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 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

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

Consider $\triangle ABC$ and $\triangle { A } _{ 1 }{ B } _{ 1 }{ C } _{ 1 }$ in such a way that $\overline { AB } =\overline { { A } _{ 1 }{ B } _{ 1 } } $ and M, N, ${ M } _{ 1 }$, ${ N } _{ 1 }$ be the mid points of AB, BC, ${ A } _{ 1 }{ B } _{ 1 }$ and ${ B } _{ 1 }{ C } _{ 1 }$ respectively, then

  1. $\overline { M{ M } _{ 1 } } =\overline { N{ N } _{ 1 } } $
  2. $\overline { C{ C } _{ 1 } } =\overline { M{ M } _{ 1 } } $
  3. $\overline { C{ C } _{ 1 } } =\overline { N{ N } _{ 1 } } $
  4. $\overline { M{ M } _{ 1 } } =\overline { B{ B } _{ 1 } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given AB = A1B1 and M, N, M1, N1 are midpoints, the triangles are congruent. The vector relationship between the midpoints of sides in congruent triangles preserves the distance between corresponding points. Specifically, M and M1 are midpoints of AB and A1B1, and N and N1 are midpoints of BC and B1C1; the displacement MM1 is equal to BB1.

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

The sides  $A B , B C , C A$  of a triangle  $A B C$  have  $3,4$  and  $5$  interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

  1. $205$
  2. $210$
  3. $315$
  4. $216$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total number of points $12$. If no three points are co-linear then total number of the   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 

$\begin{array}{l} { =^{ 12 } }{ C _{ 3 } }-\left( { ^{ 3 }{ C _{ 3 } }{ +^{ 4 } }{ C _{ 3 } }{ +^{ 5 } }{ C _{ 3 } } } \right)  \\ =220-\left( { 1+4+10 } \right)  \\ =205 \end{array}$



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

Mark the correct alternative of the following.
In which of the following cases can a right triangle ABC be constructed?

  1. $AB=5$cm, $BC=7$cm, $AC=10$cm
  2. $AB=7$cm, $BC=8$cm, $AC=12$cm
  3. $AB=8$cm, $BC=17$cm, $AC=15$cm

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

A right triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For option C, 8^2 + 15^2 = 64 + 225 = 289, which is 17^2. Thus, 8, 15, 17 forms a right triangle.

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

Mark the correct alternative of the following.
In which of the following cases, a right triangle cannot be constructed?

  1. $12$cm, $5$cm, $13$cm
  2. $8$cm, $6$cm, $10$cm
  3. $5$cm, $9$cm, $11$cm

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

Any triangle is right angled if it satisfies $a^2 + b^2 = c^2$, where c is the largest side from option (A)  we take 

$a = 12 cm$
$b = 5 cm$
$c = 13 cm$
we know $(12)^2 + (5)^2 = 169 = (13)^2$
So we can construct a right angled triangle.

Similarly in option (B) we can take 
$a = 8 cm$
$b = 6 cm$
$c = 10 cm$
with these sides also we can construct a right angled triangle.

Consider option (C) then 
$a = 5 cm$
$b = 9 cm$
$c = 11 cm$
$a^2 + b^2 = 25 + 81$   But  $c^2 = 121$
             $= 106$

So we cannot construct right angled triangle from $5 cm, 9 cm, 11 cm$.

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

Construct a right angled $\triangle ABC$ with $\angle B = 90^\circ, BC = 5\ cm$ and $AC = 10\ cm$ and find the the length of side $AB$

  1. $6.2\ cm$
  2. $5\ cm$
  3. $8.7\ cm$
  4. $7.2\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Step 1. Draw a line segment $BC=5\ \ cm$

Step 2. At $B$ draw an angle of $90^{\circ}$ and extend the ray.
Step 3. Now taking $C$ and centre draw an arc of radius $10$ cm intersecting the previous ray at $A$.
Step 4. Join $C$ to $A$.
Now using a ruler measure $AB$
$AB=8.7$ cm

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

Construct a triangle $ABC$ in which $AB = 5 cm$ and $BC = 4.6 cm$ and $AC = 3.7 cm$
Steps for the construction is given in jumbled form.Choose the appropriate sequence for the above
1) With radius as $5\ cm$ from $C$, cut an arc.
2)They arcs will intersect at point $A$. Join $AB$ and $AC$. $ABC$ is the required triangle.
3)Draw a line segment $BC = 4\ cm.$
4)With radius as $3$ cm from $B$, cut the arc. 

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

Correct sequence is:

1. Draw a line segment $AB=4$ cm.
2. With radius $5$ cm from $C$ , cut an arc.
3. With radius $3$ cm from $B$ , cut an arc.
4. The arc will intersect at point $A$, Join $AB$ and $AC$ .$ABC$ is required triangle.
So correct sequence is $3,1,4,2$
So option $C$ is correct.

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

Construct an isosceles $\triangle  XYZ,$ where $YZ=5$ units and $\angle XYZ=35^{o}$. Also, find the measure of $\angle YXZ$.

  1. $35^{o}$
  2. $70^{o}$
  3. $110^{o}$
  4. $140^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$YZ=5$ CM $,\angle XYZ=35^{\circ}$

As the triangle is isosceles therefore $\angle XZY=50^{\circ}$
Steps of construction:
1. Draw a line segment $XY=5$ cm.
2. At $Y$ draw an angle of $35^{\circ}$ and extend the arm.
3. At $Z$ draw an angle of $35^{\circ}$ and extend the ray such that it intersect the previous ray at $X$
4. Join $Y$ to $X$ and $Z$ to $X$
Now measure $\angle YXZ$
$\angle YXZ=110^{\circ}$