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 two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Let $PQR$ be a right angled isosceles triangle, right angled at $P(2, 1)$. If the equation of the line $QR$ 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

The equations of $PQ$ and $PR$ are given by
$y - 1 = \dfrac {-2\mp \tan 45^{\circ}}{1\pm (-2)\tan 45^{\circ}} (x - 2)$


$\Rightarrow y - 1 = \left (\dfrac {-2\mp 1}{1\pm 2}\right ) (x - 2)$

$\Rightarrow y - 1 = -\dfrac {1}{3} (x - 2)$ and $y - 1 = 3(x - 2)$

$\Rightarrow x + 3y = 5$ and $3x - y = 5$
The combined equation of these two lines is
$(x + 3y - 5)(3x - y - 5) = 0$
$\Rightarrow 3x^{2} - 3y^{2} + 8xy - 20x - 10y + 25 = 0$.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Let $\triangle PQR$ be a right angled isosceles triangle, right angled at $P(2, 1)$. If the equation of the side $QR$ is $2x + y = 3$, then the combined equation of sides $PQ$ and $PR$ is

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

Slopes of the line $PQ$ and $PR$ are
$\tan \left(\theta +\dfrac\pi4\right)=\dfrac{1+\tan \theta }{1-\tan \theta }=\dfrac{1-2}{1+2}=-\dfrac{1}{3}$ and $3$
$\therefore $ Equations of $PQ$ and $PR$ are $3y + x - 5 = 0$ and $y-  3x + 5 = 0$
$\therefore $ Combined equation of $PQ$ and $PR$ is
$3x^{2}+8xy-3y^{2}-20x-10y+25=0$

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

In $\triangle PQR,$ $PQ=4$ cm, $QR=3$ cm, and $RP=3.5$ cm. $\triangle DEF$ is similar to $\triangle PQR.$ If $EF=9$ cm, then what is the perimeter of $\triangle DEF: ?$

  1. $10.5$ cm
  2. $21$ cm
  3. $31.5$ cm

  4. Cannot be determined as data is insufficient
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$PQ = 4, QR = 3$ and $RP = 3.5$
Also, $EF = 9$
Now, perimeter of $\triangle PQR = PQ + QR + RP = 4 + 3 + 3.5 = 10.5$
Given, $\triangle DEF \sim \triangle PQR$
$\dfrac{EF}{QR} = \dfrac{Perimeter(\triangle DEF)}{Perimeter(\triangle PQR)}$
$\dfrac{9}{3} = \dfrac{Perimeter(\triangle DEF)}{10.5}$
Perimeter $(\triangle DEF) = 31.5$ cm

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

The correspondence $ABC\rightarrow PQR$ is a similarity in $\Delta ABC$ and $\Delta PQR$. If the perimeter of $\Delta ABC$ is $24$ and the perimeter of $\Delta PQR$ is $40$, then $AB=PQ=$

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

For similar triangles, the ratio of any pair of corresponding sides is equal to the ratio of their perimeters. Given the perimeter of ABC is 24 and PQR is 40, the ratio is 24/40, which simplifies to 3/5. Thus, the ratio AB/PQ equals 3/5.

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

The length of the sides of $\triangle DEF$ are $4,6,8$  $\triangle DEF \sim \triangle PQR$ for correspondence $DEF \leftrightarrow QPR$ if the perimeter of $\triangle PQR=36$, then the length of the smallest side of $\triangle PQR$ is_____

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sides of triangle DEF are 4, 6, and 8, giving a perimeter of 4 + 6 + 8 = 18. Triangle PQR is similar with a perimeter of 36, meaning the scale factor from DEF to PQR is 36/18 = 2. Multiplying the smallest side of DEF (which is 4) by this scale factor gives 4 * 2 = 8.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The corner of a square OPQR is folded up so that the plane OPQ is perpendicular to the plane OQR, the angle between OP and QR is 

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

When the corner of a square is folded such that two planes are perpendicular, the geometry dictates that the angle between the original lines OP and QR becomes 90 degrees.

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

If  H is orthocenter of triangle PQR then PH + QH + RH is 

  1. QR cot P + PR cot Q + PQ cot R

  2. (pq + QR + RP) (cot P + cot Q + copt R)
  3. $\dfrac{1}{2r}(cot P + cotQ + cot R)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In any triangle, the distance from the orthocenter to the vertices is given by 2R cos A, 2R cos B, and 2R cos C. Summing these and relating them to the side lengths and cotangents leads to the identity PH + QH + RH = QR cot P + PR cot Q + PQ cot R.

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

consider a triangle PQR in which the relation $ QR^2+PR^2=5*PQ^2$ holds. let G be the point of intersection of the medians PM and QN . then angle QGM is always

  1. less then 45 degree

  2. obtuse

  3. a right angle

  4. acute and larger than 45 degree
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Apollonius theorem and the properties of medians, the condition QR^2 + PR^2 = 5PQ^2 implies specific geometric constraints on the triangle, leading to the angle QGM being less than 45 degrees.

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

In $\triangle PQR$, $\angle P=30^o$, $\angle Q=60^0$, $\angle R= 90^o$ and $PQ=10 $ units. 

Find $PR$ and $QR$.

  1. $PR =$ $10$ units, $QR =$ $5\sqrt 3$ units
  2. $PR =$ $5$ units, $QR =$ $5\sqrt 3$ units
  3. $PR =$ $5\sqrt 3$ units, $QR =$ $5$ units
  4. $PR =$ $5$ units, $QR =$ $10\sqrt 3$ units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\triangle PQR$ is a $30^o-60^o-90^o$ triangle        ....given

Since, $\angle R = 90^o$, side $PQ$ is hypotenuse.

$\Rightarrow $ By $30^o-60^o-90^o$ theorem,

$\Rightarrow PR = \dfrac {\sqrt 3}{2} \times PQ$ and $QR = \dfrac 12 \times PQ$

$\Rightarrow  PR = \dfrac {\sqrt 3}{2} \times 10$ and $QR = \dfrac 12 \times 10$ 

$\Rightarrow  PR = 5 \sqrt 3$ units and $QR = 5$ units
So, option C is correct.

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

If angles P, Q, R and S of the quadrilateral PQRS, taken in order, are in the ratio 3:7:6:4 then PQRS is a

  1. rhombus

  2. parallelogram

  3. trapezium

  4. kite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the ABCD is a quadrilateral witch $\angle A, \angle B,\angle Cand \angle D$ in ratio  3:7:6:4

Sum of ratio 3+7+6+4=20
$\angle A+ \angle B+\angle C + \angle D=360^{0}$
$\therefore \angle A=\frac{3}{20}\times 360=54^{0}$
$\therefore \angle B=\frac{7}{20}\times 360=126^{0}$
$\therefore \angle c=\frac{6}{20}\times 360=108^{0}$
$\therefore \angle D=\frac{4}{20}\times 360=72^{0}$
Then  ABCD is a quadrilateral is trapezium

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures
In a kite shaped figure $ ABCD $, $ AB= AD $  and $ CB= CD $. Points $ P, Q $ and $ R $ are mid-points of sides $ AB, BC $ and $ CD $ respectively. Then, find $ \angle PQR  $
  1. $45^{\circ}$
  2. $30^{\circ}$
  3. $90^{\circ}$
  4. $60^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: ABCD is a kite shaped figure. $AB = AD$, $BC = CD$, P, Q< R are mid points of AB, BC and CD respectively


To Prove: $\angle PQR = 90^{\circ}$

Construction: Join AC and BD

In $\triangle ABC$

P is mid point of AB and Q is mid point of BC

Then, $PQ \parallel AC$ (By Mid point theorem)

In $\triangle BCD$

Q is mid point of BC and R is mid point of CD

Then, $QR \parallel BD$ (mid point theorem)

We know, diagonals of the Kite shaped figure, bisect at right angles.
thus, $AC \perp BD$ 

hence, $PQ \perp QR$ (Angle made between two lines is same as the angle between their corresponding parallel sides)
or $\angle PQR = 90^{\circ}$

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

If angles P, Q, R and S of the quadrilateral PQRS taken in order are in the ratio 3 : 7 : 6 : 4 then PQRS is a

  1. rhombus

  2. parallelogram

  3. trapezium

  4. kite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  $\angle P:\angle Q:\angle R:\angle S=3:7:6:4$    [ Given ]

$\Rightarrow$  Let $\angle P=3x,\,\angle Q=7x\,\angle R=6x$ and $\angle S=4x$.
$\Rightarrow$  In quadrilateral PQRS,
$\Rightarrow$  $\angle P+\angle Q+\angle R+\angle S=360^o$
$\therefore$   $3x+7x+6x+4x=360^o$
$\therefore$   $20x=360^o$
$\therefore$   $x=18^o$
$\Rightarrow$  $\angle P=3\times 18^o=54^o$
$\Rightarrow$  $\angle Q=7\times 18^o=126^o$
$\Rightarrow$  $\angle R=6\times 18^o=108^o$
$\Rightarrow$  $\angle S=4\times 18^o=72^o$
$\Rightarrow$  Now, $\angle P+\angle Q=54^o+126^o=180^o$
$\Rightarrow$  and $\angle R+\angle S=180^o+ 72^o=180^o$
$\Rightarrow$  If transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.
$\therefore$    $PS\parallel QR$
$\therefore$  Quadrilateral $PQRS$ is a trapezium.

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

State true or false:
In quadrilateral PQRS, $\angle P : \angle Q : \angle R : \angle S = 3 : 4 : 6 : 7$. The Quadrilateral PQRS is trapezium

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given in $\Box$ PQRS,$\angle P:\angle Q:\angle R:\angle S=3:4:6:7$

 Let $ \angle P=3x,\angle Q=4x,\angle R=6x,\angle S=7x$

Sum of interior angles of a quadrilateral$={ 360 }^{ o }$

So $\angle P+\angle Q+\angle R+\angle S=360^o$

$ \Rightarrow 3x+4x+6x+7x=360^o$

$ \Rightarrow 20x=360^o$

$ \Rightarrow x=\dfrac { 360 ^o}{ 20 } $

$ \Rightarrow x=18^o$

So $\angle P=3x=3\times 18^o={ 54 }^{ o }$

$\angle Q=4\times 18^o={ 72 }^{ o }$

$\angle R=6\times 18^o={ 108 }^{ o }$

$\angle S=7\times 18^o=126^{ o }$

Now  $\angle P+\angle S=54^o+126^o=180^o\quad \& \quad \angle Q+\angle R=72^o+108^o=180^o$

In  quadrilateral  PQRS,  $\angle P\& \angle S$ are supplementary  as  well  as  $\angle Q\& \angle R$  are supplementary.

This  is only possible when side PQ$\parallel$ SR ;  PS& QR are transversals &  the sum  of  interior  corresponding  angles  on  the  same side  of the  transversals  are  supplementary.

 So $PQ\parallel SR$.

Now $\angle P+\angle Q\neq 180 ^o\&  \angle S+\angle R\neq 180^o$

In quadrilateral PQRS, $\angle P\& \angle Q$ are not supplementary as well as $\angle S\& \angle R$ are not supplementary.

So QR is not parallel to SP.

So one pair of opposite sides are parallel.

None of the opposite angles are equal. 

None of the sides are given as equal.

The $\Box$ PQRS can only be a Trapezium.