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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.
| 51. |
The area of a triangular plot is 2.5 m2. If the base is 100 meter then find the height is(a) 10 cm.(b) 5 cm.(c) 10 meter(d) 5 meter |
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Answer» The height is 5 meter. |
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| 52. |
Find the area bounded by the x-axis, part of the curve `y=(1-8/(x^2))`, and the ordinates at `x=2a n dx=4.`If the ordinate at `x=a`divides the area into two equal parts, then find `adot` |
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Answer» Correct Answer - `4 " sq. units ; "a=2sqrt(2)` `"Given "y=1+(8)/(x^(2))`. Here y is always positive, hence, curve is lying above the a-axis. `therefore" Req. area "=int_(2)^(4)ydx=int_(2)^(4)(1+(8)/(x^(2)))dx` `=[x-(8)/(x)]_(2)^(4)=4.` If x=a bisects the area, then we have `overset(a)underset(2)int(1+(8)/(x^(2)))dx=[x-(8)/(x)]_(2)^(a)=[a-(8)/(a)-2+4]=(4)/(2).` `"or "a-(8)/(a)=0` `"or "a^(2)=8` `"or "a=pm2sqrt(2)` `"Since "agt2,a=2sqrt(2).` |
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| 53. |
The length of two perpendicular sides of a trapezium are 10 cm and 16 cm perpendicular distance between them is 8 cm. find the area of the trapezium. |
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Answer» Area of trapezium = 1/2 x (Sum of parallel sides) x height = 1/2 x (10 + 16) x 8 = 104 square cm. |
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| 54. |
The area of a trapezium is 143 cm2. Parallel sides are 15 cm. and 11 cm. respectively. The perpendicular distance between them is(a) 11 cm.(b) 15 cm.(c) 13 cm.(d) 14 cm. |
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Answer» The perpendicular distance between them is 11 cm. |
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| 55. |
The parallel sides of a trapezium are 32 m. and 20 m. respectively and distance between them is 15 m., then area of trapezium is(a) 290 m2(b) 390 m2(c) 190 m2(d) 400 m2 |
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Answer» The area of trapezium is 390 m2. |
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| 56. |
Area enclosed by the curve `y=f(x)` defined parametrically as `x=(1-t^(2))/(1+t^(2)), y=(2t)/(1+t^(2))` is equal toA. `pi` sq. unitsB. `pi//2` sq. unitsC. `(3pi)/(4)` sq. unitsD. `(3pi)/(2)` sq. units |
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Answer» Correct Answer - A Clearly t can be any real number `"Let "t=tan theta rArr x=(1-tan^(2)theta)/(1+tan^(2)theta)` `rArr" "x=cos 2theta and y =(2 tan theta)/(1+tan^(2)theta)=sin 2theta` `rArr" "x^(2)+y^(2)=1` Thus, required area is `pi` sq. units. |
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| 57. |
In the figure XY, RS are parallel. The area of ∆ ABD is 20 cm2(a) Find the area of the parallelogram ABCD. (b) Find two triangles in the diagram with same area |
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Answer» a. 20 + 20 = 40 sq. cm b. i. ABD, ABC ii. DCB, DCA |
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| 58. |
Find the area of triangle in the diagram. |
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Answer» \(\frac{1}{2}\times6\times4=12\) cm |
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| 59. |
In the figure BP : PQ : QC = 1 : 2 : 1. Area of triangle APQ is 8 sq.cm. a. Find the area of triangle ABP. b. Find the area of triangle ABC. |
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Answer» a. BP : PQ = 1:2 Area of ∆ ABP is half the area of ∆ APQ. Area of ∆ ABP = \(\frac{1}{2}\times8=4\) sq.cm b. Area of ∆ ABC = 4 + 8 + 4 = 16 sq.cm |
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| 60. |
Let A(k) be the area bounded by the curves `y=x^(2)+2x-3 and y=kx+1.` ThenA. the value of k for which A(k) is least is 2B. the value of k for which A(k) is least is `3//2`C. least value of A(k) is `32//3`D. least value of A(k) is `64//3` |
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Answer» Correct Answer - A::C `x_(1) and x_(2)` are the roots of the equation `x^(2)+2x-3=kx+1` `"or "x^(2)+(2-k)x-4=0` `{:(x_(1)+x_(2)=k-2),(x_(1)x_(2)=-4):}}` `A=overset(x_(2))underset(x_(1))int[(kx+1)-(x^(2)+2x-3)]dx` `=[(k-2)(x^(2))/(2)-(x^(3))/(3)+4x]_(x_(1))^(x_(2))` `=[(k-2)(x_(2)^(2)-x_(1)^(2))/(2)-(1)/(3)(x_(2)^(3)-x_(1)^(3))+4(x_(2)-x_(1))]` `=(x_(2)-x_(1))[((k-2)^(2))/(2)-(1)/(3)((x_(2)-x_(1))^(2)-x_(1)x_(2))+4]` `=sqrt((x_(2)+x_(1))^(2)-4x_(1)x_(2))[((k-2)^(2))/(2)-(1)/(3)((k-2)^(2)+4)+4]` `=(sqrt((k-2)^(2)+16))/(6)[(1)/(6)(k-2)^(2)+(8)/(3)]` `=([(k-2)^(2)+16]^(3//2))/(6)` which is least when `k=2 and A_("least")=32//3` sq. units |
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| 61. |
In this figure ∠x + ∠y = ……………….A) 80° B) 260° C) 180° D) 160° |
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Answer» Correct option is B) 260° |
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| 62. |
In the given figure AC = 5.4 cm, BD = 5.8 cm and AD = 8.4 cm, BC = ………………A) 2.6 cm B) 2.8 cm C) 3.0 cm D) 3.2 cm |
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Answer» Correct option is B) 2.8 cm |
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| 63. |
The probability that a randomly thrown dart hits the square board in shaded region =A) 1/4B) 1/2C) 3/4D) 1 |
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Answer» Correct option is B)1/2 |
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| 64. |
The area bounded by `y=x^(2)+2 and y=2|x|-cospi x` is equal toA. `2//3`B. `8//3`C. `4//3`D. `1//3` |
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Answer» Correct Answer - B Solving given curves `x^(2)+2=2|x|-cos pix` `rArr" "x^(2)-2|x|+2=-cos pix` `rArr" "(|x|-1)^(2)+1=-cospix` `rArr" "x=pm1` `therefore" Required Area "=int_(-1)^(1)(x^(2)+2-2|x|+cospix)dx=8//3` |
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| 65. |
Consider two curves `C_1:y =1/x` and `C_2.y=lnx` on the `xy` plane. Let `D_1`, denotes the region surrounded by `C_1,C_2` and the line `x = 1` and `D_2` denotes the region surrounded by `C_1, C_2` and the line `x=a`, Find the value of `a`A. 1 sq. unitsB. 2 sq. unitsC. `2+sqrt(3)` sq. unitsD. None of these |
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Answer» Correct Answer - B `overset(pi//6)underset(0)int((1+cosx)-(1+cos (x-(pi)/(3))))dx+overset(pi)underset(pi//6)int((1+cos (x-(pi)/(3)))-(1+cos x))dx` `=[ sin x- sin (x-(pi)/(3))]_(0)^(pi//6)+[sin(x-(pi)/(3))- sin x]_(pi//6)^(pi)` `=[((1)/(2)+(1)/(2))-(sqrt(3))/(2)]+[(sqrt(3))/(2)-(-(1)/(2)-(1)/(2))]` `=2` sq. units. |
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| 66. |
Consider the function defined implicity by the equation `y^(2)-2ye^(sin^(-1)x)+x^(2)-1+[x]+e^(2sin ^(-1)x)=0("where [x] denotes the greatest integer function").` Line x=0 divides the region mentioned above in two parts. The ratio of area of left-hand side of line to that of right-hand side of line isA. `1+pi:pi`B. `2-pi:pi`C. `1:1`D. `pi+2:pi` |
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Answer» Correct Answer - D `"Ratio "=((pi)/(2)+1)/((pi)/(2))=(pi+2)/(pi).` |
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| 67. |
Consider the function defined implicity by the equation `y^(2)-2ye^(sin^(-1)x)+x^(2)-1+[x]+e^(2sin ^(-1)x)=0("where [x] denotes the greatest integer function").` The area of the region bounded by the curve and the line `x=-1` isA. `pi+1` sq. unitsB. `pi-1` sq. unitsC. `(pi)/(2)+1` sq. unitsD. `(pi)/(2)-1` sq. units |
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Answer» Correct Answer - A `"For "-1lexlt0` `(y-e^(sin^(-1)x))^(2)=2-x^(2)` `y=e^(sin-1x)pmsqrt(2-x^(2))` `A=overset(0)underset(-1)int(e^(sin^(-1)x)+sqrt(2-x^(2)))-(e^(sin^(-1)x)-sqrt(2-x^(2)))dx` `=2overset(0)underset(-1)intsqrt(2-x^(2))dx` `=2((1)/(2)xsqrt(2-x^(2)) :|_(-1)^(0)+(2)/(2)sin^(-1)""(x)/(sqrt(2)):|_(-1)^(0))` `=[1+2(0-(-(pi)/(54)))]` `=(pi)/(2)+1` sq. units. `"For "0lexlt1,y=sin^(-1)xpmsqrt(1-x^(2))` `A=2overset(1)underset(0)intsqrt(1-x^(2))dx` `=2[(x)/(2)sqrt(1-x^(2)):|_(0)^(1)+(1)/(2)sin^(-1)""(x)/(1):|_(0)^(1)]` `=0+sin^(-1)(1)=(pi)/(2)` sq. units. `"Total area "=((pi)/(2)+1)+(pi)/(2)=pi+1.` |
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| 68. |
Find the distance between the parallel lines of a parallelogram whose base is given 12.5 cm. and area is 75 cm2 |
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Answer» Base of parallelogram = b = 12.5 cm Area of parallelogram = bh = 75 sq.cm. Height of parallelogram = 75/12.5 = 6 cm |
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| 69. |
Fill in the following table with suitable units:S NoDimension Area in units1.mm2.cm23.m24km |
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Answer»
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| 70. |
In the following figure, DE = 7 cm, BC = ……………….A) 3.5 cm B) 7 cm C) 14 cm D) 49 cm |
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Answer» Correct option is C) 14 cm |
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| 71. |
If base of a parallelogram is 18 cm and its height is 11 cm, find its area. |
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Answer» Given, base = 18 cm, height = 11 cm Area of a parallelogram = base × height = 18 × 11 = 198 sq.cm ∴ Area of the parallelogram is 198 sq.cm. |
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| 72. |
In a given figure ABCD is a trapezium in which AB || DC and DA ⊥ AB. If AB = 13 cm., AD = 8 cm. and CD = 7 cm., then find area of trapezium. |
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Answer» Area of trapezium = 1/2 x (sum of parallel sides) x height = 1/2 x (7 + 13) x 8 = 1/2 x (7 + 13) x 8 = 1/2 x 20 x 8 = 80 square cm. |
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| 73. |
If a figure A is formed by two plane figures B and C then the area of A isA) Area of B + Area of C B) Area of B – Area of C C) 2 (Area of B + Area of C) D) 1/2 (Area of B + Area of C) |
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Answer» A) Area of B + Area of C |
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| 74. |
Area is expressed in A) square units B) cubic units C) units D) circular units |
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Answer» A) square units |
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| 75. |
If a triangle and a parallelogram are on the same base and between the same parallels then, area of the triangle is ……………. the area of parallelogram. A) twice B) equal C) half D) none |
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Answer» Correct option is C) half |
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| 76. |
If the areas of two triangles are equal then they are congruent.A) True B) False , C) Can’t say D) None |
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Answer» Correct option is C) Can’t say |
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| 77. |
Some measures are given in the figure, find the area of ☐ABCD. |
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Answer» A(☐ABCD) = A(∆BAD) + A(∆BDC) In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m A(∆BAD) = x product of sides forming the right angle = \(\cfrac{1}{2}\)x l(AB) x l(AD) =\(\cfrac{1}{2}\) x 40 x 9 = 180 sq. m In ∆BDC, l(BT) = 13m, l(CD) = 60m A(∆BDC) = \(\cfrac{1}{2}\)x base x height = \(\cfrac{1}{2}\)x l(CD) x l(BT) = \(\cfrac{1}{2}\)x 60 x 13 = 390 sq. m A (☐ABCD) = A(∆BAD) + A(∆BDC) = 180 + 390 = 570 sq. m ∴ The area of ☐ABCD is 570 sq.m. |
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| 78. |
If area of a parallelogram is 29.6 sq. cm and its base is 8 cm, find its height. |
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Answer» Given, area of a parallelogram = 29.6 sq.cm, base = 8 cm Area of a parallelogram = base × height ∴ 29.6 = 8 × height ∴ height = 29.6/8 = 3.7 cm ∴ Height of the parallelogram is 3.7 cm. |
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| 79. |
If base of a parallelogram is 18 cm and its height is 11 cm, find its area. |
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Answer» Given, base = 18 cm, height = 11 cm Area of a parallelogram = base × height = 18 × 11 = 198 sq.cm ∴ Area of the parallelogram is 198 sq.cm. |
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| 80. |
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area. |
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Answer» Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm. Area of a trapezium = (1/2) x sum of lengths of parallel sides x height = (1/2) x (8.5 + 11.5) x 4.2 = (1/2) x 20 x 4.2 = 10 x 4.2 = 42 sq. cm ∴ The area of the trapezium is 42 sq. cm. |
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| 81. |
In the given figure, ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area ABCD. |
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Answer» ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, Area of a trapezium = x sum of lengths of parallel sides x height ∴ A (ABCD) = x [l(AB) + l(DC)] x l(AD) = (1/2) x (13 + 9) x 8 = (1/2) x 22 x 8 = 11 x 8 = 88 sq.cm ∴ The area of ABCD is 88 sq. cm |
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| 82. |
Fill in the blank:Area of rectangle = Length x ___ |
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Answer» Area of rectangle = Length x breadth. |
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| 83. |
A triangle and a rectangle are situated between same base and parallel lines. The base and height of triangle are 9 cm. and 6 cm. respectively. Area of rectangle is |
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Answer» The correct option is (c). |
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| 84. |
BC = 30 cm. in given figure. Calculate the area of ∆ABC. |
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Answer» In ∆ ABD, ∵ ∠BAD = ∠ABD ∴BD = AD ….(1) In ∆ ADC, ∵ ∠DAC = ∠DCA ∴ DC = AD From (1) and (2) BD = DC = AD ⇒ AD = 1/2 BC = 1/2 x 30 = 15 cm. ∴ Area of ∆ABC = 1/2 x Base x Height = 1/2 x BC x AD = 1/2 x 30 x 15 = 225 cm2 |
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| 85. |
State whether True or False:1. Area of a parallelogram, whose base is 6.5 cm. and height is 4 cm. is 26 cm2.2. 1 meter = 100 square cm.3. 1 hectare = 10000 square meter.4. If the diagonals of a rhombus are 24 cm. and 7 cm. then area of rhombus is 168 cm2. |
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Answer» 1. True |
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| 86. |
In a parallelogram ABCD, AB = 35 cm. and height is 12 cm. Corresponding to this side. BC = 25 cm. then find the corresponding height. |
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Answer» Area of parallelogram ABCD 35 x 12 = 25 x Height corresponding to BC ⇒ Height corresponding to side BC = (35 x 12)/25 = 16.8 cm |
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| 87. |
Each side of the hexagon ABCDEF has side of length 5 cm as given in the figure. Riya and Riema And the area of the region by dividing it into two parts in two different manner. Compare the area in both the case. |
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Answer» Ria’s Method Area of hexagonal shape ABCDEF = Area of ∆AFE + Area of rectangle ∆EDB + Area of ∆BCD = 1/2 x 8 x 3 + 8 x 5 + 1/2 x 8 x 3 = 12 + 40 + 12 = 64 cm2 Raima’s Method Area of hexagonal shape ABCDEF = Area of trapezium ABCF + Area of trapezium FCDE = 1/2 x (5 + 11) x 4 + 1/2 x (5 + 11) x 4 = 32 + 32 = 64 cm2 Hence, both areas are same. |
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| 88. |
Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area. |
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Answer» Lengths of the diagonals of a rhombus are 15 cm and 24 cm. Area of a rhombus = (1/2) × product of lengths of diagonals = (1/2) × 15 × 24 = 15 × 12 = 180 sq.cm ∴ The area of the rhombus is 180 sq. cm |
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| 89. |
a. Where should be the position P in CD to set the isosceles triangle with half the area of the rectangle.b. Where should be P in CD to get ∠ABP a right angle.c. If ∠APB is a right angle how can you find the position P?d. Can such right angled triangle be drawn in any of the rectangle? |
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Answer» a. P is the centre of CD b. It should be in C. c. The distance from centre of AB to P should be half the length of AB. d. No. |
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| 90. |
In the picture below, the side AC of the triangle ABC is extended to D, by adding the length of the side CB. Then the line through C parallel to DB is drawn to meet AB at E.1. Prove that CE bisects ∠C.2. Describe how this can be used to divide an 8 centimeters long line in the ratio 4 : 5.3. Can we use it to divide an 8 centimeters long line in the ratio 3 : 4? How? |
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Answer» 1. ∆ BCD is an equilateral triangle If ∠CBD = x then ∠CDB = x ∠BCD = 180 – 2x ∠BCE = x (line BC passing through parallel lines EC and BD makes equal angle) ∴ ∠ACE = x ∴ line CE bisects ∠C. 2. If AB = 8, AC = 4, and CB = 5 then the line CE divides AB in the ratio 4 : 5. 3. No. We can’t draw a triangle with sides 8cm, 3cm and 4cm. Since 3 : 4 = 6 : 8, take 6cm and 8cm as two sides then find third side. |
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| 91. |
If the curve `y=ax^(1//2)+bx` passes through the point (1,2) and lies above the x-axis for `0lexle9` and the area enclosed by the curve, the x-axis, and the line x=4 is 8 sq. units. ThenA. `a=1`B. `b=1`C. `a=3`D. `b=-1` |
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Answer» Correct Answer - C::D Since the curve `y=ax^(1//2)+bx` passes through the point (1,2) `therefore" "2=a+b" (1)"` By observation the curve also passes through (0,0). Therefore, the area enclosed by the curve, x-axis and x=4 is given by `A=overset(4)underset(0)int(ax^(1//2)+bx)dx=8 or (2a)/(3)xx8+(b)/(2)xx16=8` `or (2a)/(3)+b=1." (2)"` Solving (1) and (2), we get a =3, b=-1. |
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| 92. |
If the area enclosed by curve `y=f(x)a n dy=x^2+2`between the abscissa `x=2a n dx=alpha,alpha>2,`is `(alpha^3-4alpha^2+8)s qdot`unit. It is known that curve `y=f(x)`lies below the parabola `y=x^2+2.` |
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Answer» According to the question, `alpha^(3)-4alpha^(2)+8=overset(alpha)underset(2)int(x^(2)+2-f(x))dx` Differentiating w.r.t. `alpha` on both sides, we get `3alpha^(2)-8alpha=alpha^(2)+2-f(alpha)` `therefore" "f(x)=-2x^(2)+8x+2` |
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| 93. |
If `f(x)={{:(sqrt({x}),"for",xcancelinZ),(1,"for",x in Z):} and g(x)={x}^(2)` where {.} denotes fractional part of x then area bounded by f(x) and g(x) for `x in 0,6 ` isA. `(2)/(3)`B. 2C. `(10)/(3)`D. 6 |
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Answer» Correct Answer - B f(x) and g(x) are periodic with period 1. `therefore"Required area is "int_(0)^(6)[f(x)-g(x)]dx=6int_(0)^(1)(sqrtx-x^(2))dx=2` |
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| 94. |
Let R be the region containing the point (x, y) on the X-Y plane, satisfying `2le|x+3y|+|x-y|le4.` Then the area of this region isA. 5 sq. unitsB. 6 sq. unitsC. 7 sq. unitsD. 8 sq. units |
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Answer» Correct Answer - B The region is enclosed between two parallelograms of area `4xx2 and 2xx1` The area `=8-2=6.` |
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| 95. |
If `S_(0),S_(1),S_(2),…` are areas bounded by the x-axis and half-wave of the curve `y=sin pi sqrt(x)," then prove that "S_(0),S_(1),S_(2),…` are in A.P… |
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Answer» `y= sin pi sqrt(x)` meets x-axis when `pisqrt(x)=npi or x=n^(2), n in N.` Therefore, area of half-wave between `x=n^(2) and x=(n+1)^(2)` is `S_(n)=|overset((n+1)^(2))underset(n^(2))int sin pi sqrt(x)dx |` `"Putting "pisqrt(x)=y and pi^(2) dx =2y dy,`we get `therefore" "S_(n)=|(2)/(pi^(2))overset((n+1)pi)underset(npi)inty sin y dy |` `=|(2)/(pi^(2))[-y cos y + sin y ]_(npi)^((n+1)pi)|` `=|(2)/(pi^(2))[-(n+1)pi cos (n+1) pi +npi cos n pi ]|` `=(2(2n+1))/(pi), n in N` `"Hence, "S_(0),S_(1),S_(2),...` are in A.P.. |
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| 96. |
The area and height of the trapezium are 34 cm2 and 4 cm. One of its parallel side is 10 cm. Find the length of other parallel side. |
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Answer» Let b cm. be the required length of parallel side. Area of trapezium = 1/2 x (Sum of parallel sides) x height ⇒ 34 = 1/2 x (10 + b) x 4 ⇒ 34 x 2 = (10 + b) x 4 ⇒ 68 = (10 + b) x 4 ⇒ 10 + b = 68/4 ⇒ 10 + b = 17 ⇒ b = 17 – 10 ⇒ b = 7 Hence, the length of another parallel side is 7 cm. |
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| 97. |
Fill in the blank:Diagonals of rhombus are___to each other. |
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Answer» Diagonals of rhombus are perpendicular to each other. |
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| 98. |
If area of a rhombus is 506 cm2 and length of diagonal is 23 cm. then find the length of other diagonal(a) 11 cm.(b) 22 cm.(c) 44 cm.(d) 33 cm. |
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Answer» The length of other diagonal is 44 cm. |
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| 99. |
E, F, G and H are respectively the mid points of the sides AB, BC, CD and AD of a parallelogram ABCD. If ar (ABCD) = 24 cm2, then ar (EFGH) =A) 12 cm2 B) 48 cm2 C) 36 cm2D) none of these |
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Answer» Correct option is A) 12 cm2 |
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| 100. |
The perimeter of a rhombus is 52 cm. Distance between two parallel sides is 12 cm. Find the area of rhombus. |
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Answer» Perimeter of rhombus = 52 cm. ∴ One side of rhombus = 52/4 cm = 13 cm. Height of rhombus = 12 cm. ∴ Area of rhombus = base x height = 13 x 12 = 156 cm2 |
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