Mathematics · Quantitative Aptitude

Geometry of Triangles and Angles

758 Questions

Triangle and angle geometry covers angle sums, properties of equilateral shapes, and right triangles. These principles form the basis of advanced quantitative aptitude sections. Practicing spatial problems helps secure points in exams.

Triangle angle sumsPythagorean theoremProperties of equilateral trianglesComplementary and supplementary anglesRight triangle formulas

Geometry of Triangles and Angles Questions

Multiple choice physics motion and measurement measuring length measurement of small and large distances measurement of distance

If a star is $5.2\times 10^{16}\ m$ away. What is the parallax angle in degrees?

  1. $1.67 \times 10^{-4}$ degrees
  2. $1.67 \times 10^{-5}$ degrees
  3. $0.67 \times 10^{-4}$ degrees
  4. $2.3 \times 10^{-4}$ degrees
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given :    $1$ AU $ = 1.5\times 10^11$ m                $d = 5.2\times 10^{15}$ m

Parallax angle:     $\alpha = \dfrac{1 AU}{d} =\dfrac{1.5\times 10^{11}}{5.2\times 10^{16}} = 0.288\times 10^{-5}$  radians
$\implies$   $\alpha = \dfrac{180}{\pi} \times 0.288\times 10^{-5} = 1.67\times 10^{-4}$  degrees

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

The diagonal of rectangle $ABCD$ intersect each other at $O$. If $\angle AOB = 30^0$, then we can construct a rectangle if _________ is given.

  1. diagonal

  2. one side

  3. both sides
  4. $\angle COD$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$ABCD$ is a rectangle

$\implies AB = CD$ and $AD = BC$ ... (1)
By knowing these, we can just draw the two pair of parallel lines but the length is not fixed.
So, to  draw a rectangle we need the length of the sides.
From (1), we need only the length of two adjacent sides.
Hence, we can construct a rectangle if both sides are given.

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

$A B C$  is a triangle. The bisectors of the internal angle  $\angle B$  and external angle $\angle C$  intersect at  $D.$  if  $\angle B D C = 60 ^ { \circ }$  then  $\angle A$  is

  1. $120 ^ { \circ }$
  2. $180 ^ { \circ }$
  3. $60 ^ { \circ }$
  4. $150 ^ { \circ }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider $\triangle ABC$

Let $BC$ be extended to $E$
Since Angular bisectors Meet at $D$
$\angle ABD=\angle DBC\cdots(1)$
$\angle ACD=\angle DCE\cdots(2)$
Consider $ \triangle DBC$
By External sum property 
$\angle DCE=\angle BDC+\angle DBC$
$\implies 2\angle DCE=2(60^{\circ})+2\angle DBC$
$\implies \angle ACE=120^{\circ}+\angle ABC$
By external sum property of $\triangle ABC$
$\angle ACE=\angle BAC+\angle ABC$
$\implies \angle A=60^{\circ}$

Multiple choice maths triangles relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

In the sides $BC,CA,AB$ of a triangle $ABC$, three points $D,E,F$ are taken such that each of $BD,CE,AE$ is equal to one-third of the corresponding side, then
$\triangle DEF=\dfrac {1}{2}\triangle ABC$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If D, E, F divide the sides in 1:2 ratio, the area of triangle DEF is (1 - 3*(1/3)*(2/3)) = 1/3 of the area of triangle ABC. The statement that it is 1/2 is false.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

The angles of a triangle are in the ratio 2: 1: 3. Is the triangle right-angled triangle,

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angles of the triangle are in the ratio, 2:1: 3
Let the angles be $2x, x and 3x$
Thus, sum of the angles = 180
$2x + x+ 3x = 180$
$6x = 180$
$x = 30$ 
Hence, the angles will be 30, 60 and 90
Since, one of the angles is 90, the triangle is a right angled triangle.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In a $\triangle ABC$, $\angle A - \angle B = 30^{\circ}$ and $ \angle B -\angle C = 42^{\circ}$; find $\angle A$.

  1. $84^o$
  2. $94^o$
  3. $32^o$

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ABC$,
$\angle A - \angle B = 30$    ....(I)
$\angle B - \angle C = 42$     .....(II)
Also, sum of angles of the triangle $= 180$
$\angle A + \angle B + \angle C = 180$     ....(III)
On subtracting (I) and (II), we get

$\angle A-2\angle B+\angle C=-12^o$     ....(IV)
On subtracting (III) and (IV), we have
$3\angle B=192$
$\angle B=64^o$
From (I), we get
$\angle A=94^o$

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

If the angles of a triangle are in the ratio 2:3:4, find the three angles.

  1. $80^o, 120^o, 160^o$
  2. $20^o, 30^o, 40^o$
  3. $40^o, 60^o, 80^o$

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

The angles of a triangle are in the ratio 2:3:4 
Let $x:y:z=2:3:4$
Then $x= 2t; y= 3t; z= 4t$

Sum of all angles of a triangles is $ 180^0$.
$ 2t + 3t + 4t = 180^0 $
$ 9t = 180^0 $
$  t  = 20^0 $
$ x= 2\times 20^0= 40^0; y= 3\times 20^0= 60^0; z = 4\times 20^0= 80^0 $

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In a $\triangle ABC$, the sides AB and AC have been produced to D and E. Bisectors of $\angle CBD$ and $\angle BCE$ meet at O. If $\angle A={ 64 }^{ 0 }$, then $\angle BOC$ is 

  1. ${ 52 }^{ 0 }$
  2. ${ 58 }^{ 0 }$
  3. ${ 26 }^{ 0 }$
  4. ${ 112 }^{ 0 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: OB and OC bisect $ext. \angle B$ and $ext. \angle C$, $\angle A = 64^{\circ}$

Now, In $\triangle OBC$,
Sum of angles = 180
$\angle OBC + \angle OCB + \angle BOC = 180$
$\frac{1}{2} (ext. \angle B + ext. \angle C) + \angle BOC = 180$ (OB and OC bisect exterior angles)
$\frac{1}{2} (180 - \angle ABC + 180 - \angle ACB) + \angle BOC = 180$
$\frac{1}{2} (360 - (\angle ABC + \angle ACB)) + \angle BOC = 180$
$\frac{1}{2} (360 - (180 - \angle A)) + \angle BOC = 180$ (Angle sum property)
$\frac{1}{2} (180 + \angle A) + \angle BOC = 180$
$\angle BOC = 180 - 90 -\frac{1}{2} (64)$
$\angle BOC = 58^{\circ}$

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In $\displaystyle \triangle ABC,\angle C=30^{\circ},\angle B=90^{\circ},BC=10 cm,BD\perp AC$ then the length of AD is

  1. $\displaystyle \frac{5}{\sqrt{3}}$ cm
  2. $\displaystyle \frac{6}{\sqrt{3}}$ cm
  3. $\displaystyle \frac{7}{\sqrt{3}}$ cm
  4. $\displaystyle \frac{8}{\sqrt{3}}$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In triangle ABC $\angle C=30^{0}and \angle B =90^{0}$ and BC=10 cm and $BD\perp AC$ 

Then $Sin 30^{0}=\frac{AB}{BC}\Rightarrow \frac{1}{2}=\frac{AB}{10}\Rightarrow AB=5 cm$
And $\angle ABD =60^{0}$
Then $tan 60^{0}=\frac{AB}{AD}\Rightarrow \sqrt{3}=\frac{5}{AD}\Rightarrow AD=\frac{5}{\sqrt{3}}cm$

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

One of the exterior angle of a triangle is $ 105^0$ and the interior opposite angles are in the ratio 2 : 5 . Find the angles of the triangle.

  1. $ 30^o ; 45^o ; 105^o$
  2. $ 45^o ; 45^o ; 90^o$
  3. $ 30^o ; 75^o ; 75^o$
  4. $ 60^o ; 30^o ; 90^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have the property that, in a triangle exterior angle is equal to the sum of interior opposite angles.


Given, the interior opposite angles to the exterior angle $105^\circ$ are in the ration $2:5$

$\therefore 2x+5x=105^o$

$7x=105^o$ $\implies x=15^o$

Therefore the interior opposite angles to the angle $105^o$ are $2x=2(15)=30^o$ and $5x=5(15)=75^o$

Let the third angle of the triangle be $C$
We have the sum of interior angles of a triangle is $180^o$

$\therefore 30^o+75^o+C=180^o$
$C=180^o-75^o-30^o=180^o-105^o$
$C=75^o$

Hence, the angles are $30^o,75^o,75^o$.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

$\Delta ABC$ is a right angled at A, the value of tan B $\times$ tan C is:

  1. 0

  2. 1
  3. $- 1$

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

In $\triangle  ABC$

$\angle A+\angle B+\angle C={ 180 }^{ \circ  }\ \angle B+\angle C={ 180 }^{ \circ  }-{ 90 }^{ \circ  }={ 90 }^{ \circ  }\ \angle C={ 90 }^{ \circ  }-\angle B$
$\tan { B } \times \tan { C } \ =\tan { B } \times \tan { \left( { 90 }^{ \circ  }-\angle B \right)  } \ =\tan { B } \times \cot { B } \ =\tan { B } \times \dfrac { 1 }{ \tan { B }  } \ =1$