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 properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

If $\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be ________ similarity of postulate.

  1. SAS similarity postulate

  2. AAA similarity postulate

  3. SSS similarity postulate

  4. AAS similarity postulate
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be SAS similarity of postulate.
Because, SAS Similarity Postulate states, "If an angle of one triangle is congruent to the corresponding angle of another triangle and the sides that include this angle are proportional, then the two triangles are similar."

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

For $\triangle ABC$ and $\triangle PQR$, if $m\angle A=m\angle R $ and $m\angle C=m\angle Q$, then $ABC \longleftrightarrow $_________ is a similarity.

  1. $RQP$
  2. $PQR$
  3. $RPQ$
  4. $QPR$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For $\triangle ABC $ and $\triangle PQR$,
$m\angle A = m\angle R$
$m\angle C= m\angle Q$
$\therefore $ by AA criteria for similarity 

$ABC \longleftrightarrow RPQ $ is a similarity.

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 maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Diagonals $\overline{AC}$ and $\overline{BD}$ of quadrilateral $ABCD$ are perpendicular. $AD=DC=8, AC=BC=6, m\angle ADC = 60^o$. The area of $ABCD$ is

  1. $4\sqrt{5}+8\sqrt{3}$
  2. $16\sqrt{3}$
  3. $32\sqrt{3}$
  4. $8\sqrt{5}+16\sqrt{3}$
  5. $48$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The altitude to the base of an isosceles triangle also bisects the vertex angle, so $m\angle ADE=30$.

With the hypotenuse of the triangle having a length of $8$, $AE=4$ and $DE=4\sqrt { 3 }$.
$\triangle AEC$ is a right angle with leg $4$ and hypotenuse $6$.
Use the pythagorean theorem to determine that 
$BE=\sqrt { 6^{ 2 }-4^{ 2 } } =\sqrt { 36-16 } =\sqrt { 20 } =2\sqrt { 5 }$
The area of the quadrilateral with perpendicular diagonals is equal to half the product of the diagonals, so the area of $ABCD$ is:
$A=\dfrac { 1 }{ 2 } \times 8\times (2\sqrt { 5 } +4\sqrt { 3 } )=4(2\sqrt { 5 } +4\sqrt { 3 } )=8\sqrt { 5 } +16\sqrt { 3 }$  

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 a $\Delta ABC$, side AB has the equation $2x+3y=29$ and the side AC has the equation, $x+2y=6$. If the mid-point of BC is (5, 6), then the equation of BC is

  1. $x-y=-1$
  2. $5x-2y=13$
  3. $21x+31y=291$
  4. $3x-4y=-9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of AB and AC gives vertex A. Solving 2x+3y=29 and x+2y=6: x = 6-2y, so 2(6-2y)+3y=29, 12-4y+3y=29, -y=17, y=-17, x=40. The line BC passes through (5,6) and its slope can be found by relating it to the median or vertex properties, but checking the options, x-y=-1 passes through (5,6) since 5-6=-1.

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.