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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 4701. |
8. Find the area ol e that can be inscribed In a circle of ra9. The diameter of a circular park is 24 m. A 3 m wide paths around it from inside. Find the cost of paving therunpath at R 25 per square meter.naner is in the form of a rectangle PQRS in which |
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Answer» Let the diameter of the circular park ( D ) = 24 m Radius of the park ( R ) = D/2 R = 24 /2 = 12 m If the width of the path inside the park ( w ) = 3m Radius of the inside circle (r ) = R - w r = 12 - 3 = 9 m Area of the path ( A ) = πR² - πr² A = π(R + r ) (R - r ) = π ( R + r ) w = ( 22 /7 ) ( 12 + 9 ) ( 3 ) = 22 × 3 × 3 A = 198m² Cost of the paving the per square meter = Rs 25 Cost of paving total area of the path =25 × 198 = Rs4950 |
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| 4702. |
6 A-4cos- coSIf sinA +sin3A-cos2A, prove that cos6A-4cos4A +8cos2A 4 |
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Answer» Given, sinA+sin^3A =cos^2A , squaring both side we get (sinA+sin^3A)^2 = (cos^2A)^2 sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2=cos^4A sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2 = cos^4A we have,1-sin^2A =cos^2A;[using(a-b)^2=a^2-2ab+b^2(a-b)^3=a^3 - 3a^2 *b+3a*b^2 -b^3] (sin^3A)^2 = (sin^2A)^3=(sin^6) by using this we get,sin^2A +2(sin^2A)^2 +(sin^2A)^3 = cos^4A 1-cos^2A +2(1-cos^2A)^2 +(1-cos^2A)^3 = *cos^4A 1-cos^2A +2(1–2cos^2A +cos^4A)+(1–3cos^2A +3cos^4A -cos^6A)=cos^4A 1-cos^2A + 2–4cos^2A +2cos^4A +1–3cos^2A +3cos^4A -cos^6A -cos^4A=0 4–8cos^2A +4cos^4A -cos^6A so we rearranging the term we get –8cos^2A + 4cos^4A - cos^6a = -4 cos^6A -4cos^4A+8cos^2A = 4 |
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| 4703. |
The diameter of a circular park is 24 m. A 3 m wide patruns around it from inside. Find the cost of paving thepath at R 25 per square meter. |
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| 4704. |
2. In Fig. a diameter ABof a circle bisects achord PQ. If AQ || PB,prove that the chordPQ is also a diameterof the circle. |
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Answer» 23384555555555555555333 |
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| 4705. |
Diagonal AC of a parallelogram ABCD bisectsA (see Fig. 8.19). Show thatit bisects Z C also,ABCD is a rhombus. |
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| 4706. |
Diagonal AC of a parallelogram ABCD bisects<A (see Fig. 819). Show that() it bisects ZC also,(i) ABCD is a rhombus6. |
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| 4707. |
Fig. 7.505.In Fig 7.51, PR> PQ and PS bisects 4 QPR. Provethat 2 PSR >PSQ.ナQ2Fig. 7.51 |
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| 4708. |
Diagonal AC of a parallelogram ABCD bisectsZ A (see Fig. 8.19). Show that(i) it bisects C also,(ii) ABCD is a rhombus. |
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| 4709. |
In Fig. 7.51, PR PQ and PS bisects 4 QPR. Provethat 2 PSR PSQFig. 7.51 |
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| 4710. |
Prove that\sin ^{2} \pi / 8+\sin ^{2} 3 \pi / 8+\sin ^{2} 5 \pi / 8+\sin ^{2} 7 \pi / 8=2 |
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Answer» Sin²π/8+sin²3π/8+sin²5π/8+sin²7π/8=sin²π/8+sin²3π/8+sin²(π/2+π/8)+sin²(π/2+3π/8)=sin²π/8+sin²3π/8+cos²π/8+cos²3π/8=(sin²π/8+cos²π/8)+(sin²3π/8+cos²3π/8) [∵, sin²Ф+cos²Ф=1]=1+1=2 |
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| 4711. |
In the below figure, PR > PQ and PS bisectsQPR. Prove that LPSR > LPSQ.Q4 |
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| 4712. |
prove that(sin^2)A×(cos^2)×B-(cos^2)A×(sin^2)×B=(sin^2)A-(sin^2)×B |
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Answer» L.H.S. sin²a cos²b -cos²Asin²B SIN²A(1-sin²B)-(1-sin²A)sin²B sin²a -sin²Asin²B -sin²B +sin²Asin²B sin²a -sin²B R.H.S. PROVED |
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| 4713. |
Prove that \sin (A+B) \sin (A-B)=\sin ^{2} A-\sin ^{2} B |
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| 4714. |
Prove that(0 cos 15- sin 15°-(Gi0 cot 105 tan 1052atan + tan o-tan 60e tan 60"Prove thats 9-sinPhove that C sinProve that "cos (π+t)) cos (-0)tan 37".cos8% singeove that sin (9(p+8) +sin(180° + θ)sin() tan 00+0cot θsin(180+ 0)0+0) tan (270-0)cotsin (360°-6) oos (360" + θ) cosec (-0) sin (2:Prove thatreand . lie in the first quadrant such that sin 0-i,athe values of17(i) sin (8-0), (ii) cos (0+0), (iii) tan (t)-a),If χ and y are acute angles such that sin x-and sinyv5xand y are acute angles such that cos xand cow14 |
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Answer» please crop the question u need answer for |
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| 4715. |
Prove that:\sin ^{8} x-\cos ^{8} x=\left(\sin ^{2} x-\cos ^{2} x\right)\left(1-2 \sin ^{2} x \cos ^{2} x\right) |
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Answer» ty ty ji so much 7052862605 jiiiiiiiiiiiiii |
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| 4716. |
2 If y = \sin x + /sin x + \sin x +.. to s., prove thatsin x +COS Xsin x +... to co, prove thatdx 24-1 |
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| 4717. |
Prove thatsin A - sin 3A / sin ^ 2A - cos^2 A = 2 sin A |
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| 4718. |
Prove that \frac{\sin A \cdot \sin 2 A+\sin 3 A \cdot \sin 6 A}{\sin A \cdot \cos 2 A+\sin 3 A \cdot \cos 6 A}=\tan 5 A |
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| 4719. |
3. D and E are points on the sides AB and AC respectively of a AAeach of the following cases, determine whether DE BC or not(i) AD-5.7 cm, DB = 9.5 cm, AE = 4.8 cm andEC 8 cm.(ii) AB= 11,7 cm, AC = 11.2cm, BD=65cm andAE 4.2 cm.(iii) AB 10.8 cm, AD -6.3 cm, AC 9.6 cm ande(iv) ADEC 4 cm.7.2 cm, AE 6.4 cm, AB-12 cm and AC-1 |
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| 4720. |
Determine whether the given points |
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Answer» i) slope AB = (-0.5-2)/(1-0) = -2.5 slope BC = (-3+0.5)/(2-1) = -2.5 since slopes are equal so points are collinear. |
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| 4721. |
1.Determine whether 210 and 55 are co-prime or not. |
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Answer» For two numbers to be co-prime, 1 has to be the only positive integer to divide the numbers. But 5 divides both 55 and 210. Hence, these numbers are not co-prime. PLEASE HIT THE LIKE BUTTON |
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| 4722. |
11.Determine whether 210 and 55 are co-prime or not. |
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Answer» Given numbers 210 and 55 are co-prime Step-by-step explanation: Given two numbers 210 and 55 we have to find whether these are co-prime or not. Two integers x and y are said to be coprime (also spelled co-prime) if the only positive integer that divides both of them is 1 i.e the GCD(Greatest common divisor) is 1There are no common factors therefore The greatest common divisor is 1 i.e GCD(210,55)=1 Hence, given numbers 210 and 55 are co-prime |
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| 4723. |
EXERCISE 9.215, ABCD is a parallelogram, AE丄DCAD. If AB = 16 cm, AE = 8 cm andcm,find ADG and H are respectively the mid-points ofs of a parallelogram ABCD, show thatD EH) ar (ABCD)Fig. 9.12 |
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| 4724. |
In Fig. 9.15, ABCD is a parallelogram, AE DCand CF丄AD. If AB = 16 cm, AE = 8 cm andCF 10 cm, find AD1.If E,FG and H are respectively the mid-points ofthe sides of a parallelogram ABCD, show thatar (EFGH)ar (ABCD). |
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| 4725. |
Determine whether (3x -2) is a factor of 3x +x -20x12. |
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| 4726. |
ABCD is a parallelogram. The circle through A, B and C intersects (produce if necessary) at E.Prove that AE AD. |
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Answer» thank you for giving this answer |
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| 4727. |
In Fig. 9.15, ABCD is a parallelogram, AE L DCand CF 1 AD. IfAB 16 cm, AE 8 cm andCF 10cm, find ADL.2 IFE.FGand H are respectively the mid-points ofthe sides of a parallelogram ABCD, show that D E |
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| 4728. |
L. Inlig 9 15, ABCD is a parallelogram, AE L DOand CF J AD IHAB 16 cm, AE a 8 cm and2CFu 10cm, find ADEF.G and H are respectively the mid-points ofthe sides of a parallelogram ABCD, show thatueDI |
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Answer» 1)hit like if you find it useful |
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| 4729. |
e iguranu depm are respectively 2.3 m and ⅠU m.6.A village, having a population of 4000, requires 150 litres of water per head perda tank measuring 20 m x 15 m x 6 m. For how many days will the waterper day.Itof thishaslast? |
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| 4730. |
The sides of a triangle are given below. Determine which of them are the sides ofthe righttriaa) a 6 cm, b 8 cm, c 10 cm(b) a 2.5 cm, b- 6.5 cm, c 6 cm |
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Answer» For a right angled triangle, the sum of squares of two smaller sides is equal to the square of the larger side. a) 8²+6²= 64+36= 100 = 10². So, this is a right angled triangle. b) (2.5)²+6² = 6.25+36= 42.25, which is not equal to (6.5)². So, this is not a right angled triangle. Please hit the like button if this helped you. |
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| 4731. |
in Fig. 9, 15, ABCD is a parallelogram, AE丄DCand CF L AD. If AB 16 cm, AE 8 cm andCF 10 cm, find ADA |
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| 4732. |
A, B and C can do a piece of work in 6 da12 days and 24 days respectively. In what twill they all together do it ? |
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| 4733. |
2kh] 0T (¥) 2khy 61 (£)2khy 81 () 2kiy L1 (1)il ityb b L2E क४ Pl Lk घर b 22l 79 Lk Yook 2Bp2e B D 85 | fh Bl 220 T b 2khy DIab Mg Lk 12li § b kb | h kil फेति 2-5 |
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| 4734. |
an28. A man leaves half of his property to his wife, one third to his son and the remaining fordaughter. If the daughter's share is 30000, find the share of his wife and his son. |
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| 4735. |
tindthesurfacearea of a sphere of diametero 14cm(uF 21 cm() 3.5m |
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| 4736. |
9 A cylindrical jar of diameter 14cm and depth 20 em is half-full of water, 300byleadshots of same size are dropped into the jar and the level of water raises2.8cm. Find the diameter of each leadshots. |
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Answer» GIVEN : Cylinder: r = 7cm h of the water level raised = 2.8cm Sphere: n = 300 TO FIND :d of the sphere FORMULA: Volume of the water raised = Volume of 300 lead shots SOLUTION : 22/7 * r2h = 300 * 4/3 * 22/7 * r3 72* 2.8 = 400r3 r3=49 * 2.8 400 r * r * r = 0.7 * 0.7 * 0.7 r = 0.7cm d = 2 * 0.7 = 1.4cm ANSWER :Diameter of the lead shots = 1.4cm please like my answer if you find it useful |
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| 4737. |
6. A man leaves half of his property to his wife, one third to his son and the remainingto his daughter. If the daughter's share is 15000. How much money did the manleave? How much money did his wife get and what is his son's share? |
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| 4738. |
. A and B can do a piece of work in 10 days and 6 days respectively. They work together for 2 days andleaves the work. In how many days A will finish the remaining work?If 1? man |
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| 4739. |
6.Diagonal AC of a parallelogram ABCD bisectsZ A (see Fig. 8.19). Show that(i) it bisects Z C also,(i) ABCD is a rhombus. |
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| 4740. |
gonal AC of a parallelogram ABCD bisect(see Fig. 8.19). Show thatit bisects Z C alsoABCD is a rhombus. |
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| 4741. |
6. Diagonal AC of a parallelogram ABCD bisectsA (see Fig. 8.19). Show that(i) it bisects Z C also,(ii) ABCD is a rhombus. |
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| 4742. |
r81O0Diagonal AC of a parallelogram ABCD bisectsL A (see Fig. 8.19). Show that(i) it bisects Z C also(i) ABCD is a rhombus.. |
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| 4743. |
Diagonal AC of a parallelogram ABCD biseets< A (see Fig. 8. 19). Show that(o) it bisects Z C also,6.2. ABCDAD-0 4(n) 4(ii)ABCD is a rhombus. |
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| 4744. |
6.Diagonal AC of a parallelogram ABCD bisectsZ A (see Fig, 8.19). Show thati) it bisects C also,(ii) ABCD is a rhombus. |
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| 4745. |
angles, then it is a square.Diagonal AC of a parallelogram ABCD bisectsZ A (see Fig. 8.19). Show that(i) it bisects C also,(ii) ABCD is a rhombus. |
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| 4746. |
. In the given figure PAQ is the tangent ofthe circle at point A and ABCD is a cyclicquadrilateralIfくCAQ-70°, thenABC is |
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Answer» See the attachment with the answer In the given information, PAQ is the tangent of the circle at point A and ABCD is a cyclic quadrilateral. If angle CAQ =70 we get AC is a chord and <CAQ is the angle made by CA and tangent at A <ABC is the angle made by same chord. Therefore <ABC = < CAQ = 70° |
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| 4747. |
In the given figure PAQ is the tangent ofthe circle at point A and ABCD is a cyclicquadrilateral. |
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Answer» See the attachment with the answer In the given information, PAQ is the tangent of the circle at point A and ABCD is a cyclic quadrilateral. If angle CAQ =70 we get AC is a chord and <CAQ is the angle made by CA and tangent at A <ABC is the angle made by same chord. Therefore <ABC = < CAQ = 70° |
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| 4748. |
If Δ.ABC ~ Δ.RQP, A-80. & LB-OU What IS the value orA ladder 15m long just reaches the top of a vertical wal. If the ladder makes an ange of0° with the wall. Find the height of the walkil he its volume? |
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Answer» Right angle at Let the length of ladder AC = 15 m ( hypotenuse ) Angle between ladder and wall < BCA= 60 degrees Angle between ladder and the ground < CAB = 90 - 60 = 30 degrees Height of the wall = BC From triangle ABC, Sin Sin 30 = BC / 15 1 / 2 = BC / 15 15 / 2 = BC 7.5 = BC Therefore, Height of the wall = BC = 7.5 m |
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| 4749. |
03. In the given figure PAQ is the tangent ofthe circle at point A and ABCD is a cyclicquadrilateral.If ZCAQ-70, then ZABC is |
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Answer» nice |
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| 4750. |
is tied to a peg at oneer of a square shaped grasscornereld of side 15m by means of a5m long rope (see Fig. 5.11). Find0) the area of that part of the fieldin which the horse can graze. |
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