Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60^{o}. If the area of the quadrilateral is $$4\sqrt 3 $$, then the perimeter
of the quadrilateral is :

A

12.5

B

13.2

C

12

D

13

Here; cos$$\theta $$ = $${{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

and $$\theta $$ = 60

$$ \Rightarrow $$ cos 60

$$ \Rightarrow $$ 10 = 29 $$-$$ c

$$ \Rightarrow $$ c

$$ \Rightarrow $$

also; cos$$\theta $$ = $${{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

and $$\theta $$ = 120

$$ \Rightarrow $$ $$-$$ $${1 \over 2}$$ = $${{{a^2} + {b^2} - 19} \over {2ab}}$$

$$ \Rightarrow $$ a

$$ \Rightarrow $$ a

$$ \therefore $$ Area = $${1 \over 2} \times 2 \times 5$$ sin 60 + $${1 \over 2}$$ ab sin 120

$$ \Rightarrow $$ $${{5\sqrt 3 } \over 2} + {{ab\sqrt 3 } \over 4}$$ = $$4\sqrt 3 $$

$$ \Rightarrow $$ $${{ab} \over 4}$$ = 4 $$-$$ $${5 \over 2}$$ = $${3 \over 2}$$

$$ \Rightarrow $$

$$ \therefore $$ a

$$ \Rightarrow $$ a = 2, b = 3

Perimeter = Sum of all sides

= 2 + 5 + 2 + 3 = 12

2

Let the orthocentre and centroid of a triangle be A(-3, 5) and B(3, 3) respectively. If C is the circumcentre
of this triangle, then the radius of the circle having line segment AC as diameter, is :

A

$${{3\sqrt 5 } \over 2}$$

B

$$\sqrt {10} $$

C

$$2\sqrt {10} $$

D

$$3\sqrt {{5 \over 2}} $$

In a triangle, orthocentre, centroid and circumcenter are collinear and centroid divides orthocenter and circumcenter in 2 : 1 ratio.

Here AB = 2 BC

$$AB = \sqrt {{{\left( {3 + 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}} $$

$$ = \sqrt {36 + 4} $$

$$ = \sqrt {40} $$

$$ = 2\sqrt {10} $$

$$\therefore\,\,\,\,$$ BC $$ = \sqrt {10} $$

$$\therefore\,\,\,\,$$ AB + BC

$$ = 2\sqrt {10} + \sqrt {10} $$

$$ = 3\sqrt {10} $$

AC is the diameter of the circle.

$$\therefore\,\,\,$$ Radius $$ = {1 \over 2}\,\,$$ AC

$$ = {1 \over 2} \times 3\sqrt {10} $$

$$ = 3\sqrt {{5 \over 2}} $$

Here AB = 2 BC

$$AB = \sqrt {{{\left( {3 + 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}} $$

$$ = \sqrt {36 + 4} $$

$$ = \sqrt {40} $$

$$ = 2\sqrt {10} $$

$$\therefore\,\,\,\,$$ BC $$ = \sqrt {10} $$

$$\therefore\,\,\,\,$$ AB + BC

$$ = 2\sqrt {10} + \sqrt {10} $$

$$ = 3\sqrt {10} $$

AC is the diameter of the circle.

$$\therefore\,\,\,$$ Radius $$ = {1 \over 2}\,\,$$ AC

$$ = {1 \over 2} \times 3\sqrt {10} $$

$$ = 3\sqrt {{5 \over 2}} $$

3

If the tangent at (1, 7) to the curve x^{2} = y - 6

touches the circle x^{2} + y^{2} + 16x + 12y + c = 0, then the value of c is :

touches the circle x

A

95

B

195

C

185

D

85

Equation of tangent at (x

xx

Now equation of tangent at (1, 7) to x

$$x\,.\,1 = 4\,.\,{1 \over 4}\left( {{{y + 7} \over 2}} \right) - 6$$

$$ \Rightarrow 2x = y + 7 - 12$$

$$ \Rightarrow 2x - y + 5 = 0$$

This tangent touches the circle.

So, perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

For the circle,

$${x^2} + {y^2} + 16x + 12y + C = 0$$

center is ($$-$$8, $$-$$6)

and radius (r) = $$\sqrt {{8^2} + {6^2} - c} $$

$$ = \sqrt {100 - c} $$

Distance of the tangent from the center of the circle

d = $$\left| {{{2\left( { - 8} \right) - \left( { - 6} \right) + 5} \over {\sqrt {{2^2} + {1^2}} }}} \right|$$

$$ = \left| {{{ - 16 + 6 + 5} \over {\sqrt 5 }}} \right|$$

And we know d = r

$$\therefore\,\,\,$$ $$\left| {{{ - 16 + 11} \over {\sqrt 5 }}} \right| = \sqrt {100 - c} $$

$$ \Rightarrow \left| {{{ - 5} \over {\sqrt 5 }}} \right| = \sqrt {100 - c} $$

$$ \Rightarrow \left| { - \sqrt 5 } \right| = \sqrt {100 - c} $$

$$ \Rightarrow 5 = 100 - c$$

$$ \Rightarrow c = 95$$

4

A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line, $$y - 4x + 3 = 0,$$ then its radius is equal to :

A

2

B

$$\sqrt 5 $$

C

$$\sqrt 2 $$

D

1

Equation of the line passing through the points (2, 3) and (4, 5) is

y $$-$$ 3 = $$\left( {{{5 - 3} \over {4 - 2}}} \right)$$ x $$-$$ 2 $$ \Rightarrow $$ x $$-$$ y + 1 = 0 . . . . . (1)

Equation of the perpendicular line passing the midpoint

(3, 4) is x + y $$-$$ 7 = 0 . . . .(2)

Lines (1) and (2) intersect at the center of the circle .

So, the center of the circle is (3, 4)

Therefore, the radius is

$$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$ = $$\sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( {3 - 4} \right)}^2}} $$ = $$\sqrt 2 $$ units.

y $$-$$ 3 = $$\left( {{{5 - 3} \over {4 - 2}}} \right)$$ x $$-$$ 2 $$ \Rightarrow $$ x $$-$$ y + 1 = 0 . . . . . (1)

Equation of the perpendicular line passing the midpoint

(3, 4) is x + y $$-$$ 7 = 0 . . . .(2)

Lines (1) and (2) intersect at the center of the circle .

So, the center of the circle is (3, 4)

Therefore, the radius is

$$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$ = $$\sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( {3 - 4} \right)}^2}} $$ = $$\sqrt 2 $$ units.

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (4) *keyboard_arrow_right*

AIEEE 2003 (2) *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*