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 mapping your way mapping mapping space around us bearing and drawings

$ \Delta ABC ~ \Delta PQR $ for the correspondence $ABC \leftrightarrow PQR $ . If the perimeter of $ \Delta ABC $ is $12$ and the perimeter of $ \Delta PQR $ is $20 $ , then $AB : PQ = $ ______

  1. $\dfrac {6}{5}$
  2. $\dfrac {2}{5}$
  3. $\dfrac {3}{5}$
  4. $\dfrac {1}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$ABC\leftrightarrow PQR$
$\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}=k(say)$

$AB=k(PQ);AC=k(PR);BC=k(QR)$
$[AB+BC+AC]=k[PQ+RQ+PR]$
Perimeter $(\Delta ABC)$ = perimeter $(\Delta PQR)\times k$
$12=20\times k$

$k=\dfrac{12}{20}=\dfrac{3}{5}$

$\therefore \dfrac{AB}{PQ}=k=\dfrac{3}{5}$
Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

$PQ$ is the diameter of a semicircle with radius $4\ cm$ and $\angle PRQ$ is the angle on the semicircle. If $QR = 2\sqrt {7} cm$, then length of $PR$ is :

  1. $8\ cm$
  2. $6\ cm$
  3. $5\ cm$
  4. $2\sqrt {11} cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\angle PRQ = 90^{\circ}\Rightarrow PR = \sqrt {(PQ)^{2} - (RQ)^{2}}$

                                       $= \sqrt {64 - 28} $

                                       $= 6$.
$\therefore$ The solution is $6$

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Let $P(-1, 0), Q(0, 0)$ and $R(3, 3\sqrt{3})$ be three points. The equation of the bisector of the angle PQR is?

  1. $x+\sqrt{3}y=0$
  2. $\sqrt{3}x+y=0$
  3. $x+\dfrac{\sqrt{3}}{2}y=0$
  4. $\dfrac{\sqrt{3}}{2}x+y=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

P=(-1, 0), Q=(0, 0), R=(3, 3*sqrt(3)). The angle at Q is 120 degrees. The bisector of angle PQR will have a slope that splits the angle between the lines QP and QR.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

In triangle $PQR$, an exterior angle at $A$ measures $160^o$, and $\angle$ $Q = 70 ^o$. Which is the longest side of the triangle?

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

Exterior angle $A = 180 -$ $\angle$ $P = 180-160$
$\angle$ $P = 20$
$\angle$ $Q = 70$
Interior angle $= 20 + 70 +$ $\angle$ $R = 180$
$\angle$ $R = 90$
In a triangle inequality theorem,the largest side is across from the longest angle.
So, $90^o$ is the longest angle in the triangle, $\overline{PQ}$, across from it, is the largest side.
Therefore, $\overline{PQ}$ is the largest side.

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The crescent shaded in the diagram, is like that found on many flags. $PSR$ is an arc of a circle, centre $O$ and radius $24.0$ cm. Angle POR $=$ $48.2^{\circ}$.
$PQR$ is a semicircle on $PR$ as diameter, where $PR$ $=$ $19.6$ cm
$[\pi = 3.14] [\cos 24.1 = 0.91]$

The area of the crescent is

  1. $116.4$ cm$^2$
  2. $123.4$ cm$^2$
  3. $112.2$ cm$^2$
  4. $23.4$ cm$^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Length PQR $= \displaystyle  \frac{1}{2} \times 2 \pi r = \frac{1}{2} \times 2 \times 3.14 \times 9.8 = 30.772 cm$
Length PSR $ = \displaystyle \frac{\theta}{360} \times 2 \pi = \frac{48.2}{360}\times 2 \pi r= \frac{48.2 }{360} \times 2 \times 3.14 \times24$
$= 20.1797 cm$
Perimeter of crescent $= 30.772 + 20. 1797 = 51$
Area of sector POR $= \displaystyle \frac{\theta}{360} \times \pi \times r^2 = \frac{48.2}{360} \times 3.14 \times 14^2$
$= 242.16 cm^2 = 242 cm^2$
In triangle POR height, $ h = 24 cos24.1 = 21.91 cm$
Area $= \displaystyle \frac{1}{2} bh = \frac{1}{2} \times 19.6 \times 21.91= 214.70 cm^2$
$\therefore $ Area of shape PSR remaining $= 242.16 - 214.70 = 27.46 cm^2$
Area of semicircle $= \displaystyle \frac{1}{2} \times \pi \times 9.8^2 = 150.859$
$\therefore$ Area of crescent $=150.859 - 27.46 = 123.4 cm^2$

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

If $\Delta ABC \sim \Delta QRP, \displaystyle \frac{ar (ABC)}{ar (PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC=15 cm$; then PR is equal to

  1. $10\ cm$
  2. $12\ cm$
  3. $\displaystyle \frac{20}{3}\ cm$
  4. $8\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given :

Area of ∆ ABCArea of ∆QRP = 94


AB = 18 cm , BC = 15 cm So PR = ?

We know when two triangles are similar then " The areas of two similar triangles are proportional to the squares of their corresponding sides.

Area of ∆ ABCArea of ∆ QRP = $AB^2QR^2$ = $Bc^2PR^2$ = $AC^2QP^2$
So, we take 
Area of ∆ ABC Area of ∆ QRP = $BC^2PR^2$

Now substitute all given values and get

94 = 152PR2

Taking square root on both hand side, we get

32 = 15PR

PR = 10 cm

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

If $\triangle ABC\sim \triangle QRP,\dfrac{Ar(ABC)}{Ar(QRP)}=\dfrac{9}{4}$,$AB=18\ cm$ and $BC=15\ cm$; then $PR$ is equal to:

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

Given $ \triangle  ABC \sim  \triangle  QRP $
Therefore, $ \frac { Area\triangle ABC\quad  }{ Area\triangle QRP\quad }=\frac { { BC }^{ 2 } }{ { PR }^{ 2 } }  $
or $ \frac { 9 }{ 4 }  = \frac { { 15 }^{ 2 } }{ { PR }^{ 2 } }  $
or $ { PR }^{ 2 }\quad =\quad \frac { { 15 }^{ 2 }\quad \times \quad 4 }{ 9 }  $ cm.
Therefore, $ { PR }=\frac { { 15 }\times \quad 2 }{ 3 }  = $ 10 cm.

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