter of...
Solution: Given height of the cone, h = 11 cm Radius of the cone, r = 2.5 cm Volume of the cone = (1/3)r2h = (1/3)×2.52×11 = (11/3)×6.25 cm3 When lead shots are dropped into vessel, (2/5) of water...
Solution: Given radius of metallic cones, r = 2 cm Height of cone, h = 3 cm Volume of cone = (1/3)r2h = (1/3)×22×3 = 4 cm3 Radius of the solid sphere, R = 6 cm Volume of the solid sphere = (4/3)R3 =...
Solution: Given radius of the metallic sphere, R = 10.5 cm Volume of the sphere = (4/3)R3 = (4/3)×10.53 = 1543.5 cm3 Radius of cone, r = 3.5 cm Height of the cone, h = 3 cm Volume of the cone =...
Solution: Given radius of the metal cylinder, r = 14 cm Height of the metal cylinder, h = 21 cm Radius of the sphere, R = 3.5 cm Volume of the metal cylinder = r2h = (22/7)×142×21 = 22×2×14×21 =...
Solution: Given height of the circular cylinder, h = 10 cm Diameter of circular cylinder = 4.5 cm So radius, r = 4.5/2 = 2.25 cm Volume of circular cylinder = r2h = ×2.252×10 = 50.625 cm3 Base...
Solution: Given dimensions of the cuboidal solid = 9 cm× 11 cm× 12 cm Volume of the cuboidal solid = 9×11×12 = 1188 cm3 Diameter of shot = 3 cm So radius of shot, r = 3/2 = 1.5 cm Volume of shot =...
Solution: Radius of the solid circular cylinder, r = 14 cm Height, h = 12 cm Volume of the cylinder = r2h = ×142×12 = ×196×12 = 2352 = 2352×22/7 = 7392 cm3 Edge of the cube, a = 2 cm Volume of cube...
Given height of the cone, h = 20 cm Diameter of the cone = 16.8 cm Radius of the cone, r = 16.8/2 = 8.4 cm Volume of water in the vessel = (1/3)r2h = (1/3)×8.42 ×20 = (1/3)×(22/7)×8.4 ×8.4 ×20 =...
Solution; Given radius of the glass jar, R = 8 cm Diameter of the sphere = 12 cm Radius of the sphere, r = 12/2 = 6 cm When the sphere is removed from the jar, volume of water decreases. Let h be...
Solution; Given internal diameter of cylindrical can = 21 cm Radius of the cylindrical can, R = 21/2 cm Diameter of sphere = 10.5 cm Radius of the sphere, r = 10.5/2 = 21/4 cm Let the rise in water...
Solution; Given inner diameter of the well = 6 m Radius of the well, r = 6/2 = 3 m Depth of the well, H = 22 m Volume of the soil dug out of well = r2H = ×32×22 = 198 m3 Width of the embankment = 5...
Solution: Given internal diameter of hollow sphere = 4 cm Internal radius, r = 4/2 = 2 cm External diameter = 8 cm External radius, R = 8/2 = 4 cm Volume of the hollow sphere, V = (4/3)(R3-r3) V =...
Solution: Given internal radius of the tube, r = 3 cm Thickness of the tube = ½ cm = 0.5 cm External radius of tube = 3+0.5 = 3.5 cm Height of the tube, h = 21 cm Volume of the tube = (R2-r2)h =...
Solution: Given radius of the sphere, r = 6 cm Volume of the sphere = (4/3)r3 = (4/3)×63 = 288 cm3 Let r be the internal radius of the hollow cylinder. External radius of the hollow cylinder, R = 4...
Solution: Given inner diameter of the pipe = 6 cm So inner radius, r = 6/2 = 3 cm Outer diameter = 10 cm Outer radius, R = 10/2 = 5 cm Let h be the height of the pipe. Volume of pipe = (R2-r2)h =...
Solution: Given height of the cylinder, h = 2.5 mm = 0.25 cm Radius of the cylinder, r = 12 cm Volume of the cylinder = r2h = ×122×0.25 = ×144×0.25 = 36 cm3 Let R be the radius of the sphere. Volume...
Solution: Given radius of each sphere, r = 2 cm Volume of a sphere = (4/3)r3 = (4/3)×23 = (4/3)×8 = (32/3) cm3 Volume of 8 spheres = 8×(32/3) = (256/3) cm3 Let R be radius of new sphere. Volume of...
Solution: Given height of the cylinder, h = 9 cm Diameter of the cylinder = 40 cm Radius of the cylinder, r = 40/2= 20 cm Volume of the cylinder = r2h = ×202×9 = ×400×9 = 3600 cm3 Height of the...
Solution: Given dimensions of the cylindrical vessel = 22 m × 20 m Let the rainfall be x cm. Volume of water = (22×20×x)m3 Diameter of the cylindrical base = 2 m So radius of the cylindrical base =...
Solution: Radius of the sphere, r = 9 cm Volume of the sphere, V = (4/3)r3 = (4/3)××93 = 12××81 = 972 cm3 Diameter of the wire = 2 mm So radius of the wire = 2/2 = 1 mm = 0.1 cm Since the sphere is...
Solution: Given diameter of the metallic sphere = 6 cm Radius of the sphere, r = 6/2 = 3 cm Volume of the sphere, V = (4/3)r3 = (4/3)××33 = 4××9 = 36 cm3 Length of the wire, h = 36 m = 3600 cm Since...
Solution: Given height of the cylinder, h = 8 cm Radius of the cylinder, r = 3 cm Radius of hemisphere , r = 3 cm Scale = 1:200 Hence actual radius, r = 200×3 = 600 Actual height, h = 200×8 = 1600...
Given height of the cylinder, H = 13 cm Radius of the cylinder, r = 5 cm Radius of the hemisphere, r = 5 cm Height of the cone, h = 12 cm Radius of the cone, r = 5 cm Slant height of the cone, l =...
Solution; Given height of the cylinder, H = 10 cm Height of the cone, h = 6 cm Common diameter = 3.5 cm Common radius, r = 3.5/2 = 1.75 cm Volume of the solid = Volume of the cone + Volume of the...
Solution: Given common radius, r = 7 cm Height of the cone, h = 4 cm Height of the cylinder, H = 4 cm Volume of the solid = Volume of the cone + Volume of the cylinder + Volume of the hemisphere =...
Solution; Given diameter of the cylinder = 6 cm Radius of the cylinder, r = 6/2 = 3 cm Height of the cylinder, H = 12 cm Slant height of the cone, l = 5 cm Radius of the cone, r = 3 cm Height of the...
Solution: Let the radius of the dome be r. Internal diameter = 2r Given internal diameter is equal to total height. Total height of the building = 2r Height of the hemispherical area = r So height...
Let the radius of the hemisphere be r. Inner diameter = 2r Given greatest height equal to inner diameter. So total height of the hall = 2r Height of the hemispherical part = r Height of cylindrical...
Solution; Given radius of the cone, r = 3.5 cm Radius of hemisphere, r = 3.5 cm = 7/2 cm Volume of hemisphere = (2/3)r3 = (2/3)×(22/7)×(7/2)3 = (2/3)×(22/7)×(7/2)×(7/2)×(7/2) = (22/3)×(7/2)×(7/2) =...
Solution: Given radius of the hemisphere, r = 5 cm Radius of cone, r = 5 cm Height of the cone, h = 7 cm Volume of the solid = Volume of the hemisphere + Volume of the cone = (2/3)r3 + (1/3)r2h =...
Solution; (i) Given height of the rocket = 26 cm Height of the cone, H = 6 cm Height of the cylinder, h = 26-6 = 20 cm Diameter of the cone = 5 cm Radius of the cone, R = 5/2 = 2.5 cm Diameter of...
Solution; Given height of the cylinder, H = 30 cm Radius of the cylinder, r = 7 cm Height of cone, h = 24 cm Radius of cone, r = 7 cm Slant height of the cone, l = √(h2+r2) l = √(242+72) l =...
Given diameter of the cylindrical part of tent, d = 24 m Radius, r = d/2 = 24/2 = 12 m Height of the cylindrical part, H = 11 m Since vertex of cone is 16 m above the ground, height of cone, h =...
Given radius of the cone, r = 3.5 cm Radius of hemisphere, r = 3.5 cm Total height of the toy = 15.5 cm Height of the cone = 15.5 – 3.5 = 12 cm Slant height of the cone, l = √(h2+r2) l = √(122+3.52)...
Solution: Given height of the cylinder, h = 10 cm Radius of the cylinder, r = 3.5 cm Radius of the hemisphere = 3.5 cm Total surface area of the article = curved surface area of the cylinder +...
Given edge of the cube, a = 7 cm Diameter of the hemisphere, d = 7 cm Radius, r = d/2 = 7/2 = 3.5 cm Surface area of the hemisphere = 2r2 = 2×(22/7)×3.52 = 44×12.25/7 = 539/7 = 77 cm2 Surface area...
Solution: Dimensions of the cuboid = 15 cm× 10 cm × 3.5 cm Volume of the cuboid = 15×10×3.5 = 525 cm3 Radius of each depression, r = 0.5 cm Depth, h = 1.4 cm Volume of conical depression = (1/3)r2h...
Solution: Given dimensions of the box = 16 cm ×8 cm ×8 cm So volume of the box = lbh = 16×8×8 = 1024 cm3 Radius of the glass sphere, r = 2 cm Volume of the sphere = (4/3)r3 = (4/3)×(22/7)×23 =...
Given dimensions of the block of wood = 20 cm × 10 cm× 10 cm Volume of the block of wood = 20×10×10 = 2000 cm3 [Volume = lbh] Diameter of the cone, d = 10 cm Radius of the cone , r = d/2 = 10/2 = 5...
Solution: Given side of the cube, a = 28 cm Radius of the cylinder, r = 28/2 = 14 cm Height of the cylinder, h = 28 cm Volume of the cube = a3 = 283 = 28×28×28 = 21952 cm3 Volume of the cylinder =...