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 mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

In $\Delta ABC$, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is 

  1. $(b-c)sin(\frac{B-C}{2}) =a cos(\frac{A}{2}) $
  2. $(b-c)cos(\frac{A}{2})= a sin(\frac{B-C}{2}) $
  3. $(b+c)sin(\frac{B+C}{2})=a cos(\frac{A}{2}) $
  4. $(b-c)cos(\frac{A}{2}) = 2a sin(\frac{B+C}{2}) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Law of Sines and Mollweide's formulas, the correct relation is (b-c)cos(A/2) = a*sin((B-C)/2).

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

In triangle $XYZ$, $XZ=YZ$. If the measure of angle $Z$ has ${a}^{o}$, how many degrees are there in the measure of angle $X$?

  1. $x^o=\dfrac {180^o-2a}{2}$
  2. ${ x }^{ o }=\cfrac { { 180 }^{ o }-{ a }^{ o } }{ 2 } $
  3. $x^o=\dfrac {180^o-3a}{3}$

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

Given, $XZ=YZ$ which implies angles $ZXY$ and $ZYX$ are equal and let it be $\theta$.
We have $ZXY+ZYX+XZY = 180$ , which implies $\theta+\theta+a=180$
Which implies $\theta =\dfrac { (180-a)}{2}$

Multiple choice maths mapping your way mapping mapping space around us bearing and drawings

Perpendicular AL, BM are drawn from the vertices A,B of a triangle ABC to meet BC, AC at L, M. by proving the triangles ALC, BMC similar, or otherwise, then CM.CA=CL.CB

  1. True

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

In right triangles ALC and BMC, we have ∠ALC = ∠BMC = 90° and ∠ACM = ∠BCL (same angle at C). By AA similarity, ΔALC ~ ΔBMC. This gives the proportion: CM/CA = CL/CB, which can be rewritten as CM·CA = CL·CB. This is a standard power of a point result.

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

The line segments joining the midpoints of the sides of a triangle form four triangles each of which is:

  1. similar to the original triangle

  2. congruent to the original triangle

  3. an equilateral triangle

  4. an isosceles triangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $\triangle ABC$, D, E and F are mid points of AB, BC, CA respectively.

In $\triangle ABC$
F is mid point of AC and D is mid point of AB. 
Thus, By Mid point theorem
$FD = \dfrac{1}{2} CB$ and $FD = CB$ or $FD = CE$ and $FD \parallel CE$ (1)

Similarly,
$DE = FC$ and $DE \parallel FC$ (2)
$FE = DB$ and $FE \parallel DB$ (3) 
From (1), (2) and (3)
$\Box ADEF$, $\Box DBEF$ and $\Box DECF$ are parallelograms
The diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Hence, $\triangle DEF \cong \triangle ADF$
$\triangle DEF \cong \triangle DBE$
$\triangle DEF \cong \triangle FEC$
or, $\triangle DEF \cong \triangle ADF \cong \triangle ECF \cong \triangle ADF$
thus, mid points divide the triangle into 4 equal parts.

Thus, the smaller triangles are congruent to each other and similar to the original triangle.

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

$D, E, F$ are the mid points of the sides $AB, BC,CA$ respectively of $\triangle ABC$. Then $\triangle DEF$ is congruent to 

  1. $\triangle ABC$
  2. $\triangle AEF$
  3. $\triangle BDF , \triangle CDE $
  4. $\triangle ADF $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given: $\triangle ABC$, $D, E \ and \ F$ are mid points of $AB, BC, CA$ respectively.

In $\triangle ABC$

$F$ is mid point of $AC \ and \ D$ is mid point of $AB.$ 

Thus, By Mid point theorem

$FD = \dfrac{1}{2} CB$ and $FD = CB$ or $FD = CE$ and $FD \parallel CE$         ....(1)

Similarly,
$DE = FC$ and $DE \parallel FC$            ...(2)

$FE = DB$ and $FE \parallel DB$             ...3) 

From (1), (2) and (3)

$\Box ADEF$, $\Box DBEF$, $\Box DECF$ are parallelograms

The diagonal of a parallelogram divides the parallelogram into two congruent triangles.

Hence, $\triangle DEF \cong \triangle ADF$

$\triangle DEF \cong \triangle DBE$

$\triangle DEF \cong \triangle FEC$

$\triangle DEF \cong \triangle ADF \cong \triangle ECF \cong \triangle ADF$

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

In $\triangle ABC$, $AB=3cm, AC=4cm$ and $AD$ is the bisector of $\angle A$. Then $BD:DC$ is:

  1. $9:16$
  2. $16:9$
  3. $3:4$
  4. $4:3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In $\triangle ABC$,
AD bisects $\angle A$
By angle bisector theorem,
$\dfrac{AB}{AC} = \dfrac{BD}{DC}$
$\dfrac{BD}{DC} = \dfrac{3}{4}$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

What will be the critical angle of water if $ _a\mu _w=\frac{4}{3}$

  1. $\displaystyle { 42 }^{ \circ }$
  2. $\displaystyle { 49 }^{ \circ }$
  3. $\displaystyle { 22 }^{ \circ }$
  4. $\displaystyle { 1 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ _a\mu _w=\frac{4}{3}$

$sin(i _c)=\frac{1}{ _a\mu _w}=\frac{3}{4}$
So, $i _c=49^0$

Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

The critical angle of a medium B with respect another medium A is 45$^{\circ}$ and the critical angle of a medium C with respect the medium B is 30$^{\circ}$. The critical angle of medium C with respect to A is :

  1. Less than 30$^{\circ}$
  2. Greater than 30$^{\circ}$
  3. 30$^{\circ}$

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \mu _{B} sin 45^{\circ}=\mu _{A} $


$ \mu _{C} sin 30^{\circ}=\mu _{B} $

$ \mu _{C} sin 30 sin 45=\mu _{A} $

$ \mu _{C} \dfrac{1}{2\sqrt{2}}=\mu{A} $

$ \therefore critical \ angle = sin ^{-1} \left ( \dfrac{1}{2\sqrt{2}} \right )<30^{\circ} $

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

If the measures of the sides of a triangle are ________, then it is not a right angled triangle.

  1. $3,4,5$
  2. $5,12,13$
  3. $8,24,26$
  4. $7,24,25$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} 8,24,26 \ \sin  ce, \ { \left( 8 \right) ^{ 2 } }+{ \left( { 24 } \right) ^{ 2 } }\ne { \left( { 26 } \right) ^{ 2 } } \end{array}$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

$\angle B$ is a right angle is in $\Delta ABC$v and P,Q are points of trisection of hypotenuse $\bar{AC}.$ then $BP^{2}+BQ^{2}=\frac{5}{9}AC^{2}.$

  1. True

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

Using Pythagorean theorem: In right triangle ABC, let BC be hypotenuse. If P, Q trisect AC, then AP = PQ = QC = AC/3. Using Pythagoras: BP^2 + BQ^2 = (AB^2 + AP^2) + (AB^2 + AQ^2) = 2AB^2 + (AC/3)^2 + (2AC/3)^2. With AB^2 + BC^2 = AC^2 for right triangle, this simplifies to 5/9 AC^2.