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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.
| 1. |
The cost if a harvesting a square field at Rs. 900 per hectare is Rs. 8100. find the cost of putting a fence around it at Rs. 18 per metre . |
| Answer» Correct Answer - Rs. 21600 | |
| 2. |
The area of a parallelogram is 392 `m^(2)` . If its altitude is twice the corresponding base , determine is `392 m^(2)`. If its altitude is twice the corresponding base , determine the base and the altitude . |
| Answer» Correct Answer - Base = 14 m , altitude = 28 m | |
| 3. |
the length of the diagonal of a square is `10sqrt(2)` cm .Its area isA. `200 m^(2)`B. `100 m^(2)`C. `150m^(2)`D. `100sqrt(2) cm^(2)` |
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Answer» Correct Answer - B Area `=(1)/(2)xx("diagonal")^(2)=(1)/(2) xx(10 sqrt(2) )^(2)cm^(2) = 100 cm^(2)` |
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| 4. |
the area of an equilateral triangle is `4 sqrt(3) cm^(2)` Its perimeter isA. `9 cm`B. 12 cmC. `12 sqrt(3)` cmD. `6sqrt(3) cm` |
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Answer» Correct Answer - B `(sqrt(3))/(4) a^(2)=4sqrt(3)implies a^(2)=16implies a=4. ` `therefore ` perimeter `=3a =(3xx4) cm = 12.` |
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| 5. |
Find the are of an equilateral triangle having each side of length 10 cm . `["Take"sqrt(3)=1.732.]` |
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Answer» Here ,a = 10 cm `"(i) Area of the triangle"=((sqrt(3))/(4)xxa^(2))sq " units"` `=((sqrt(3))/(4)xx10xx10)cm ^(2)= (25xxsqrt(3))cm^(2)` `= (25xx1.732) cm ^(2)= 43.3 cm ^(2) ` (ii) Height of the triangle `=((sqrt(3))/(2)xxa)"units"` `=((sqrt(3))/(2)xx10)cm =(5xxsqrt(3))cm` `=(5xx1.732) cm = 8.66 cm.` |
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| 6. |
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle. |
| Answer» Correct Answer - `750 m^(2)` | |
| 7. |
The lengths of thesides of a triangle are in the ration 3:4:5 and its perimeter is `144 c mdot`Find the area of thetriangle and the height corresponding to the longest side. |
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Answer» On dividing 144 cm in the ratio 3:4:5 , we get `A=(144xx(3)/(12))cm = 36 cm , b=(144xx(4)/(12))cm = 48cm` `and C=(144xx(5)/(12))cm = 60 cm.` `therefore s=(1)/(2) (36+48+60)cm =72 cm ` `(s-a)=(72-36) cm =36 cm .` `(S-b) =(72-48) cm = 24 cm ` `and (s-c) =(72-60) cm = 12 cm .` (i) Area of the triangle `=sqrt(s(s-a)(s-b)(s-c))` `=sqrt(72xx36xx24xx12)cm^(2)` `=72xx12cm^(2)=864 cm^(2).` (ii) Let base = 60 cm and the corresponding height = h cm . `therefore 30h= 864implies h=(864)/(30)=28.8.` Longest side = 60 cm , corresponding height = 28.8 cm . |
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| 8. |
A lawn is in the form of a rectangle whose sides are in the ratio 5:3 the area of the lawn is 3375 `m^(2)` .Find the cost of fencing the lawn at Rs. 65 per metre. |
| Answer» Correct Answer - Rs.15600 | |
| 9. |
The lengths of the two sides of a right triangle containing the right angle differ by `2cm.` If the area of the triangle is `24cm^2.` Find the perimeter of the triangle. |
| Answer» Correct Answer - 24cm | |
| 10. |
The length of the sides of a triangle field are 20 m , 21 m and 29 m the cost of cultivating the field at Rs. 9 per `m^(2)` isA. Rs.2610B. Rs.3780C. Rs.1890D. Rs. 1800 |
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Answer» Correct Answer - C `s=(1)/(2) (20+21+29)=35 ,(s-a) = 15 ,(s-a)=14 and (s-c) =6.` `therefore "area"=sqrt( 35xx15xx14xx6)m^(2)=(7xx5xx3xx2)m^(2)=210 m^(2)`. Required cost `=Rs. (210xx9)=Rs. 1890.` |
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| 11. |
The base and height of a triangle are in the ratio 3:4 and its area is `216 cm^(2) .` The height of the triangle isA. 18 cmB. 24 cmC. 21 cmD. 28 cm |
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Answer» Correct Answer - B Let the base be 3x cm and height be 4x cm. then , `(1)/(2) xx3x xx4x=216implies x^(2)=36implies x=6.` `therefore "height " =(4xx6) cm = 24 cm .` |
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| 12. |
The Hypotenuse of a right triangle is 65 cm and its base is 60 cm . Find the length of perpendicular and the area of the triangle . |
| Answer» Correct Answer - `25 cm, 750 cm^(2)` | |
| 13. |
of a square field is `6050 m^(2)` . The length of its diagonal isA. 135mB. 120mC. 112mD. 110m |
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Answer» Correct Answer - D `(1)/(2) xx("diagonal")^(2) =6050implies ("diagonal")^(2) =12100 =(110)^(2)implies "diagonal "=110 m.` |
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| 14. |
The cost of painting if four walls of a room 12 m long at Rs. 30 per `m^(2) ` is Rs. 7560 and th cost of covering the floor with mat at Rs. 25 per `m^(2) ` is Rs. 2700 .Find the dimensions of the room . |
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Answer» Correct Answer - Length = 12 m, breadth =9m and height = 6m Area of walls `=("total cost painting")/("rate per "m^(2))=((7560)/(30))m^(2)=252m^(2).` Area of the floor `=("total cost of matting ")/("rate per "m^(2))=((2700)/(25))m^(2)=108m^(2)`. Let breadth =b metres and height =h metres , then , Also , `2(l+b)xxh=252implies h=252impliesh=6m.` |
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| 15. |
The area of a square field is 8 hectares , How long would a man take to cross it diagonally by walking at the rate of 4 km per hour ? |
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Answer» Correct Answer - 6 minutes Area `=(8xx10000)m^(2)implies(1)/(2) xx("diagonal")^(2)=8xx10000` `implies("diagonal")^(2)=160000implies "diagonal "=400 m. ` |
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| 16. |
किसी समबाहु त्रिभुज की ऊंचाई 15 cm है। त्रिभुज का क्षेत्रफल ज्ञात करें? |
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Answer» Let each side of the triangle be a cm . Then , its height `=((sqrt(3))/(2)xxa)cm.` `therefore (sqrt(3))/(2)xxa = 15 implies a=((30)/(sqrt(3))xx(sqrt(3))/(sqrt(3)))=10sqrt(3).` Thus , each side of the triangle `= a cm = 10 sqrt(3)`cm . Area of the triangle `=((sqrt(3))/(4)xxa^(4))=((sqrt(3))/(4)xx10sqrt(3)xx10sqrt(3))cm^(2)` `=(75sqrt(3))cm^(2)=((75xx173)/(100))cm^(2)` `=(519)/(4)cm ^(2) =129.75 cm^(2)` |
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| 17. |
किसी समबाहु त्रिभुज की माध्यिका `6sqrt(3)`cm है तब त्रिभुज का परिमाप ज्ञात करें?A. 8 cmB. 9 cmC. `3sqrt(3) cm `D. `6 cm` |
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Answer» Correct Answer - B Area of the triangle `=((sqrt(3))/(4) xxa^(2) )={(sqrt(3))/(4) xx(6sqrt(3))^(2)}cm^(2)` `=((sqrt(3))/(4)xx108)cm^(2)=(27sqrt(3)) cm^(2).` `(1)/(2) xx 6sqrt(3) xxh=27 sqrt(3) implies h =(27sqrt(3))/(3sqrt(3))=9 cm . ` |
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| 18. |
Find the area of an isosceles triangle having each of whose equal sides is 13 cm and whose base is 24 cm. |
| Answer» Correct Answer - `60 cm^(2)` | |
| 19. |
Find the area of the triangle whose sides are 42 cm , 34 cm and 20 cm Also, find the height corresponding to the longest side . |
| Answer» Correct Answer - `336 cm^(2)`,16 cm | |
| 20. |
The length of the diagonal of a square is 24 cm. Find its area . |
| Answer» Correct Answer - `288 mc^(2)` | |
| 21. |
Find tha area of a triangle whose sides are 42 cm, 34cm and 20 cm. |
| Answer» Correct Answer - `336 cm^(2)` | |
| 22. |
Find the area of the rhombus , the lengths of whose diagonals are 30 cm and 16 cm .Also , find the perimeter of the rhombus . |
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Answer» Correct Answer - `240 cm^(2)` , 68 cm We know that the diagonals of a rhombus bisect each other at right angles . So ,`OA=15 cm ,OB=8cm and angle AOB=90^(@)` `therefore AB^(2)=(OA^(2)+OB^(2))=(15)^(2)+(8)^(2)=225+64=289` `implies AB=sqrt(289)=17 cm ` So perimeter `=(4xx17)cm =68 cm` |
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| 23. |
The diagonals of a rhombus are 48 cm and 20 cm long . Find the the perimeter of the rhombus . |
| Answer» Correct Answer - `480 cm^(2)` | |
| 24. |
A parallelogram and a rhombus are equal in area.The diagonals of the rhombus measure 120m and 44m .If one of the sides of the parallelogram measures 66m ,find its corresponding altitude. |
| Answer» Correct Answer - 40 m | |
| 25. |
The adjacent sides of a parallelogram are 32 cm and 24 cm . If the distance between the longer sides is 17.4 cm, find the distance between the shorter sides . |
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Answer» Correct Answer - `23.2 cm ` Area of the parallelogram `=(32xx17.4) cm^(2).` `24xxh=32xx17.4.` Find h. |
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| 26. |
One side of a rhombus is 20 cm long and one of its digonals measures 24 cm. the area of the rhombus is |
| Answer» Correct Answer - `384 cm^(2)` | |
| 27. |
A room 4.9 m long and 3.5 m broad is covered with carpet leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, find its cost at 40 per metre. |
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Answer» Correct Answer - Rs. 1320 Area to be carpeted `=(4.9-0.5)(3.5-0.5)m^(2)=13.2m^(2)`. lt brgt Length of the carpet`=((13.2)/(0.80))m=16.5m.` |
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| 28. |
The shape of the cross section of a canal is a trapezium . If the canal is 10 m wide at the top , 6 m wide at the bottom and the area of its cross section is `640 m^(2),` find the depth of the canal . |
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Answer» Correct Answer - `80 m` `(1)/(2)(10+6)xxd=640implies d=80 cm.` |
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| 29. |
Find the area of a trapezium parallel sides are 11 cm and 25 cm long and nonparallel sides are 15 cm and 13 cm . |
| Answer» Correct Answer - `216 cm^(2)` | |
| 30. |
Find the area of a trapezium whose parallel sides are 11 m and 25 m long, and the nonparallel sides are 15 m and 13 m long. |
| Answer» Correct Answer - `216 m^(2)` | |
| 31. |
The parallel sides of a trapezium are 9.7 cm and 6.3 cm , and the distance between them is 6.5 cm. The area of the trapezium isA. `104 cm^(2)`B. `78 cm^(2)`C. `52cm^(2)`D. `65 cm^(2)` |
| Answer» Correct Answer - C | |
| 32. |
Find the area of a tranpezium whose parallel sides are 35 cm and 23 cm long and the distance between them is 15 cm . |
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Answer» Area of the trapezium `=(1)/(2) xx ("Sum of parallel sides ")xx("distance between them")` `={(1)/(2) xx(35+23)xx15}cm^(2) = 435 cm^(2).` |
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| 33. |
Find the area of the quadrilateral ABCD in which AB= 42 cm. `Bc=21 cm ,CD=29 cm ,DA = 34 cm and ` diagonal BD= 20 cm . |
| Answer» Correct Answer - `546 cm^(2)` | |
| 34. |
In the given figure ABCD is a trapezium in which AB = 40 m, BC=15 m ,CD= 28 m ,AD=9 m and `CE bot AB.` area of trap.ABCD is A. `306 m^(2)`B. `316 m^(2)`C. `296 m^(2)`D. `284 m^(2)` |
| Answer» Correct Answer - A | |
| 35. |
In the given figure ABCD is a quadrilateral in which `angle ABC=90^(@),angle BDC=90^(@),AC=17 cm ,BC=15 cm ,BD=12 cm and CD=9 cm .` The area of quad. ABDC is A. `102 cm^(2)`B. `114 cvm^(2)`C. `95 cm^(2)`D. `57 cm^(2)` |
| Answer» Correct Answer - B | |
| 36. |
Find the perimeter and area of the quadrilateral ABCD in which AB= 17 cm `AD= 9 cm, CD=12 cm angle ACB=90^(@)` and AC=15 cm. |
| Answer» Correct Answer - 46 cm,`114 cm^(2)` | |
| 37. |
The height of an Equilateral triangle is `6sqrt3 cm` Find its area |
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Answer» Correct Answer - `20.76 cm^(2)` Height `=(sqrt(3)a)/(2) implies (sqrt(3)a)/(2) =6 cm implies a=(12) /(sqrt(3))cm =(12)/(sqrt(3))xx(sqrt(3))/(sqrt(3))cm = 4sqrt(3)cm.` `therefore area =(sqrt(3))/(4) xxa^(2)=(sqrt(3))/(4)xx(4sqrt(3))^(2)cm^(2)=((sqrt(3))/(4)xx48)cm^(2)=(12xx1.73)cm^(2).` |
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| 38. |
Each side of an equilateral triangle is 10 cm. Find (i) the area of the triangle and (ii) the height of the triangle . |
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Answer» Correct Answer - `(i) 43.3 cm ^(2) `(ii) 8.66 cm ` (i) Area of the triangle`=((sqrt(3))/(4)xxa^(2))=((1.732)/(2)xx10xx10) cm ^(2)=43.3 cm^(2)`. `(ii) (1)/(2) xxaxx"height "=43.3 cm ^(2)implies (1)/(2) xx10cm xxh = 43.3 cm^(2)implies h=(43.3 cm^(2))/(5 cm) = 8.66 cm.` |
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| 39. |
The base of an isosceles triangle measures 24 cm and its area is `192 cm^(2)` , Find its perimeter. |
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Answer» Here , base ,b=24 cm am and let each equal side be a cm . Then , `area =(1)/(4)b sqrt(4a^(2)-b^(2))sq " units"=(1)/(4) xx24xxsqrt(4a^(2)-576)cm^(2)` `= 12xxsqrt(a^(2)-144)cm^(2),` `"but area " =192 cm^(2) ["given "]`. `therefore 12xxsqrt(a^(2)-144)=192implies sqrt(a^(2)-144)=16` `implies a^(2) -144 = 256implies a^(2) = 400 implies a=20.` `therefore ` perimeter of the triangle `=(2a+b)cm ` `=(2xx20 +24)cm =64 cm.` |
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| 40. |
The perimeter of a right is 40 cm and its hypotenuse measures 17. cm find the area of the area of the triangle . |
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Answer» Correct Answer - `60 cm^(2)` Sum of two sides `=(40-17) cm = 23 cm.` Let these sides be x cm and (23-x) cm . `therefore x^(2)+ (23-x)^(2)=(17)^(2)implies x^(2)+x^(2)-46 x+ 529= 289` `implies 2x^(2)- 46x+240 =0implies x^(2)-23 x+120=0` `implies (x-15) (x-8) =0implies x=15 or x=8` ` therefore ` base = 15 cm and height = 8 cm .find the area of the triangle. |
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| 41. |
Find the length of the hypotenuse of an isosceles right - angled triangle whose area is 200 `cm^(2)`.Also , find its perimeter.`["Given ",sqrt(2)=1.41.]` |
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Answer» Correct Answer - `28.2 cm , 68.2 cm` Let base = height =a cm . Then , `(1)/(2) xxaxxa=200cm^(2)implies a^(2)= 400cm^(2)implies a=sqrt(400)cm^(2)= 20 cm `. `therefore " hypotenuse "=sqrt(a^(2)+a^(2))=sqrt(2a^(2))=sqrt(2xx400) cm ` `=20sqrt(2) cm = (20xx1.41) cm.` |
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| 42. |
The perimeter of a rectangular plot of land is 80 m and its breadth is 16 m . Find the length and area of the plot . |
| Answer» Correct Answer - Length `=24m ,area = 384m^(2)` | |
| 43. |
The length of a rectangle is thrice its breadth and the length of its diagonal is `8sqrt(10)` cm . The perimeter of the rectangle isA. `15sqrt(10)cm`B. `16sqrt(10)cm`C. `24 sqrt(10) cm `D. 64 cm |
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Answer» Correct Answer - D Let the breadth be x cm, then , length = 3x cm . `therefore "diagonal "=sqrt((3x)^(2)+x^(2))cm = xsqrt(10)cm.` Thus ,`xsqrt(10) = 8 sqrt( 10)implies =8.` perimeter `=2(3x+x)cm = 8x cm =(8xx8) cm = 64 cm.` |
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