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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.
| 78351. |
8. A cyclist completes 15 lapminutes. Findcompletes 15 laps around an oval racing track 400 m long in 8 minutospeed of the cyclist in km/h. |
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| 78352. |
9 cmABCD rectangle with AB = 8 cm, BC = 16 cm and AE=4 cm. Find the area of ABCE.15 marca of ABEC equal to the sum of the area of ABAE and ACDE. Why?4 cm E8 cm16 cmGovernment |
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| 78353. |
Q.4. In the given figure, ABCD is a rectangle, whereAB = 8 cm, BC = 6 cm and the diagonals bisecteach other at O. Find the area of the shadedregion by Heron's formula.[Board Term I, 2012, Set-66]6 cm8 cm |
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| 78354. |
8. A circle is inscribed in aAABC with sides AB-8 cm,BC 10 cm and CA 12 cm(see figure). Find AD, BEand CF |
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| 78355. |
/In a trapezium ABCD, AB 11 DC and DC-2AB. If EF is drawn parallel to AB cuts ADin F and BC in E such that BE:EC -3:4. Diagonal DB intersects EF at G. Prove that:7 EF 10AB. |
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Answer» See diagram. given DC = 2 ABDraw BH parallel to AD. H will be the middle point of DC as AB = DH.Sin EF || AB, FI = AB TheΔBHC andΔBIE are similar as the correspondingsides are parallel.So IE / HC = BE / BC IE = HC * BE / BC = y * 4x/7x = 4 y /7 EF = FI + IE= y + 4 y /7 = 11 y / 7 so 7 EF = 11 AB HIT THE LIKE BUTTON ✔️ 👍 |
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| 78356. |
In a trapezium ABCD, AB |DC and DC-2AB. If EF is drawn parallel to ABBE 3cuts AD in F and BC in E such that Bc-4 Diagonals DB intersects EF at GProve that 7 EF 10AB |
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| 78357. |
In a trapezium ABCD, AB | | DC and DC = 2AB. EF | | AB, where E and F lie on BC and AD respectivelysuch thatDiagonal DB intersects EF at G. Prove that,BE 4EC 37EF-11AB. |
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Answer» given DC = 2 ABDraw BH parallel to AD. H will be the middle point of DC as AB = DH.Sin EF || AB, FI = AB TheΔBHC andΔBIE are similar as the correspondingsides are parallel.So IE / HC = BE / BC IE = HC * BE / BC = y * 4x/7x = 4 y /7 EF = FI + IE= y + 4 y /7 = 11 y / 7 so 7 EF = 11 AB |
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| 78358. |
Ef sin «=, then the value of 3 cos -4cos « |
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Answer» answer 0 right answer hoga |
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| 78359. |
In ADEF, find EF8x 32x6x +5Work: |
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Answer» Since, angle DEF = angle DFE=>DF=DE(sides opposite to equal angles are also equal)=>6x+5=8x+3=>2x=2=>x=1 EF=2x=2 units. |
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| 78360. |
Fig. 7.505. In Fig 7.51, PR> PQ and PS bisects Z QPR. Provethat L PSR > L PSQFig. 7.51 |
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| 78361. |
8. In the given figure, BM and DN are both perpendiculars to the segmentAC and BMDN. Prove that AC bisects BDlg ndCoco ohtuse Points X and Y |
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Answer» See ∆DNP and ∆BMP, DN=MB (given)angle NPD = angle MPB (opposite angles)angle DNP = angle BMP =90° (given perpendiculars) So, ∆DNP and ∆BMP are congruentTherefore, DP=PB |
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| 78362. |
5. In Question 4, point C is called a mid-point of line segment AB. PProve that every linesegment has one and only one mid-point. |
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| 78363. |
ABCD is a rhombus. Show that diagonal ACbisects ZAas well as zC and diagonal BD bisectsB as well as 4D |
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| 78364. |
5. The area of a square park is the same as of a rectangular park. If the side of thesquare park is 60 m and the length of the rectangular park is 90 m, find the breadth ofthe rectangular park. |
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Answer» sin 90 cos 12233445555 |
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| 78365. |
Mrs Jyoti walks around a square park of side 50 m daily. if she walks around the park 5 time every day, find the distance she covers in 30 days (in km) |
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Answer» Perimeter of square= 4*50= 200distance walked in one day= (5*200)=1000 mdistance covered in 30 days= (1000*30) m= 30000 m perimeter of park= 4a=4*50=200mno.of rounds =5distance covered in one day=5*2001000mDistance covered in 30 days= 1000*3030000m |
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| 78366. |
5. Sanju completes 12 rounds arounda square park every day. If oneside of the park is 120 m, find out inkilometres and metres the distance thatSanju covers daily. |
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Answer» Perimeter of square park = 4×side = 120×4 = 480m area covered by sanju daily = 480×12 = 5760m |
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| 78367. |
15. In the figure, D, Eand F are themidpoint of sidesAB, AC and BCrespectivelyIf perimeter of BÎDEF is 15 cm.determine the lengthsof sides AB, BC and AC. |
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Answer» thanks |
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| 78368. |
3. Ja Fig. 6.30, if AB 11 CD, EF丄CD andGED-126 find ZAGE, GEF andFGEAGFig. 630 |
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| 78369. |
A train travels 225 km with a constant speed of 60 km/h and another 230 km with a constantspeed of 120 km/h. What is its average speed for the total trip? |
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Answer» ✓ |
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| 78370. |
48.In figure, AB-AC,<C, AB8 cm, BE-6 cm and AE-3 cm, then find perimeter ofADC.(1) 14 cm(2) 15 cm(3) 17 cm(4) 19 cm |
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Answer» thank u |
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| 78371. |
जज ¥ & 4B ) CD, CD | EF Ry :2=3:17 है, तो 2 का मान ज्ञात कीजिए। की 3 e |
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| 78372. |
3. In Fig 630, it ABI CD, EF 1 CD and 4 GED- |
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| 78373. |
(d) If AB II EF, DE II BC. AC I DF , find x and y60°110 |
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| 78374. |
36.Line segment AB is parallel to another line scgment CD.of AD. Show that i) ΔΑΟΒ ΔDOC ii)''O' is the midpointo' is also the midpoint of BC |
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| 78375. |
In Fig. 3.29, if AB || CD, CD | EF andyz 3:7,find x. |
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Answer» question is incomplete without figure.. |
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| 78376. |
O42.LiIn Fig. 6.29, if AB l CD, CD II EF and y: z-3:7,find x.are coinferioang lFig. 6.29 |
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| 78377. |
2. In the adjoining figure, ABCD is a llgm inpFwhich E and F are the midpoints of AB andCD respectively. If GH is a line segment thatcuts AD, EF and BC at G, P and Hrespectively, prove that GP - PH. |
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| 78378. |
18) If it takes 3 hours 20 minutes to paint 5beuchchairs, how long will it take to paint 8chairs?sel to make 12 |
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| 78379. |
5. In the adjoining figure, BM LAC and DDN AC. If BM-DN, prove that ACbisects BD. |
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| 78380. |
2 hours 10 minutes from 15 hours. |
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Answer» Answer is 12 hour 50 minutes |
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| 78381. |
A person walks at the rate of 4 km/h. How long will he take to go round a square park 5 times, whosearea is 2500 m?10. |
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| 78382. |
10. A person walks at the rate of 4 km/hr. How long will he take to go round a square park 5 times, whose areais 2500 m? |
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| 78383. |
A motorbike travels 120 km in 2 litres of petrol. How many litres of petrolwill it require to travel 300 km? |
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Answer» 120km in 2 litres.1km in 2/120=1/60 lts.So 300 km in 300/60 =5 L |
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| 78384. |
* in 12 days. in how many days can dih.6 men or 10 women can do a piece of work in 12 days. In how many days can the(2017)same work be done by 3 men and 5 women?(1) 6 days (2) 12 days (3) 18 days (4) 24 days |
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Answer» so the 12+12= 24days |
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| 78385. |
In a triangle ABC, AB-AC 15 cm, BC- 18 cm. find 0) cosLABC and (i) tan |
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| 78386. |
3 In Fig. 6.30, if AB II CD, EF L CD2 GED = 126", find < AGE, < GEF and L FGE |
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Answer» hit like if you find it useful |
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| 78387. |
Find the area of the trapezium whose parallel sides are 18 cm and 36 cm, respectively, and non-parallel sides are equal and of 15 cm each. |
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| 78388. |
3.In Fig. 6.30, if AB II CD, EF 1 CD andGED = 126", find < AGE. < GEF and ZFGE.eAG F, |
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| 78389. |
2 In Fig 6.29, ifAB CD, CD EF andy3:7,find rFig. 6 20 |
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Answer» hit like if you find it useful |
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| 78390. |
e figure 10.15, show that:ABII CDCD II EFAB II EFfy your answer.407 B1602020°Fig. 10.15 |
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Answer» since angle ABC = 40 = Angle ( BCE+ECD) , so these makes alternate interior angle..so AB || CD now angle ECD + angle CEF = 20°+ 160° = 180° (adjacent angles) so EF||CD and because CD is parallel to both AB and EF SO, AB||EF |
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| 78391. |
3.In fig. AB CD EFIf AC = 5.4, CE= 9, BD= 7.5then find DFCD EFSOL: ABACE |
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Answer» AB||CD||EFTherefore,AC/CE=BD/DF (BY Intercept Theorem)5.4/9=7.5/DF5.4×DF=7.5×9DF=(7.5×9)/5.4Therefore,DF=12.5 |
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| 78392. |
30°F24. In the given figure AB, CD and EF are three towers. The angleof elevation of the top of thetower CD from the top of thetower AB is 60° and that from EFis 30°. BD = 2/3 m, CD: EF = 5:4and DF = 4 m. What is the heightof the tower AB ?(a) 6 m(b) 12 m(c) 7 m(d) none of the aboveB60° Pl 30° |
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Answer» correct answer is (c) 7m CD/EF=x+3/x+1=5/4=x=7 |
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| 78393. |
2. In Fig. 6.29, irABICD, CD IEF andy z 3.7,find x.h |
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| 78394. |
9. The length of a line segment AB is of the length of the line segment CD. The length ofanother line segment EF is of the length of the line segment AB. What fraction of the lengthof CD is equal to the length of EF? If CD = 4 cm then find the length of EF. |
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Answer» =>AB=2/3CD=>AB=2/3 (4)=8/3cm=> EF=3/8AB=> EF=3/8 (8/3)=1cmso CD=4EF an examination is being held in two parts and three are 572 examinèès in all. 3/11 of the total number of examinees appears in parts onely and 3/13 appears in parts 2 only .the rest appears in both the parts .find the number of candidates appearing in both the parts Length of 1st line segment = AB 2/3 CDLength of 2nd line segment = EF 3/8 ABThe fraction of the CD equal to Length EF Length of CD = 4 cmThe Length of EF =AB = 2/3 of CDAB = 2/3 = 3/8EF = 3/8 of AB EF = 3/8 (8/3) 1cm CD = 4 of EF So, 1 CM = 1/4 CM of EF |
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| 78395. |
CB, AB CD and EF bisects BD at G. Prove that G is mid-point of EF.CDg In the figure,ABA F |
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| 78396. |
Cave tor school?me will be 5 hours 10 minutes after 7:30 a.m.?e will be 7 hours 15 minutes before 10:35 p.m |
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Answer» 5 hours 10 minutes after 7:30 am = 12:40 pm 7 hours 15 minutes after 10:35 pm =5:50 am |
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| 78397. |
7.A company manufactures two types of novelty souvenirs made of plywoodSouvenirs ofassembling. Souvenirs of type B requireeach for assembling. There are 3 hours 20 minutes available for cutting and 4hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for typeB souvenirs. How many souvenirs of each type should the company manufacturetype A require 5 minutes each for cutting and 10 minutes each8 minutes each for cutting and 8minutesin order to maximise the profit? |
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| 78398. |
A polyhedron can have,(A) 10 faces, 20 edges, 15 vertices(B) 10 faces, 22 edges, 15 vertices(C) 10 faces, 20 edges, 12 vertices(D) 10 faces, 22 edges, 12 vertices22.6มา |
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| 78399. |
The cost of pioughin8.50 per sq. m.3. Aperson walks at 3 km/hr. Find how long will it take him to go round a square ground 5 times,the artaof which is 2025 m2 |
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Answer» total area will be 5*2025=10125m^2hence in km =10125=10.125kmhence time =distance /speed=10.125/3=3.375hr |
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| 78400. |
SEL-Ze drawn from the vertices of ABC to the opposite sides are equal, prove that the triangle is17, Ifthe altitudequilateral. |
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Answer» Given :AD = BE = CF and angles A = B = C= 90° To prove :Triangle ABC is equialteral. Proof : In triangle ABE and ACF Angle AEB = AFC ( 90° each ) BE = CF ( Given ) Angle A = A ( Common ) Hence Triangle ABE ~ ACF by AAS congruency AB = AC ( c.p.c.t ) In triangle AOE and EOC OE = OE ( Common ) Angle E = E ( 90° each ) AO = OC [ :. AD = FC, their halfs OA = OC ] Hence triangle AOE ~ EOC by RHS congruency AE = EC ( c.p.c.t ) In triangle ABE and BCE Angle E = E ( 90° each ) BE = BE ( common ) AE = EC ( proved above ) Hence, triangle ABE ~ BCE by SAS congruency. AB = BC ( c.p.c.t ) As AB = AC and AB = BC, so AC = BC. Hence, AB = BC = CA. HENCE PROVED. |
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