Mathematics · Quantitative Aptitude

Geometry and Coordinate Geometry

84 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 triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

If a triangle $PQR$ has been constructed taking $QR = 6 $ cm, $PQ = 3 $ cm and $PR = 4 $ cm, then the correct order of the angle of triangle is

  1. $\displaystyle \angle P< \angle Q< \angle R $
  2. $\displaystyle \angle P> \angle Q< \angle R $
  3. $\displaystyle \angle P> \angle Q> \angle R $
  4. $\displaystyle \angle P< \angle Q> \angle R $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, in $\triangle PQR$, $QR=6$ cm, $PQ=3$ cm, $PR=4$ cm

We know, 
(i) the shortest side is always opposite the smallest interior angle.

(ii) the longest side is always opposite the largest interior angle.
Here, $QR=6$ cm is the largest side, therefore $\angle P$ is the greatest.

And $PQ=3$ cm is the smallest side, therefore $\angle R $ is the smallest angle.
Therefore, the correct order is $\angle P>\angle Q>\angle R$.

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

In a $\displaystyle \Delta PQR$ PQ = PR and $\displaystyle \angle Q$ is twice that of $\displaystyle \angle P$ Then $\displaystyle \angle Q$__

  1. 75$\displaystyle ^{\circ}$
  2. 65$\displaystyle ^{\circ}$
  3. 72$\displaystyle ^{\circ}$
  4. 100$\displaystyle ^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \because $ PQ = PR 
$\displaystyle \Rightarrow $ $\displaystyle \angle Q=\angle R$
Given that $\displaystyle \angle Q=2\angle P$
We have
$\displaystyle \angle P+\angle Q+\angle R=180^{\circ}$
$\displaystyle \frac{\angle Q}{2}+\angle Q+\angle Q=180^{\circ}$
$\displaystyle \frac{5}{2}\angle Q=180^{\circ}$
$\displaystyle \angle Q=72^{\circ}$        

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 construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

To construct a quadrilateral minimum of its _________ elements are required.

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

To construct a unique quadrilateral, we will be need a minimum of $5$ dimensions.

If we have five dimensions, we can draw a side first then mark angle on both ends then we can construct a quadrilateral uniquely.
Or if we have three sides and two included angles then also we can construct a unique quadrilateral.
Unless we are constructing any one of the special quadrilaterals.

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.