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

In triangle $ ABC $, $ M $ is mid-point of $ AB $ and a straight line through $ M $ and parallel to $ BC $ cuts $ AC $ in $ N $. Find the lenghts of $ AN $ and $ MN $ if $ BC= 7 $ cm and $ AC= 5 $ cm.

  1. $ AN= 2.5 $ cm and $ MN= 3.5 $ cm
  2. $ AN= 1.5 $ cm and $ MN= 3.5 $ cm
  3. $ AN= 2.5 $ cm and $ MN= 4.5 $ cm

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

M is the mid point of AB and MN II BC. Thus, N is the mid point of AC and


$MN = \dfrac{1}{2} BC$ (Mid point theorem)

$MN = \dfrac{1}{2} (7) $

$MN = 3.5 cm$

Also, $AN = \dfrac{1}{2} AC$

$AN = \dfrac{1}{2} (5)$

$AN = 2.5 cm$

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem
State true or false:

In triangle  $ ABC  $,  $ P  $ is the mid-point of side  $ BC  $. A line through $ P  $ and Parallel to  $ CA  $ meets  $ AB  $ at point  $ Q  $; and a line through  $ Q  $ and parallel to  $ BC $ meets median  $ AP  $ at point  $ R  $. Can it be concluded that,
$ AP= 2AR $ ?

  1. True

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

Given: $\triangle ABC$, P is mid point of BC, $QR \parallel BC$ and $PQ \parallel AC$

Since, $ PQ \parallel AC$ and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.

Now, In $\triangle ABP$
Since, $QR \parallel BP$ and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence, $AP = 2 AR$

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

State true or false:


In triangle $ ABC $, angle $ B $ is obtuse. $ D $ and $ E $ are mid-points of sides $ AB $ and $ BC $ respectively and $ F $ is a point in side $ AC $ such that $ EF $ is parallel to $ AB $. Then, $ BEFD $ is a parallelogram. 

  1. True

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

Given: $D$ is mid point of $AB$ and $E$ is mid point of $BC$, $F$ is any point on $AC$ and $EF \parallel AB$

Now, in $\triangle ABC$,
E is mid point of BC and $EF \parallel AB$
By Mid point Theorem, $F$ is mid point of $AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, $DF \parallel BE$
Since, $DF \parallel BE$ and $EF \parallel AB or BD$
Hence, BEFD is parallelogram.

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

In $\bigtriangleup : ABC$ , $E$ and $F$ are mid-points of sides $AB$ and $AC$ respectively. If $BF$ and $CE$ intersect each other at point $O$, then the $\bigtriangleup :OBC$ and quadrilateral $AEOF$ are equal in area.

  1. True

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

Given that $E$ and $F$ are mid points. We have to prove that ${a} _{1}(BOC)={a} _{1}(AEOF)$

As $EF\parallel BC$ (By midpoint theorem.)
${ a } _{ 1 }(\triangle BEF)={ a } _{ 1 }(\triangle ACFE)$  ($\because $ Between the parallel lines.)
${ a } _{ 1 }(\triangle BEF)-{ a } _{ 1 }(\triangle EOF)={ a } _{ 1 }(\triangle CFE)-{ a } _{ 1 }(\triangle EOF)\ \therefore { a } _{ 1 }(\triangle BOE)={ a } _{ 1 }(\triangle FOC)\longrightarrow (i)$
${a} _{1}(AEOF)={a} _{1}(BOC)$
$\therefore$ The statement is true.

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

If A B C D E F is a regular hexagon with A B = a and B C = b, then CE equals

  1. b-a

  2. -b

  3. b-2a

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

In a regular hexagon ABCDEF, the vector CE can be found using vector addition. Since AB = a and BC = b, the vectors for the sides are related by the geometry of the hexagon, leading to CE = b - a.

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