Mathematics · Quantitative Aptitude

Geometry of Triangles and Angles

846 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 business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

The equation $ \displaystyle 3x^{2}-8xy-3y^{2}=0 $ and $ \displaystyle x-2y=3 $ represents the sides of a triangle which is

  1. equilateral

  2. isosceles

  3. right angled triangle

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

Given equation of pair of lines $\displaystyle 3{x}^{2}-8xy-3{y}^{2}=0$,
Let $\displaystyle \frac{y}{x}=m$, we get $\displaystyle 3{m}^{2}+8m-3=0$ Let $\displaystyle {m} _{1}$ and ${m} _{2} $ are slopes of lines,we get $\displaystyle {m} _{1}\times{m} _{2}=-1$

$\displaystyle \therefore $ lines are perpendicular to each other.
Then triangle is Right angled triangle . 

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

Find the radian measure corresponding to the degree $-47^{o}30'$

  1. $\dfrac {-19\ \pi}{72}rad$
  2. $\dfrac {19\ \pi}{72}rad$
  3. $\dfrac {13\ \pi}{72}rad$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$-47^{o} 30'$
$\Rightarrow - (47+ \dfrac{30}{60}) (\because 1^{o} =60')$
$\Rightarrow  -\left( 47+ \dfrac{1}{2} \right)$
$-\left( \dfrac{95}{2} \right)$
Radian measure $\Rightarrow \dfrac{\pi}{180} \times \dfrac{-95}{2}$
$\Rightarrow \pi x - \dfrac{19}{72} \Rightarrow - \dfrac{19 \pi}{72}$ radian
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

Can we construct a rhombus $ABCD$ with $AB=4\ cm$? Its diagonal intersect at the point $O$ and $\angle OAB = 60^0$.

  1. Yes

  2. No

  3. Sometimes yes

  4. Can't say
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given : $AB=4$cm

Diagonal intersect at $O$ and $\angle OAB=60^{o}$ ....... $(1)$
Draw side $AB$ of $4$cm.
In a rhombus, all sides are equal and diagonals bisect the opposite angles
From $(1)$ we get, $\angle A=120^{o}$
$\implies \angle B=60^{o}$ ........... (Adjacent angles are supplementary)
Draw a side $AD$ from A of $4$cm such that $\angle BAD=120^{o}$
Now, from $D$, draw side $DC = 4$cm such that $\angle ADC=60^{o}$
And then join $B-C$ such that $BC=4$cm and $\angle DCB=120^{o}$.
At last we get a rhombus $ABCD$ with length of each side is $4$ cm and diagonals $AC$ and $BD$.
Hence, we can construct a rhombus with $AB=4\ cm$.

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

Construct a rectangle $ABCD$, where $AB=10$ cm and $BC=8$ cm.Steps for its construction is given in a jumbled form. Identify its correct sequence.
1) Join these cuts with a line $CD$ and rectangle $ABCD$ is formed
2) Draw a straight line $AB$ of length $10$ cm
3) Draw perpendicular lines at $A$ and $B$ using protractor.
4) Using compass cut arc at the perpendicular from $A$ and $B$ of lengths $8$ cm

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

Correct sequence for constructing rectangle $ABCD$ is:

Draw a straight line $AB$ of $10 $ cm.
Draw perpendicular lines at $A$ and $B$ using proctor.
Using compass cut arc at the perpendicular from $A$ and $B$ of lengths $8$ cm.
Join these cuts with a line $CD$ and rectangle $ABCD$ is formed.
Correct sequence is $2,3,4,1$.

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)
$\frac{1}{2} (180 + \angle A) + \angle BOC = 180$
$\angle BOC = 180 - 90 -\frac{1}{2} (64)$
$\angle BOC = 58^{\circ}$