Mathematics · Quantitative Aptitude

Geometry and Coordinate Geometry

72 Questions

Geometry questions test your knowledge of triangle properties, congruence, similarity, and quadrilaterals. They form a significant portion of the mathematics section in competitive exams. Practicing these builds spatial reasoning and theorem application skills.

triangle similaritycongruence rulesquadrilateral propertiesright angle properties

Geometry and Coordinate Geometry Questions

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

In a $\Delta$ PQR, if $3\sin P+4\cos Q=6$ and $4 \sin Q+3\cos P=1$, then the angle $R$ is equal to :

  1. $\dfrac{3\pi}{4}$
  2. $\dfrac{5\pi}{6}$
  3. $\dfrac{\pi}{6}$
  4. $\dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given trignometric equations are:

$3 \sin{P} +4 \cos{Q} =6$ -------(1)
$4 \sin{Q} +3 \cos{P} =1$ -------(2)

Squaring both equations (1) and (2) and adding them, we get
$\Rightarrow 9\left( \sin ^{ 2 }{ P } +\cos ^{ 2 }{ P }  \right) +16\left( \sin ^{ 2 }{ Q } +\cos ^{ 2 }{ Q }  \right) +24\left( \sin { Q } \cos { P } +\cos { Q } \sin { P }  \right) =36+1$

$ \Rightarrow 9+16+24\sin { \left( P+Q \right)  } =37$

$ \therefore \sin { \left( P+Q \right)  } =\cfrac { 37-25 }{ 24 } =\cfrac { 12 }{ 24 } =\cfrac { 1 }{ 2 } $

$\therefore P+Q=30°$

Hence, angle $R=180°-30°=150°=\cfrac { 5\pi  }{ 6 } $radian

Multiple choice mathematics and statistics angle and its measurement directed angles

In a $\Delta$PQR, if $\angle P - \angle Q = 42^{\circ}$ and $\angle Q - \angle R = 21^{\circ}$, find $\angle P, \angle Q$ and $\angle R$.

  1. $\angle P = 105^{\circ} \angle Q = 53^{\circ} \angle R = 32^{\circ}$
  2. $\angle P = 25^{\circ} \angle Q = 53^{\circ} \angle R = 32^{\circ}$
  3. $\angle P = 95^{\circ} \angle Q = 53^{\circ} \angle R = 32^{\circ}$
  4. $\angle P = 75^{\circ} \angle Q = 53^{\circ} \angle R = 32^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,
$\angle P - \angle Q = 42$ (I)
$\angle Q - \angle R = 21$ (II)
Sum of angles = 180
$\angle P + \angle Q + \angle R = 180$ (III)
Add all the three equations:
$2 \angle P + \angle Q = 180 + 42 + 21$
$2 \angle P + \angle Q = 243$ (IV)

Add I and IV,
$2 \angle P + \angle P = 243 + 42$
$3 \angle P = 285$
$\angle P = 95^{\circ}$
From IV, $\angle Q = 243 - 2\times 95$
$\angle Q = 53^{\circ}$
From II,
$\angle R = 53 - 21 = 32^{\circ}$

Multiple choice mathematics and statistics angle and its measurement directed angles

In $\displaystyle \angle ROP,$ the vertex is at:

  1. $R$
  2. $P$
  3. $O$

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

Here, the angle is written as $\angle{ROP}$ and the angle is made by the intersection of two lines. 

So, here $RO$ and $OP$ are two lines which makes the angle at point $O$.
Hence, the vertex is at $O$.

Multiple choice mathematics and statistics angle and its measurement directed angles

In $\displaystyle \angle PRQ $, the two arms are:

  1. $\displaystyle \overrightarrow{PR} $ and $\displaystyle \overrightarrow{RQ} $
  2. $\displaystyle \overrightarrow{RP} $ and $\displaystyle \overrightarrow{PQ} $
  3. $\displaystyle \overrightarrow{QR} $ and $\displaystyle \overrightarrow{QP} $

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An angle is made by the intersection of two lines and that lines are also called as arms.
In $\angle{PRO}$, the angle is formed by the intersection of $\overrightarrow{PR}$ and $\overrightarrow{RO}$ at $R$.
Hence, the two arms are $\displaystyle \overrightarrow{PR} $ and $\displaystyle \overrightarrow{RQ} $.
Multiple choice maths drawing of different geometrical figures constructing perpendicular lines perpendicular to a line from an external point constructing an perpendicular line constructing a perpendicular bisector construction of a perpendicular bisector construction of penpendicual bisector set squares

$\overset \leftrightarrow{PQ}$ is perpendicular to $\overset \leftrightarrow{RS}$ is symbolically written as

  1. $\overset \leftrightarrow{PQ}\, \perp \, \overset \leftrightarrow{RS}$
  2. $\overset \leftrightarrow{PQ}\, \parallel \, \overset \leftrightarrow{RS}$
  3. $\overset \leftrightarrow{PQ}\, \neq \, \overset \leftrightarrow{RS}$
  4. $\overset \leftrightarrow{PQ}\, = \, \overset \leftrightarrow{RS}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \overleftrightarrow { PQ } $ is perpendicular to $ \overleftrightarrow { RS } $ is symbolically written as $ \overleftrightarrow { PQ } \bot  \overleftrightarrow { RS }  $

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

Let O(0, 0), P(3,4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR,PQR, OQR are of equal area. The coordinates of R are 

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

Let coordinate of $R = (a,b)$

Given area of triangle OPR, PQR, and OQR are same. 

$\cfrac{1}{2}\begin{vmatrix} 0&0&1\3&4&1\a&b&1\end{vmatrix}=\cfrac{1}{2}\begin{vmatrix} 3&4&1\6&0&1\a&b&1\end{vmatrix}=\cfrac{1}{2}\begin{vmatrix} 0&0&1\6&0&1\a&b&1\end{vmatrix}$

$\Rightarrow 3b-4a=24-4a-3b=6b$.

Solving this equation be get $a=3, b =\cfrac{4}{3}$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If $P=(x _{1}, y _{1}), Q=(x _{2}, y _{2})$ and $R=(x _{3}, y _{3})$ are three points of a triangle in $\mathbb{R}^{2}$. Then, area of a $\triangle PQR$ in terms of determinant of matrix $M=\begin{bmatrix} 1& 1 & 1 \ x _{1} & x _{2} & x _{3} \ y _{1} & y _{2} & y _{3}\end{bmatrix}$ is

  1. $-|det(M)|$
  2. $|det(M)|$
  3. $\dfrac{1}{2}|det(M)|$
  4. $2|det(M)|$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P=\left( { x } _{ 1 },{ y } _{ 1 } \right) ,Q=\left( { x } _{ 2 },{ y } _{ 2 } \right) ,R\left( { x } _{ 3 },{ y } _{ 3 } \right) $
$\left| M \right| =\triangle =\begin{vmatrix} 1 & 1 & 1 \\ { x } _{ 1 } & { x } _{ 2 } & { x } _{ 3 } \\ { y } _{ 1 } & { y } _{ 2 } & { y } _{ 3 } \end{vmatrix}$
Area of triangle$=\cfrac { 1 }{ 2 } det(M)$
Answer: Option C
Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

If $\triangle ABC \cong \triangle QPR$ and $\dfrac {ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac {9}{4}$, $AB=18\ cm$ and $BC=15\ cm$, then $PR$ is equal to________ $cm$

  1. $10$
  2. $12$
  3. $20/3$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of areas is 9/4, so the ratio of corresponding sides is sqrt(9/4) = 3/2. Since ABC ~ QPR, AB/QP = BC/PR = 3/2. 15/PR = 3/2, so 3*PR = 30, PR = 10.

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

Let $\Delta$ PQR be a right angled isoceles triangle which is right angled at $P(2,1)$. lf the equation of the line OR is $2x+y=3$, then the equation representing the pair of lines PQ and PR is 

  1. $3x^{2}-3y^{2}+8xy+20x+10y+25=0$
  2. $3x^{2}-3y^{2}+8xy-20x-10y+25=0$
  3. $3x^{2}-3y^{2}+8xy-10x+15y+20=0$
  4. $3x^{2}-3y^{2}+8xy-10x-15y+20=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the triangle is isosceles, and right angled at P, therefore 
$PQ=PR$
And 
$PQ$ is perpendicular to $PR$.
The the combined equation of both PQ and PR should be of the form,
$ax^{2}-ay^{2}+bxy+cx+dy+e=0$
Since they are perpendicular and the coefficients of $x^{2}$ and $y^{2}$ will be equal and opposite.
Now the point P must satisfy the equation.
From the above options, the point P (2,1) only satisfies the equation in Option B.
Hence the correct Option is B.

Multiple choice maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

If we have to construct a square $PQRS$ whose diagonal is $8 \sqrt 2$ cm then its side is equal to ?

  1. $8$ cm
  2. $4\sqrt2$ cm
  3. $4$ cm
  4. $8\sqrt2$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the diagonal of square is $a$, then its side $=\dfrac{a}{\sqrt2}$

If diagonal is $8\sqrt2 $ cm, then its side $=\dfrac{8\sqrt2}{\sqrt2}=8$ cm.

Multiple choice maths geometrical construction constructing a perpendicular bisector construction of a perpendicular bisector construction of penpendicual bisector set squares

For drawing the perpendicular bisector of $PQ$, which of the following radii can be taken to draw arcs from $P$ and $Q$?

  1. $\dfrac{PQ}2$
  2. $\dfrac{PQ}3$
  3. $\dfrac{2PQ}3$
  4. $\dfrac{PQ}4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To draw a perpendicular bisector of a given side, take any length that is greater than half the length of the side. Draw the arcs from the edges of the base. The point where arcs meet is on the perpendicular bisector.


From the given options,

$\dfrac{2PQ}{3}$ can be considered to draw to draw arcs from edges $P, \ Q$

Remaining options has the value $\leq \dfrac{PQ}{2}$

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

$P, Q, R$ and $S$ are the mid-points of sides $AB. BC, CD$ and $DA$ respectively of rhombus $ABCD$. Show that $PQRS$ is a rectangle.
Under what condition will $PQRS$ be a square ?

  1. When $ABCD $ is a square.
  2. When $ABCD$ is a parallelogram
  3. When $ABCD$ is a rectangle
  4. When $ABCD$ is a square or a rectangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $ABCD$ is a rhombus. $P, Q, R, S$ are mid points of $AB, BC, CD, DA$ respectively.
Join: $AC$ and $BD$

In $\triangle ABC$
$P$ is mid point of $AB$ and $Q$ is mid point of $AC$.
Thus, by mid point theorem, $PQ \parallel AC$ and $PQ = \dfrac{1}{2} AC$
Similarly, In $\triangle ACD$,
$S$ is mid point of $AD$ and $R$ is mid point of $CD$.
Thus, by mid point theorem, $SR \parallel AC$ and $SR = \dfrac{1}{2} AC$
Hence, $PQ \parallel SR$ and $PQ = SR$

Similarly, $PS = QR$ and $PS \parallel QR$
Thus, the opposite sides of $PQRS$ are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since $AC \perp BD$ thus, $PS \perp PQ$ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of $PQRS$ are equal and parallel and the sides meet each other at right angles. Hence, $PQRS$ is a rectangle.

If $PQRS$ had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if $ABCD$ is a square.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

For construction of a $\triangle PQR$, where $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the third step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass. 
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

  1. $3$
  2. $4$
  3. $2$
  4. $5$
  5. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $p$ to $Q$. $PQR$ is required triangle.
So the third step is $2$
Option $C$ is correct.