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 maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

Suppose the triangle ABC has an obtuse angle at C and let D be the midpoint of side AC Suppose E is on BC such that the segment DE is parallel to AB. Consider the following three statements
i) E is the midpoint of BC
ii) The length of DE is half the length of AB
iii) DE bisects the altitude from C to AB

  1. only (i) is true

  2. only (i) and (ii) are true

  3. only (i) and (iii) are true

  4. all three are true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The triangle ABC has height h and base l,
E is the midpoint of BC, line parallel to the base will be interect the triangle at the midpoint of the opposite side.
Triangle ADE is similar to triangle ABC, as all 3 angles are equal.
Therefore, the altitude of ADE is half of ABC, and DE will be half of BC.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

Let $ABC$ be a triangle and let $P$ be an interior point such that $\angle BPC = 90$, $\angle BAP = \angle BCP$. Let $M, N$ be the mid-points of $AC, BC$ respectively. Suppose $BP = 2PM$. Then $A, P, N$ are collinear ?

  1. True

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

By geometric construction and properties of the point P, the conditions given satisfy the requirements for A, P, and N to be collinear.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

If $\displaystyle \Delta ABC$ is an isosceles triangle and midpoints $D, E,$ and $F$ of $AB, BC,$ and $CA$ respectively are joined, then $\displaystyle \Delta DEF$ is:

  1. Equilateral

  2. Isosceles

  3. Scalene

  4. Right-angled
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: In $\triangle ABC, D, E$ and $F$ are midpoints of sides $AB, BC$ and $CA$.

$BE=EC$
$\therefore DF=\dfrac { 1 }{ 2 } BC$
$\therefore \dfrac { DF }{ BC } =\dfrac { 1 }{ 2 }$ ....... $\left( 1 \right) $

Similarly, $ \dfrac { DE }{ AC } =\dfrac { 1 }{ 2 } $ and $ \dfrac { EF }{ AB } =\dfrac {1 }{ 2 } $

$\Rightarrow \dfrac { DF }{ BC } =\dfrac { DE }{ AC } =\dfrac { EF }{ AB } =\dfrac { 1 }{ 2 } $
$\triangle ABC$ is propotional to $\triangle DEF$ as the sides of the triangles are proportional. 
So corresponding angles are equal.
Hence, $\triangle ABC\sim \triangle EDF$ [by SSS similarly theorm]
$\therefore \triangle DEF$ is isosceles.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

In a $\triangle ABC$, if $D, E, F$ are the midpoints of the sides $BC, CA, AB$ respectively then $\overline {AD} + \overline {BE} + \overline {CF} =$

  1. $\overline {0}$
  2. $\overline {AE}$
  3. $\overline {BD}$
  4. $\overline {CE}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The vectors AD, BE, and CF are medians of the triangle. The sum of the vectors from the vertices to the midpoints of the opposite sides is zero.

Multiple choice physics measurements and units summary of si units fundamental quantities system of units

The solid angle subtended at any point inside the surface due to small area ds is given by 

  1. $r^2ds \cos \theta$
  2. $\dfrac{ds \cos \theta }{r^2}$
  3. $\dfrac{ds\cos \theta }{r}$
  4. $\dfrac{r^2}{ds \cos \theta}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The solid angle $\Omega=\dfrac{A}{r^2}$

where, $A=$ area
When solid angle subtended at any point inside the surface due to small area $ds$
$A=ds$ $cos\theta$
Solid angle $\Omega=\dfrac{dscos\theta}{r^2}$
The correct option is B.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

In $\Delta ABC$, there are 35 lines drawn parallel to the base BC such that each line divides the other side into, equal parts. 
If BC =1.8 m find the length of $P _7 Q _7$.

  1. 1.8 m

  2. 3.5 m

  3. 0.35 m

  4. 0.18 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the properties of parallel lines in a triangle, the length of the segments follows an arithmetic progression. With 35 lines, the segments divide the side into 36 equal parts; the 7th line corresponds to a ratio of 7/36 of the base, but the calculation 1.8 * (7/36) = 0.35 m is correct.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If $D$ is the midpoint of side $BC$ of a triangle $ABC$ and $AD$ is perpendicular to $AC$ then

  1. $3{a}^{2}={b}^{2}-3{c}^{2}$
  2. $3{b}^{2}={a}^{2}-{c}^{2}$
  3. ${b}^{2}={a}^{2}-{c}^{2}$
  4. ${a}^{2}+{b}^{2}=5{c}^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ACD, \cos{C}=\dfrac{b}{\left(\dfrac{a}{2}\right)}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{2b}{a}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2b}=2b$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=4{b}^{2}$
$\Rightarrow {a}^{2}-{c}^{2}=3{b}^{2}$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If the angles of a triangle are in the ratio $2:3:7,$ then the sides opposite to these angles are in the ratio

  1. $\sqrt{2}:2:\left(\sqrt{3}+1\right)$
  2. $2:\sqrt{2}:\left(\sqrt{3}+1\right)$
  3. $1:\sqrt{2}:\dfrac{\sqrt{2}}{\left(\sqrt{3}-1\right)}$
  4. $\dfrac{1}{\sqrt{2}}:1:\left(\dfrac{\sqrt{3}+1}{2}\right)$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Let $A=2\alpha, B=3\alpha, C=7\alpha$
$\therefore A+B+C={180}^{0}$
$\Rightarrow 2\alpha+3\alpha+7\alpha={180}^{0}$
$\Rightarrow 12\alpha={180}^{0}$
$\Rightarrow \alpha={15}^{0}$
$\therefore A=2\alpha=2\times{15}^{0}={30}^{0}$
$B=3\alpha=3\times{15}^{0}={45}^{0}$
$C=7\alpha=7\times{15}^{0}={105}^{0}$
$a:b:c=\sin{{30}^{0}}:\sin{{45}^{0}}:\sin{{105}^{0}}$
       $=\dfrac{1}{2}:\dfrac{1}{\sqrt{2}}:\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$a:b:c=\sqrt{2}:2:\sqrt{3}+1$
        $=\dfrac{1}{\sqrt{2}}:1:\dfrac{\sqrt{3}+1}{2}$
or $1:\sqrt{2}:\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}$
or $1:\sqrt{2}:\dfrac{2\sqrt{2}}{2\left(\sqrt{3}-1\right)}$
$\therefore a:b:c= 1:\sqrt{2}:\dfrac{\sqrt{2}}{\sqrt{3}-1}$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

In a triangle $ABC, \cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$ then the triangle is

  1. isosceles

  2. right-angled

  3. equilateral

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

$\cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$
Using transformation angle formula, we have
$\left(\cos{A}+\cos{B}\right)+\cos{C}=\dfrac{3}{2}$
$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+\cos{C}=\dfrac{3}{2}$
Using sub-multiple angle formula to $\cos{C}=1-2{\sin}^{2}{\dfrac{C}{2}}$ we get
$\Rightarrow  2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+1-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}$ 
Since $A+B+C=\pi\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$
$\Rightarrow  2\cos{\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}-1$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{C}{2}\right)}\right]=\dfrac{1}{2}$
Again
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{\pi}{2}-\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\cos\left(\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{C}{2}\right)}\left[2\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{A}{2}\right)}\right]=\dfrac{1}{4}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{8}$
$\therefore \sin{\left(\dfrac{A}{2}\right)}=\dfrac{1}{2}, \sin{\left(\dfrac{B}{2}\right)}=\dfrac{1}{2},\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{2}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}=\sin{\dfrac{\pi}{6}}, \sin{\left(\dfrac{B}{2}\right)}=\sin{\dfrac{\pi}{6}},\sin{\left(\dfrac{C}{2}\right)}=\sin{\dfrac{\pi}{6}}$
$\therefore {\left(\dfrac{A}{2}\right)}={\left(\dfrac{B}{2}\right)}={\left(\dfrac{C}{2}\right)}=\dfrac{\pi}{6}$
$\Rightarrow \angle{A}=\angle{B}=\angle{C}=2\times\dfrac{\pi}{6}=\dfrac{\pi}{3}$
Hence, the triangle is equilateral.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

In a triangle if the sum of two sides is $x$ and this product is $y ,\left(x\ge 2\sqrt{y}\right)$ such that $\left(x+z\right)\left(x-z\right)=y$ where $z$ is the third side of the triangle.
On the basis of the above information, answer the following questions:
The sides of the triangle are:

  1. $\dfrac{x\pm\sqrt{\left({x}^{2}-4y\right)}}{2},z$
  2. $\dfrac{y\pm\sqrt{\left({y}^{2}-4z\right)}}{2},z$
  3. $\dfrac{z\pm\sqrt{\left({z}^{2}-4x\right)}}{2},z$

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

$\because b+c=x,bc=y$
$\therefore b,c$ are the roots of ${t}^{2}-\left(b+c\right)t+bc=0$
or ${t}^{2}-xt+y=0$
$\therefore t=\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2}$
Hence, sides are  $\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2},z$ 

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

The steps for construction of $\triangle DEF$ with $DE = 4\ cm, EF=6.5\ cm$ and $DF = 8.6\ cm$ are given below in jumbled order:
1. Draw arcs of length $4\ cm$ from $4\ cm$ from $D$ and $6.5\ cm$ from $F$ and mark the intersection point as $E$.
2. Join $D-E$ and $F-E$.
3. Draw a line segment of length $DF = 8.6\ cm$.

The correct order of the steps is:

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

Correct sequence is

Step 1: Draw a line segment of length $DF=8.6 cm$.
Step2 :Draw arcs of length $4 cm$ from $4 cm$ from $D$ and $6.5 cm$ from $F$ and mark the intersection point as $E$
Step 3: Join $D-E$ and $F-E$.
So the sequence is $3-1-2$

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

In $\triangle ABC$, $AB=5\ cm, BC= 6\ cm ,AC=4\ cm$. Identify the type of triangle.

  1. Right angled triangle

  2. Isosceles triangle

  3. Equilateral triangle

  4. Scalene triangle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$5cm+6cm>4cm$
$\Rightarrow AB+BC>AC$

$6cm+4cm>5cm$
$\Rightarrow BC+AC>AB$

$4cm+5cm>6cm$
$\Rightarrow AC+AB>BC$

Sum of any two sides taken in pair is greater than the third side. So a triangle can be formed.

Now all the sides of triangle are unequal . So the triangle is Scalene triangle.

Further ${(LargestSide)}^{2}$ is not equal to $({Side1})^{2} + ({Side2})^{2}$. Hence, not right angled.

Option $D$ is correct
Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

The lengths of the sides of some triangles are given, which of them is not a right angled triangle?

  1. $5$ cm , $12$ cm, $13$ cm
  2. $7$ cm, $24$ cm, $25$ cm
  3. $5$ cm, $8$ cm, 1$0$ cm
  4. $3$ cm, $4$ cm, $5$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a right angled triangle.Sum of squares of two sides of a triangle=square of third i.e.,square of hypotenuse.

${ 5 }^{ 2 }+{ 12 }^{ 2 }=25+144=169$
$ { 13 }^{ 2 }=169$
$\implies\quad { 5 }^{ 2 }+{ 12 }^{ 2 }={ 13 }^{ 2 }$
 A)is a right angled triangle.
$ Now,{ 24 }^{ 2 }+{ 7 }^{ 2 }=576+49=625$
$\ implies\quad { 25 }^{ 2 }=625$
$\ implies\quad { 24 }^{ 2 }+{ 7 }^{ 2 }={ 25 }^{ 2 }$
B)is also a right angled triangle.
$ { 3 }^{ 2 }+{ 4 }^{ 2 }=9+16=25$
$ { 5 }^{ 2 }=25$
$\ implies\quad { 3 }^{ 2 }+{ 4 }^{ 2 }=5$
${ 5 }^{ 2 }=25$
 D)is also a right angled triangle.
$In\quad C)$
$ { 5 }^{ 2 }+{ 8 }^{ 2 }=25+64=89$
$ { 10 }^{ 2 }=100$
$ { 5 }^{ 2 }+{ 8 }^{ 2 }\neq { 10 }^{ 2 }$
$\therefore $C)is not a right angled triangle.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

The sides $AB, BC, CA$ of a trinagle $ABC$ have $3, 4$ and $5$ interior point on them. The number of triangles that can be constructed using these points  as vertices are

  1. $220$
  2. $205$
  3. $190$
  4. $85$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total number of points $12.$ If no three points are co-linear$,$

Th en$,$ total number of triangles would be $^{12}{C _3}.$ 
But $3$ points on $AB,4$ points on $BC$ and $5$ points on $CA$ are co-linear$.$
So$,$ total number of triangles formed should be$:$
${ = ^{12}}{C _3} - \left( {^3{C _3}{ + ^4}{C _3}{ + ^5}{C _3}} \right) = 220 - \left( {1 + 4 + 10} \right) = 205$  
Hence,
option $(B)$ is correct answer.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

If $b=3, c=4, \angle B=\dfrac{\pi}{3}$, then the number of triangles that can be constructed is

  1. $0$
  2. $1$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $b=3,c=4,\angle B=\dfrac{\pi}{3}$

We know that as per Sine rule

$\dfrac{\sin \dfrac{\pi}{3}}{3}=\dfrac{\sin C}{4}$

$\dfrac{\sqrt 3}{2\cdot3}=\dfrac{\sin C}{4}$

$\sin C = 2\sqrt{3}$  which is greater than 1

and sin lies between -1 and 1 that means angle C is not possible

Thus Zero triangles can be construct.

Hence A is the correct option