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 congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

A triangle cannot be drawn with the following three sides:

  1. $2m, 3m, 4m$
  2. $3m, 4m, 8m$
  3. $4m, 6m, 9m$
  4. $5m, 7m, 10m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A triangle with three sides a,b and c will be possible when:

\$a+b>c\$
\$ b+c>a\$
\$ a+c>b\$
\$ Here,a=2,b=3,c=4\$
\$ 2+3>4\$
\$ 3+4>2\$
\$ 2+4>3\$
\$ \therefore A)is\quad possible.\$
\$ Here,a=3,b=4,c=8\$
\$ 3+4=7\$
\$7<8\$
\$ \therefore a+b>c\quad is\quad not\quad satisfied.\$
\$ C)&amp; D)\quad will\quad also\quad be\quad possible.\$
\$ \therefore B)Correct\quad answer.\$

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

The angle measuring $\displaystyle \frac{\pi ^{c}}{4}$ when expressed in centesimal system is ___ 

  1. $\displaystyle 50^{g}$
  2. $\displaystyle 60^{g}$
  3. $\displaystyle 75^{g}$
  4. $\displaystyle 100^{g}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
Given angle $ = \dfrac {{\pi}^c}{4} = \dfrac {{180}^{0}}{4} = {45}^{0} $

We know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $


$ \Rightarrow {45}^{0} = {\dfrac {10}{9}} \times 45^{g}  ={50}^{g} $

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The equation $x-y = 4$ and $x^2 + 4xy + y^2 = 0$ represent the sides of

  1. an equilateral triangle

  2. a right angled triangle

  3. an isosceles triangle

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

$(y+m _{1}x)(y+m _{2}x)=0$


$y^{2}+(m _{1}+m _{2})xy+m _{2}m _{1}x^2=0$

Comparing coefficients we get

$m _{1}+m _{2}=4$

$m _{1}m _{2}=1$

This implies

$m _{1}^{2}+1=4m _{1}$

$m _{1}^{2}-4m _{1}+1=0$

Therefore $m _{1}=2-\sqrt{3}$ and $m _{1}=2+\sqrt{3}$

$\tan A=2-\sqrt{3}$ and $\tan A=2+\sqrt{3}$

Hence $A=15^0$ and $A=75^0$

Corresponding values of $m _{2}=75^{\circ}$ and $15^{\circ}$

Therefore the angle between the lines is $75^0-15^0$
$=60^0$

The equation of the angle bisectors of the lines is $x=\pm y$

The line $x=-y$ is perpendicular to $x-y=4$

Hence the above is an isosceles triangle with the vertical angle being $60^0$

Hence the triangle is an equilateral triangle.

Multiple choice maths introduction to euclid's geometry conditional statements and converse euclid's postulates axioms, postulates and theorems euclid's fifth postulate

$\angle A=\angle B$ and $\angle B=\angle C$, According to which axiom of Euclid the relation between $\angle A$ and $\angle C$ is established?

  1. I

  2. II

  3. III

  4. IV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that $\quad \angle A=\angle B\quad &amp; \quad \angle B=\angle C.\quad $

Then, according to Euclid's first axiom, which states that 
"things which are   equal to the same thing are also equal to each other",
 $\quad \angle A=\angle C\quad $
Ans- Option A.

Multiple choice maths areas related to circles area of a sector of a circle sector and arc of a circle area of sectors and segments

If the angle subtended by the arc of a sector at the center is $90$ degrees, then the area of the sector in square units is

  1. $2\pi r^2$
  2. $4\pi r^2$
  3. $\dfrac{\pi r^2}{4}$
  4. $\dfrac{\pi r^2}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the Central angle is $90^{\circ}$, it means it is a Quad-circle.

So the Area of this sector is $\dfrac{1}{4}$ th of the Circle's Area $= \dfrac{1}{4}* \pi r^2$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

ABC is a right angel triangle right angled at vertex A. A circle is drawn to touch sides AB and AC at points P and Q respectively such that other end points of diameters passing through P and Q lie on side BC. If AB = 6. then the area of circular sector which lies outside the triangle is :

  1. $\pi -2$
  2. $\pi -3$

  3. 4
  4. $\pi +2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a geometry problem involving a circle inscribed in a right triangle. The sector area outside the triangle is calculated based on the geometry of the circle touching the sides. The result is pi - 2.

Multiple choice business maths basic concepts in geometry conditional statements and converse euclid's postulates introduction to euclid's geometry

The converse of "If in a triangle $ABC, AB=AC$, then $\angle B=\angle C$", is

  1. lf in a triangle $ABC, \angle B=\angle C$, then $AB=AC$.
  2. lf in a triangle$ABC, AB\neq AC$, then $\angle B\neq\angle C$.
  3. lf in a triangle $ABC, \angle B\neq\angle C$, then $AB\neq AC$.
  4. lf in a triangle $ABC, \angle B\neq\angle C$, then $AB=AC$ .
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Take $p:AB=AC$

and $q:\angle B=\angle C$
So the given statement is symbolically represented as $p\rightarrow q$
Now by definition, Converse of a conditional statement $p\rightarrow q$ is $q\rightarrow p$
So $q\rightarrow p$ is given by 
"If in a triangle $ABC, \angle B=\angle C,$ then $AB=AC.$"

Multiple choice business maths basic concepts in geometry conditional statements and converse euclid's postulates introduction to euclid's geometry

The converse of "if in a triangle $ABC, AB>AC$, then $\angle C=\angle B$", is

  1. lf in a triangle $ABC, \angle C=\angle B$, then $AB>AC$.
  2. lf in a triangle$ABC, AB\not\simeq AC$, then $\angle C\not\simeq \angle B$.
  3. lf in a triangle $ABC, \angle C\not\simeq \angle B$, then $ AB\not\simeq AC$.
  4. lf in a triangle $ABC, \angle C\not\simeq \angle B$, then $AB>AC$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Take $p:AB>AC$

and $q: \angle C=\angle B$
So the given statement is symbolically represented as $p\rightarrow q$
Now by definition, Converse of a conditional statement $p\rightarrow q$ is $q\rightarrow p$
Thus $q\rightarrow p$ is given by
"If in a $\triangle ABC, \angle C=\angle B$ then $AB>AC$."

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

Let  $ABC$ be triangle. Let $A$ be the point $(1,2),y=x$be the perpendicular bisector of $AB$ and $x-2y+1=0$ be the angle bisector of $\angle C$. If equation of $BC$ is given by $ax+by-5=0$, then the value of $a+b$ is 

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

$\begin{array}{l} { m _{ AB } }=-1 \ y-2=-1\left( { x-1 } \right)  \ y-2=-x+1 \ x+y=3 \ \underline { x-y=0 }  \ 2x=3 \ \therefore x=\frac { 3 }{ 2 } \, \, \, \, \, ,y=\frac { 3 }{ 2 }  \ \frac { { 1+h } }{ 2 } =\frac { 3 }{ 2 } \, \, \,  \ h=2 \ \frac { { 2+k } }{ 2 } =\frac { 3 }{ 2 }  \ k=1\, \, \, \, \, \, \, \, \, \, \, B\left( { 2,1 } \right)  \ As\, \, image\, \, through\, \, x-2y+1=0 \ { m _{ AD } }=-2 \ y-2=-2\left( { x-1 } \right)  \ y-2=-2x+2 \ 2x+y=4\times 2\, \, \, \, \, \, \, x-2y+1=0 \ 4x+2y=8 \ \underline { x-2y=-1 }  \ 5x=7 \ x=\frac { 7 }{ 5 } \, \, \, \, \, \, y=\frac { { x+1 } }{ 2 } =\frac { { 12 } }{ { 5\times 2 } } =\frac { 6 }{ 5 }  \ \frac { { m+1 } }{ 2 } =\frac { 7 }{ 5 } \, \, \, \, \, \, \, \, \frac { { n+1 } }{ 2 } =\frac { 6 }{ 5 }  \ m+1=\frac { { 14 } }{ 5 } \, \, \, \, \, \, \, n+2=\frac { { 12 } }{ 5 }  \ m=\frac { 9 }{ 5 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n=\frac { { 12 } }{ 5 } -2=\frac { 2 }{ 5 }  \ E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ BC=BE\, \, \, \, \, \, B\left( { 2,1 } \right) \, \, \, E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ y-1=\left( { \frac { { 1-\frac { 2 }{ 5 }  } }{ { 2-\frac { 9 }{ 5 }  } }  } \right) \left( { x-2 } \right)  \ 3x-y=5 \ a=3\, \, \, \, b=-1 \ a+b=2 \end{array}$