REVISION - X
SIMILARITY
1) In the adjoining figure, LM is parallel to BC, AB =6cm. AL= 2cm and AC=9cm. Find
a) the length of CM. 6cm
b) the value of Area of ∆ALM/Area of trapezium LBCM. 1/8
2) In figure, ABCD is a trapezium with AB || DC and AB= (1.2) DC.
Prove that O is the point of trisection of diagonals AC and BD.
3) In the figure alongside PQRC is a parallelogram PQ =16cm, QR=10cm. L is a point on PR such that RL : LP = 2:3. QL produced meets RS at M and PS produced at N.
a) prove that ∆RLQ ~ ∆ PLN. Hence, find PN. 15
b) Name a triangle similar to triangle RLM. Evaluate RM as a fraction. ∆PLQ, 32/3 cm
4) In the figure, DE || BC.
a) Prove that ∆ ADE and ABC are similar.
b) Given that ADA= (1/2) BD, calculate DE, if BC= 4.5 cm. 1.5cm
5) In a ∆ PQR , L and M are two points on the base QR such that angle LOQ= angle QRP and angle RPM= angle RQP, Prove that
a) ∆ PQR ~ ∆ RPM
b) QL. RM = PL. PM
c) PQ²= QR. QL.
6) In the figure, the median BD and CE of a ∆ ABC meet at G. Prove that
a) ∆ EGD ~ ∆ CGB
b) BG = 2GD
7) In the figure, BC || DE, area of ∆ ABC= 25 cm², area of trapezium BCED= 24cm², DE= 14 cm.
calculate the length of BC . 10cm
8) in the figure, P is a point on AB such that AP: PB= 4:3. PQ is parallel to AC.
a) Calculate the ratio PQ: AC, giving reasons for your answer.
b) In ∆ ARC, angle ARC= 90° and in ∆ PQS, angle PSQ = 90°.
Given QS = 6cm. Calculate the length of AR. 3:7
9) In the figure, ABCD is a parallelogram. P is a point on BC such that BP: PC= 1:2, DP produced meet AB produced at Q. Given area of triangle CPQ = 20cm², calculate
a) area of the triangle CDP . 40cm²
b) area of the parallelogram ABCD. 120cm²
10) In the figure, ABC is a triangle, DE || BC and AD/DB = 3/2.
a) Write down AD/AB. 3/5
b) Prove that ∆ ADE ~ ∆ ABC
And write down the ratio of DE/BC. 3/5
c) Prove that ∆ DEF ~ ∆ CFB.
Write down the ratio of Area of ∆ DFE/Area of ∆ DEC. 3/8
11) Calculate the remaining sides of a triangle whose shortest side is 5 cm and which is similar to a Triangles with sides 4cm, 7cm, 8cm. 8.75,10,
12) A vertical pole 4m high casts a shadow 1.6m long at the same moment as a flagstaff casts a shadow 12m long. Calculate the height of the flagstaff. 30m
13) In a ∆ ABC , D and E are points on AB and AC respectively such that DE|| BC , Prove that ∆ ADE ~ ∆ ABC.
If BC= 12cm, BD= 7cm and AD= 8cm, find the length of DE. 6.4cm
14) ∆ ABC ~ ∆ PQR and angle C =90°, if AC= PQ= 5cm and QR= 4cm, calculate the length of AB and BC. 17/2, 20/3
15) In a ∆ ABC, D is a point on BC such that angle BAD = angle ACB, AB= 6cm, AD = 5cm and BD= 4cm. If CD= x and AC = y, find x and y. 5, 6
16) In the following figure, Find
a) AB. 2.5
b) BC. 6
17) In the figure, AB and CD intersect at X. If AX/BX= DX/CX. prove that AD|| BC.
18) The perimeters of two similar triangles are 40cm and 25cm. If a side of the first triangle is 15cm, find the length of the corresponding side of the second Triangle. 75/8 cm
19) in the figure, angle RPS= angle SRQ. Prove that ∆ PQR ~ ∆ RQS.
20) Two isosceles triangle have their vertex angle equal. Prove that the triangle are similar.
21) The perimeter of two similar triangles ABC and PQR are respectively 32cm and 24 cm. If PQ = 12cm. find AB. 16cm
22) D is a point on side BC of ∆ ABC such that angle ADC and angle BAC are equal. Prove that CA² = DC . CB.
23) In figure, ST || QR, PS= 4cm, QS= 5cm and PR= 8cm. Find
a) PT. 32/9cm
b) TR. 40/9 cm
c) QR/ST. 8/4
d) Area of ∆ PQR/Area of ∆ PST. 81/16
24) If the areas of two similar triangles are equal, prove that they are congruent.
25) ∆ ABC ~ ∆ DEF, area of ∆ ABC= 25cm² and area of ∆ DEFA = 36cm², if EF= 7.2 cm, find the length of 2 BC . 6cm
26) Two isosceles triangle have equal vertical angles and their areas are in the ratio 4:9. Find the ratio of their heights. 2:3
27) In figure , DE|| BC and BD|| FE, AD = 5cm, DB= 6cm and AE= 3cm, calculate
a) EC. 3.6cm
b) area of ∆ ADE: area of ∆ EFC. 25:36
c) area of ∆ ADE: area of ∆ ABC. 25:121
d) area of ∆ EFC: area of trapezium BFEA. 3;8
28) ∆ ABC ~ ∆ DEF. If BC= 4cm, EF= 5cm area of ∆ ABC= 32 cm², find the area of ∆ DEF. 50cm²
29) In a triangle ABC, D and E are points on AB and AC such that DE || BC. If DE: BC= 4:5, Calculate the ratio of the areas of ∆ ADE and trapezium BCED . 16:9
30) In the figure, D and E are midpoints of AB and AC respectively. Find
a) the ratio of the areas ∆ ADE and ∆ ABC. 1:4
b) the ratio of the areas of ∆ ADE and trapezium BCED . 1:3
c) the ratio of the areas of ∆ ABC and trapezium BCED. 4:3
31) if one diagonal of a trapezium divides the other diagonal in the ratio 1:2. Prove that one of the parallel sides will be half of the other.
32) P and Q are points on the sides AB and AC respectively of ∆ ABC such that PQ|| BC and divides ∆ ABC into two parts, equal in area. Find PB: AB. (√2-1):√2
33) X, Y and Z are mid points of the sides of PQ, QR, and PR respectively of ∆ PQR. If the area of ∆ PQR= 64 cm², find the area of ∆ XYZ. 16
34) PQRS is a parallelogram M is the midpoint of PS. RM intersect SQ in X. RM produced to meet QP produced at Y. Prove that XY= 2RX.
35) ABCD is a parallelogram. A straight line is drawn through A meeting straight line BD at L and BC at M and DC produced at N. Prove that AL/LN = AM/AN.
36) in4 figure AB|| DC, area ∆ OAD = 12 cm² and area ∆ BCD= 18cm². Calculate
a) area ∆ OCD. 6cm²
b) OB: OD. 2:1
c) area ∆ OAB. 24cm²
37) ABCD is a parallelogram. E is any point on BC. AE and BD intersect at O.
Prove that AO. OB = OD. OE.
38) ABCD is a trapezium in which AD || BC and BD intersect at E. If ∆ AED~ ∆ BEC, prove that AB = CD.
39) In figure , ∆ ABCD is a square ∆ AEB and AFC are equilateral triangles . Prove that area of ∆ AEB = (1/2) area of ∆ AFC.
40) PQR is an equilateral triangle. The base QR is produced both ways to X and Y such that QR = QX= RY. If PX and PY are joined, show that the triangles PQX and XPY are equiangular and hence, prove PX²= 3QR².
41) AF is the median of ∆ ABC. FD and FE are bisectors of angle AFB and angle AFC meeting AB and AC at E respectively. Show that DE || BC.
42) in the figure, ABC is a right triangle with angle ABC= 90°. BD perpendicular AC, DM perpendicular BC and DN perpendicular AB. Prove
a) DM²= DN. MC
b) DN²= DM. AN
43) On a map drawn to a scale 1:2500, a rectangular plot of land ABCD has the following measurements . AB =12 cm, BC=16 cm. Angles A,B,C,D are all 90° each. Calculate
a) the diagonal distance of the plot in km. 5
b) the area of the plot in km². 12
44) The model of a ship is made to a scale 1:250. Find
a) the length of the ship , if the length of its model is 1.2m. 300m
b) the aea of the deck of the ship, if the area of the decy of its model is 1.6 m². 100000 m²
c) the volume of its model, when the volume of the ship is 1 cubic kilometres. 64m³
45) what a mad drone is still one is to 250000 triangular plot of land as the following measurement 3493 that's why length of the area of the plot the scale of the map one stupid 1000 upload a plan area 20 to be represent on the map find the number of the kilometres in the ground is represent by on the map the area in that can represented by the map of represent the plot of land is made to scale 1:200 the length of the model is calculated system the volume of the model 200 calculate the volume to 6 which sites to weight 40 medicine larged such that smallest side is 12 find the scale fashion used to find the length of the sides of the image Triangle with size 12 21 10.5 is reduced such that the longest side is 7 find the scale factor and use it to find the length of the other sides area 90 it is reduced by the scale factor of 23 find the area Trianglesquare roots are trees and large scale factor what is the rear the image of the square a rectangular website 85 red due to the scale factor 0.5 find the area of the image of the rectangle two similar jugs of height 4 and 6 respectively the capacity of the smaller journalist 48 find the capacity of the largest of a model of CP is made up scale 1 institute 200 the length of the model is 5 calculate the length of the C they are the Deck of the city is 18 find the area of the day of the model of the volume of the modern history of 5000 the length of on the map the area of the lamb scale of a map is one is to 2000 a plot of land area 15 is to be represented on the map find the length of the on ground represent by on the map area that can be represent on the map the area on the map represent the plot of land the dimension of the model of the building are 1.575 1.75 if the model is on to scale 1 is 280 find the actual time in the building also calculate the floor in the model 200 the volume of the room in the modelthe length of the model is 4.5 calculate the length of the sea area of the Deck of the 625000 calculate theory of the day of the model the volume of the model is 150 calculate the volume of the sheep in litres in a scale drawing a building one of the drawing equals to actual building rectangular Courtyard at the building calculate the length of the breath angular present the courtyards in the scale drawing the building area in the drawing is 3 calculate theory of the wall in the building
TRIGONOMETRY
1) (cos³x + sin³x)/(cosx + sinx)+ (cos³x - sin³x)/(cosx - sinx)= 2.
2) cosA/(1- tanA) + sinA/(1- cotA)= cosA + sinA.
3) sinA/(cotA + cosecA) = 2+ sinA/(cotA- cosecA)
4) sinx/(1+ cosx) + (1+ cosx)/sinx= 2cosecx.
5) √{(1+ cosx)/(1- cosx)}= cosecx + cotx.
6) 1/(sinx + cosx) + 1/(sinx - cosx)= 2sinx/(1- 2cos²x)
7) 1 - cos²θ/(1+ sinθ)= sinθ.
8) √{(1- cos θ)/(1+ cosθ)= sinθ/(1+ cosθ)
9) sin²θ(1+ cot²θ)= 1
10) (1- tan²θ)cos²θ= 1.
11) cot²A - 1/sin²A= 1
12) tan²θ/(secθ -1)= 1+ secθ
13) (secx -1)/(sex +1)= (1- cosx)/(1+ cosx).
14) cos²x + 2sinx cosx + sin²x = 1+ 2sinx cosx.
15) (sinx tanx)/(1- cosx)= 1+ secx.
16) (1+ cosx)(1- cosx)(1+ cot²x)= 1.
17) 1/(1+ sinx) + 1/(1- sinx)= 2sec²x
18) cosecx/(cosecx -1) + cosecx/(cosecx +1)= 2sec²x
19) sec²θ+ cosec²θ= sec²θ cosec²θ
20) tan²x - sin²x = tan²x sin²x.
21) sin⁴θ + sin²θ cos²θ = sin²θ
22) sin⁴x cosec²x + cos⁴x sec²x =1.
23) tan⁴x + tan²x = sec⁴x - sec²x.
24) cos⁴θ - sin⁴θ = 2 cos²θ -1.
25) (secx - cosx)(secx + cosx)= sin²x + tan²x
26) sec²x cosec²x = tan²x + cot²x +2.
27) 1/(cosecx + cotx)= cosecx - cotx
28) (secx + tanx)/(secx - tanx)= {(1+ sinx)/cosx}².
29) tan²θ/(secθ -1)² = (1+ cosθ)/(1- cosθ).
30) sinx/(1- cosx)= cosecx + cotx.
31) sin²x/(sinx - cosx) + cosx/(1- tanx)= sinx + cotx.
32) (secx - tanx)/(secx + tanx)= 1- 2secx tanx + 2 tan²x
33) tanA/(1- cotA) + cotA/(1- tanA)= secA cosecA +1.
34) (sinx - 2sin³x)/(2cos³x - cosx)= tanx.
35) (cotx + tany)/(cot y+ tanx)= cot x tany
36) sin²x cos²y - cos²x sin²y = sin²x - sin²y.
37) tan²x sec²y - sec²x tan²y = tan²x - tan²y.
38) √(secx + tanx) √(secx - tanx)= 1
39) √(sin²x+ cos²x + tan²x) = secx.
40) (cosA cosecA - sinA secA)/(cosA + sinA)= cosecA - secA
41) √{(1+ cosx)/(1- cosx) + √{(1- cosx)/(1+ cosx) = 2 cosecx.
42) √{(secx -1)/(secx +1) + √{(secx +1)/(secx -1) = 2 cosecx
43) (sinx + cosx)/(sinx - cosx) + (sinx - cosx)/(sinx +cosx) = 2/(1- 2 cos²x).
44) (tan²x - sec²x)/(cot²x - cosec²x)= 1.
45) (tanx + 1/cosx)²+ (tanx - 1/cosx)²= 2 {(1+ sin²x)/(1- sin²x)}
46) (cotx + cosecx -1)/(cotx - cosecx +1)= (1+ cosx)/sinx.
47) 1/(cosecx - cotx) - 1/sinx = 1/sinx - 1/(cosec x + cotx).
48) sin⁶x + cos⁶x = 1- 3 sin²x cos²x.
49) If sinx + cosx = m and secx + Cosecx = n then show n(m²-1)= 2m.
50) If x= r cosa sinb , y= r cos a cos b and z= r sin a, show x²+ y²+ z²= r²
51) If m= tanx + sinx and n= tanx - sinx show m²- n²= 4 √(mn).
52) m= cosx/cos y and n= cosx/sin b, show (m²+ n²) cos²y = n².
53) If a secx + b tanx = m and a tanx + b secx = n, show that m²- n²= a²- b²
54) If cosx + sinx =√2 cosx, show cosx - sinx =√2 sinx.
1) If 2cosx = 2/5, find sinx.
2) If sinx =3/5 and cosy = 12/13, evaluate
a) sec²x
b) tanx + tany.
3) If 2 sinx -1=0, show that sin3x = 3sinx - 4 sin³x
4) Given 5cosx - 12x = 0 then find (sinx + cosx)/(2cosx - sinx).
5) Given A is an acute angled and 13 sinA = 5, evaluate (5sinA- 2cosA)/tanA.
6) If cosx = 5/13, find sinx, tanx. Cotx, cosecx.
7) If sinx = 4/5 then find the value of secx + cotx
8) If 3tanx = 4, find the value of (sinx tanx -1)/2tan²x.
9) If tany = 3/4, find the value of (4siny - 2cosy)/(4siny + 3cos y).
10) If 2 sinx = 1, find(tanx + cotx)².
11) If cosx = 1/25, show (1- cos²x)/(2- sin²x)= 3/5.
12) Solve:
a) sinx = cosx.
b) 2sin²x = 1/2.
c) 2cos²x -1=0
d) 4tan²x = 12.
1) tan10 tan20 tan30 tan70 tan80= 1/√3
2) sinx/sin(90-x) + cosx/cos(90-x)= sec(90- x) cosec(90- x)
3) (sin²20+ sin²70)/(cos²20+ cos²70) + (sin(90-x) sinx)/tanx + (cos(90-x) cosx)/cotx= 2.
4) 14 sin30+ 6 cos60 - 5 tan45
5) cos75/sin15 + sin12/cos78 - cos18/sin72.
6) 3 cos80 cosec10 + 2 cos59 cosec31
7) sec²27- cot²63.
8) sin(90- x) cos(90- x)= tanx/(1+ tan²x).
9) 1- (cosx cos(90- x))/cotx = cos²x.
10) cosx cos(90-x)- sinx sin(90- x)= 0
11) sin(90-x) cos(90- x)= tanx cos²x.
12) cos(65+ x)- sin(25- x)= 0
13) sin(50+ x) - cos(40- x)= 0
14) (sin(90-x) cos(90-x))/tanx = 1- sin²x.
15) sin(90- x) cos(90- x)= tanx/(1+ cot²(90- x)).
16) (sinx cos(90-x) cosx)/sin(90-x) + (cosx sin(99- x)/cos(90- x)= 1
17) cosec²(90- x) - tan²x = cos²(90- x)+ cos²x.
18) cot(90- x)/(1+ tan(90-x)) + tan(90-x)/(1- cot(90-x))= secx cosecx +1.
19) (cos(90-x) sec(90- x) tan(99-x) cot(90-x))/(sun²(90-x)+ cos²(90-x))= 1
20 If A, B, C interior angles of ∆ ABC prove that
a) sin{(B+ C)/2}= cos(A/2).
b) tan {(A+ B)/2} = cot(A/2)
TYPE -1
TYPE -1
1) 8x²+15=26x. 5/2,3/4
2) x(2x +5)=25. -5, 5/2
3) (x -3)/(x+3) + (x +3)/(x-3) =5/2, x≠-3, x≠ 3. 9,-9
4) 2x -3= √(2x²- 2x +21. 6
5) (x²-5x)/2= 0. 0,5
6) 3x +35= 2x². 5, -7/2
7) 6x²+ x = 35. -5/2,7/3
8) 6x(3x -7)= 7(7- 3x). 7/3,-7/6
9) 3(y⅖-6)= y(y+7)-3. 5, -3/2
10) x²- 4x -12=0, when x belongs to N. 6
11) 2x²- 9x +10=0, when a) x∈N b) x ∈Q. 2, 2 and 5/2
12) a²x²+ 2ax +1=0, a≠ 0. -1/a, -1/a
13) 1/x - 1/(x +2)= 1/24. -8,6
14) (x +1)/(x-1) = (3x -7)/(2x-5). 3,4
15) (3x +1)/(7x+1) = (5x +1)/(7x+5). 1,-2/7
16) 5/(2x +1)+ 6/(x+1) = 3. 2,-2/3
17) (x +3)/(x-2) - (1- x)/x = 17/4. 4,-2/9
18) a/(ax -1)+ b/(bx-1) = a+ b, a+ b≠ 0, ab≠ 0. (a+ b)/ab, 2/(a+ b)
19) 2²ˣ⁺³ - 9. 2ˣ +1= 0. 0,-3
20) Frame the quadratic equation whose roots are:
a) -2,1. x²+ x -2=0
b) -3,-4. x²+7x +12=0
c) a,-b. x²-(a - b)x - ab=0
d) -2/3, 4/5. 15x²- 2x -8=0
e) -3,2/5. 5x²+ 13x -6=0
TYPE -2
1) 3x²- x -7= 0 (two decimal places. 1.70 or -1.37
2) 1/(x +1) + 2/(x+2) = 4/(x +4).
3) 3/(x+1) - 1/(x +2) = 1/(x+3). Two decimal places. -2.27 or -5.73
4) 2(3x²-1)= x. 2/3,-1/2
5) (x +3)/(2x+3) = (x +1)/(3x+2). (-3+√6),(-3-√6)
6) (x -2)/(x+2) + (x +2)/(x-2) = 4. 2√3,-2√3
7) 1/(x +2)+ 2/(x+2) = 4/(x +4). (2+2√3),(2-2√3)
8) a(x²+1)= (a²+1)x, a≠ 0. a, 1/a
9) 4x²- 4ax - (a²- b²)= 0. (a+b)/2, (a- b)/2
TYPE -3
1) An Aeroplane travelled a distance of 400 km at an average speed of x kmph . On the return journey the speed was increased by 40 kmph. Write down an expression for the time taken for:
a) the onward journey
b) the return journey
If the return journey took 30 minutes less than the onward journey, write an equation in x and find the value of x.
2) Ca A travels x km for every litres of petrol, while car B travel (x+5) km for every litre of petrol.
a) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.
b) If car A uses 4 liters of petrol more than car B in covering the 400 km. Write down an equation in terms of x and solve it to determine the number of liters of petrol used by car B for the journey.
3) In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find :
a) the number of a rows in the original arrangement.
b) the number of seats in the auditorium after rearrangement.
4) a hotel bill for a number of people for overnight stay is Rs4800. If there were 4 people more, the bill each person had to pay would have a reduced by Rs200. Find the number of people staying overnight.
5) A trader buys x articles for a total cost of Rs600.
a) write down the cost of 1 articles in terms of x.
If the cost per article were Rs 5 more, the number of articles that can be bought for Rs600 would be four less.
b) write down the equation in x for the above situation and solve it for x.
6) A train covers a distance of 600 km at x kmph . Had the speed been (x+20) kmph, the time taken to cover the distance would have been reduced by 5 hours. Write down an equation in terms of x and solve it to evaaluate x.
7) The sum of two number is 18 and their product is 56. Find the numbers .
8) The cost of 2x articles is Rs(5x + 54) while the cost of (x + 2) articles is Rs(10-4). Find x.
9) The difference of the square of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.
10) There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers.
11) A years ago the father was 8 times as old as his son. now his age is the square of his son's age. Find their present ages.
12) The length of a rectangle exceeds its width by 8cm and the area of the rectangle is 240 sq.cm. Find the dimensions of the rectangle.
13) The area of right angled triangle is 600 sq.cm. if the base of the triangle exceeds the altitude by 10cm, find the dimension of the triangle.
14) The length of the hypotenuse of a right-angle exceeds the length of the base by 2cm and exceeds twice the length of the attitude by 1cm. Find the length of each side of the triangle.
15) Two squares have sides x cm and (x + 5)cm. The sum of their areas is 697sq.cm.
a) Express this as an algebraic equation in terms of x.
b) solve this equation to find the sides of the squares .
16) The sum of the numerator and denominator of a fraction is 8. If 1 is added to both the numerator and denominator, the fraction is increased by 1/15. Find the fraction.
17) Rs 6500 were divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs30 less. Find the original number of persons .
18) A piece of cloth costs Rs200. If the piece were 5 m longer, and each metre of cloth cost Rs 2 less, the cost of piece would have remained unchanged . How long is the piece and what is its original rate per metre ?
19) The length of a rectangle is 8 metres more than its breadth and its area is 425m².
a) taking x metres as the breadth of the rectangle, write an equation in x that represents the above statement.
b) solve the above equation and find the dimensions of the rectangle.
20) The perimeter of a rectangle plot of land is 114 m and its area is 810 square metre.
a) take the length of the plot as x metres . Use the perimeter and 114m to write the value of the breadth in terms of x.
b) use the values of length, breadth and area to write an equation in x.
c) solve the equation to find the length and breadth of the plot.
21) Rs6400 were divided equally among x persons. Had this money been divided equally among (x +14) persons, each would have got Rs28 less. Find the value of x.
22) A shopkeeper buys x books for Rs720.
a) write the cost of one book in terms of x.
b) if the cost per book were Rs5 less, the number of books that can be bought for Rs720 were 2 more. Write down the equation in x for the above situation and solve it to find x.
23) A fruit bought x apples for Rs1200.
a) write the cost price of each apple in terms of x.
b) if 10 of the apples were rotten and he sold each of the rest of Rs3 more than the cost price of each, write the selling price of (x-10) apples.
c) if he made a profit of Rs60 in this transaction, from an equation in x and solve it to find x.
24) An aeroplane flying with a wind of 30kmph takes 40 minutes less to fly 3600 km, than what would have taken to fly against the same wind. Find the plane's speed of flying in still air.
25) Some students planned a picnic. The budget for food was Rs500. But 5 of these failed to go and thus the cost of the food for each member increased by Rs5. How many students attended the picnic ?
26) One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill it in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
27) In a flight of 2800km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 100 kmph and timing increased by 30 minutes. Find the original duration of light.
28) Two circles touch externally. The sum of their areas is 130π cm² and the distance between their centres is 14cm. Find the radii of the circles.
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