Mathematics

Geometry and Trigonometry

115 Questions

Geometry and trigonometry questions cover circles, tangents, and trigonometric ratios. Problems often combine algebraic geometry with angle properties to test spatial reasoning. This forms a core component of the mathematics section in engineering and civil services exams.

Circle tangentsTrigonometric ratiosHyperbola propertiesEllipse tangentsSecant construction

Geometry and Trigonometry Questions

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

To draw a pair of tangents to a circle which are inclined to each other at an angle of $60^0$, it is required to draw tangents at endpoints of those two radii of the circle, the angle between them should be

  1. $135^{0}$
  2. $90^{0}$
  3. $60^{0}$
  4. $120^{0}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given:

$ PA$ & $PB$ are two tangents drawn from P to a circle with centre O at A & B respectively. 
$\angle APB={ 60 }^{ o }$ 
To find out-
$ \angle AOB$ 
Solution-
$ PA$ & $PB$ are two tangents drawn from P to the circle at A & B, respectively.
$ \therefore  PA=PB\Longrightarrow \Delta PAB$ is isosceles. 
i.e $\angle PAB=\angle PBA.$ 
or $\angle PAB+\angle PBA=2\angle PAB$  .....(i)
Now, $\angle PAB+\angle PBA+\angle APB={ 180 }^{ o }$    ....(angle sum property of triangles)
$ \Longrightarrow \angle PAB+\angle PBA{ +60 }^{ o }={ 180 }^{ O }$
$\Longrightarrow 2\angle PAB={ 120 }^{ o }$     ...(from i)
$ \therefore  \angle PAB={ 60 }^{ o }=\angle PBA$  .........(ii)
Again, $\angle OAP={ 90 }^{ o }$    ....(angle between a radius and the tangent at the point of contact.)
$ \therefore \angle OAB=\angle OAP-\angle PAB={ 90 }^{ o }-{ 60 }^{ o }$    .... (from ii)    .........(iii)
Since, $OA=OB$     ...(radii of the same circle)
$ \therefore  \Delta OAB$ is isosceles
$\Longrightarrow \angle OAB=\angle OBA={ 30 }^{ o }$   ....(from iii)
So, $\angle AOB={ 180 }^{ o }-(\angle OAB+\angle OBA)={ 180 }^{ o }-{ 30 }^{ o }-{ 30 }^{ O }={ 120 }^{ O }$      ...(angle sum property of triangles)

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

If two tangents inclined at an angle of $60^{\circ}$ are drawn to a circle of radius 3 cm, then length of the tangent is equal to :

  1. $\sqrt{3}cm$
  2. $2\sqrt{3}cm$
  3. $\frac{2}{\sqrt{3}}cm$
  4. $3\sqrt{3}cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Tangent is perpendicular to radius at the point of contact.

By symmetry with respect to the line joining the center and the point from which tangents are drawn, we have the length of tangent $= \dfrac{3}{\tan 30^o} = 3\sqrt{3} cm$.
So option D is the right answer.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The equation of tangent to the circle ${x^2} + {y^2} = 36$ which are incline at the angle of  ${45^ \circ }$ to the $x-$axis are 

  1. $x + y = \pm \sqrt 6 $
  2. $x = y \pm 3\sqrt 2 $
  3. $y = x \pm 6\sqrt 2 $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Tangents to the circle x^2 + y^2 = r^2 inclined at an angle of 45 degrees to the x-axis have a slope of m = tan(45 degrees) = 1 or -1. Using the formula y = mx + r*sqrt(1 + m^2), with r = 6, we get y = x plus or minus 6 times square root of 2.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A tangent drawn from the point (4, 0) to the circle $\displaystyle x^{2}+y^{2}=8 $ touches it at a point A in the first quadrant. The coordinates of another point B on the circle such that $AB$ = 4 are

  1. $(2, -2)$
  2. $(-2, 2)$
  3. $\displaystyle \left ( -2\sqrt{2},0 \right ) $
  4. $\displaystyle \left ( 0,-2\sqrt{2} \right ) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^2  + y^2 = (2\sqrt 2)^2 , C = (0,0) , r = 2 \sqrt 2$

Let $y = mx + 2\sqrt 2 \sqrt{m^2 + 1}$ be a tangent

To find the tangent through $(4,0)$ substitute into the equation

$\implies 0 = 4m +2\sqrt 2 \sqrt{m^2 + 1}$

$\implies 16m^2 = 8(m^2 + 1)$

$\implies m = -1$

Equation is

$y = -x + 4$

Substituting in circle equation

$x^2 + (-x + 4)^2 = 8$

$\implies 2x^2 – 8x + 8 = 0$

$\implies x = 2 \implies y = 2$

$A = (2,2)$

Any point on the circle is given be$ (2\sqrt2 \cos \theta, 2\sqrt2 \sin \theta )$

Let B = $(2\sqrt2 \cos \theta _1, 2\sqrt2 \sin \theta _1)$

$AB = 4 \implies AB^2 = 16$

$\implies (2\sqrt2 \cos \theta _1, -2)^2 + (2\sqrt2 \sin \theta _1 - 2)^2 = 16$

$\implies 8 + 4 + 4 – 4\sqrt 2(\cos \theta _1 + \sin \theta _1) = 16$

$\implies \sin \theta _1 + \cos \theta _1 = 0$

$\theta _1 = \dfrac{-\pi}{4}$

$B = (2\sqrt 2 \times\dfrac{1}{\sqrt 2} 2\sqrt 2 \times\dfrac{-1}{\sqrt 2})  = (2,-2)$

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

From a point $R(5, 8)$ two tangents $RP$ and $RQ$ are drawn to a given cirlce $S = 0$ whose radius is $5$. If circumcentre of the triangle PQR is $(2, 3)$, then the equation of circle $S= 0$ is

  1. $x^2 + y^2 + 2x + 4y - 20 = 0$
  2. $x^2 + y^2 + x + 2y - 10 = 0$
  3. $x^2 + y^2 - x - 2y - 20 = 0$
  4. $x^2 + y^2 - 4x - 6y - 12 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circumcenter of triangle PQR, where P and Q are points of tangency from R, is the midpoint of the chord of contact and the center of the circle. Using the given circumcenter and radius, the equation can be derived.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

A tangent to the circle $x^{2}+y^{2}=a^{2}$ meets the axes at points A and B. The locus of the mid point of AB is 

  1. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{a^{2}}$
  2. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{a^{2}}$
  3. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=4a^{2}$
  4. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{a^{2}}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the tangent to the circle x^2 + y^2 = a^2 be written in intercept form as x/p + y/q = 1. Since it is tangent to the circle centered at the origin with radius a, the perpendicular distance from the origin to the line equals a, which gives 1/p^2 + 1/q^2 = 1/a^2. The intercepts on the axes are A(p, 0) and B(0, q), so the midpoint (h, k) of AB is (p/2, q/2), meaning p = 2h and q = 2k. Substituting these into the tangent condition yields 1/(4h^2) + 1/(4k^2) = 1/a^2, which simplifies to 1/x^2 + 1/y^2 = 4/a^2.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

R is the midpoint of the segment $\bar{PT}$, and $Q$ is the midpoint of line segment $\bar{PR}$. If $S$ is a point between $R$ and $T$ such that the length of segment $\overline{QS}$ is $10$ and the length of segment $\overline{PS}$ is $19$, what is the length of segment $\overline{ST}$?

  1. $13$
  2. $14$
  3. $15$
  4. $16$
  5. $17$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Given that $PR=RT$ and $PQ=QR$

Let $QR=PQ=x$ , we get $PR=RT=2x$
Given that $S$ is a point between $R$ and $T$
Given $QS=10$ , $PS=19$
$PS=PQ+QS=PQ+10=19$
$\Rightarrow PQ=9$
Therefore we get $PT=4x=36$
$\Rightarrow ST=PT-PS=36-19=17$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Assume that, $\Delta RST \sim \Delta XYZ$. Complete the following statement.


$\displaystyle \frac{RT}{XY} = \frac{- -}{YZ}, \frac{RS}{XY} = \frac{ST}{- -}, \frac{XY}{ - -} = \frac{YZ}{ST}$

  1. ST, YZ, RT

  2. ST, YZ, RS

  3. YT, YS, RZ

  4. ST, YZ, RZ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given triangle RST similar to triangle XYZ, the ratios of corresponding sides are equal: RS/XY = ST/YZ = RT/XZ. The provided option B correctly completes the ratios.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The number of tangents to the circle ${x}^{2}+{y}^{2}=3$ that are normals to the ellipse $\cfrac{{x}^{2}}{9}+\cfrac{{y}^{2}}{4}$ is

  1. one

  2. two

  3. three

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $y=mx+c$ is tangent to $x^2+y^2=3$ 
Then by condition of tangency
$\left|\dfrac{c}{\sqrt{m^2+1}}\right|=\sqrt{3}$
$\Rightarrow c^2=3(m^2+1)$        ...(i)
$y=mx+c$ is Normal to $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
in $c=\dfrac{(b^2-a^2)m}{\sqrt{a^2+b^2m^2}}$

$\Rightarrow c=\dfrac{(4-9)m}{\sqrt{9+4m^2}}$

$\Rightarrow c^2=\dfrac{25m^2}{4m^2+9}$

$\Rightarrow 3(m^2+1)(4m^2+9)=25m^2$
let $m^2=t$
$\Rightarrow 3(t+1)(4t+9)=25t$
$\Rightarrow$ since, $D<0$.
Hence, There is no rout:
Hence, there is no tangent to circle which is Normal to ellipse.
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Tangents are drawn to the ellipse $ \displaystyle \frac{x^2}{a^2}+\displaystyle \frac{y^2}{b^2}=1 $ at points where it is intersected by the line $ \ell x+my+n=0 $. Find the point of intersection of tangents at these points.

  1. $ \displaystyle \frac{-a^2}{n},\displaystyle \frac{-b^2m}{n} $
  2. $ \displaystyle \frac{-a^2\ell}{n},\displaystyle \frac{-b^2m}{n} $
  3. $ \displaystyle \frac{-a^2\ell}{n},\displaystyle \frac{-b^2}{n} $

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the point of intersection of the
Line $lx+my+n=0$ and the ellipse $\cfrac { { x }^{ 2 }

}{ {a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of tangent at $P$ is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } +\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the point of intersection of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { -{ b }^{ 2 }m }{ n } $
Therefore, the point of intersection of given line and  the given ellipse is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $
Hence, option 'B' is correct.

Multiple choice position of point wrt ellipse ellipse maths

A tangent to the ellipse $4x^2+9y^2=36$ is cut by tangent at the extremities of the major axis at $T$ and $T'$. The circles on $TT'$ as diameters passes through the point 

  1. $(0,-\sqrt5)$
  2. $(\sqrt5,0)$
  3. $(0,0)$
  4. $(3,2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ellipse 4x^2 + 9y^2 = 36 is x^2/9 + y^2/4 = 1. The tangents at the extremities of the major axis are x = 3 and x = -3. A tangent to the ellipse is y = mx + sqrt(9m^2 + 4). The intersection points T and T' with x=3 and x=-3 are found, and the circle with diameter TT' is constructed. This circle passes through the foci of the ellipse, which are at (+/- sqrt(9-4), 0) = (+/- sqrt(5), 0).

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

P, Q, R are the points of intersection of a line 1 with sides BC, CA, AB of a $\Delta$ ABC 
respectively, then $\dfrac{BP}{PC} \dfrac{CQ}{QA} \dfrac{AR}{RB}$

  1. 1

  2. 2

  3. -1

  4. -2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a direct application of Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

In a trapezium, the lengths of its parallel sides are $a$ and $b$. The length of the line joining the midpoints of its non-parallel sides is :

  1. $\dfrac{a-b}{2}$
  2. $\dfrac{a+b}{2}$
  3. $\dfrac{ab}{2}$
  4. $\dfrac{ab}{a+b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line segment joining the midpoints of the non-parallel sides of a trapezium is known as its median. The length of this line segment is always equal to half the sum of the lengths of the parallel sides, expressed as (a + b) / 2.

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

The line joining the mid points of the diagonals of a trapezium has length $3$cm. If the longer base is $97$cm then the shorter base is:

  1. $94$cm
  2. $92$cm
  3. $91$cm
  4. $90$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line joining the mid point of the diagonals of a trapezium is half the length of the difference between the two sides.
Let the smaller side be $x$
Then, $3 = \dfrac{97 -x}{2}$
$6= 97 - x$
$x = 91$ cm

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

The parallel sides of a trapezium are $x$ and $y$ in length. The length of the line segment joining the mid points of the non parallel sides is:

  1. $\dfrac{x+y}{2}$
  2. $x+y$
  3. $\dfrac{2x+3y}{2}$
  4. $\dfrac{xy}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line segment joining the mid points of non parallel sides of a trapezium is the average of sum of the parallel sides.
Hence, $= \dfrac{x+y}{2}$