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) 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