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 properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

D is the mid point of the base BC of a triangle ABC. DM and DN are perpendiculars on AB and AC respectively. If $DM=DN$, the triangle  is

  1. Isosceles

  2. Equilateral

  3. Right angled

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

Given: D is mid point of BC. $DM \perp AB$ and $DN \perp AC$, $DM = DN$
Now, In $\triangle DMB$ and $\triangle DNC$,
$DM = DN$ (Given)
$\angle DMB = \angle DNC$ (Each $90^{\circ}$)
$BD = DC$ (D is mid point of BC)
Thus, $\triangle DMB \cong \triangle DNC$ (SAS rule)
Thus, $\angle B = \angle C$ (By cpct)
hence, $\triangle ABC$ is an Isosceles triangle.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

If ABC and DEF are similar triangles such that $\displaystyle \angle A=47^{\circ}$ and $\displaystyle \angle B=83^{\circ}$ then $\displaystyle \angle F$ is

  1. $\displaystyle 60^{\circ}$
  2. $\displaystyle 70^{\circ}$
  3. $\displaystyle 50^{\circ}$
  4. $\displaystyle 100^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since triangle ABC and DEF are similar 

$\therefore \angle A=\angle D,\angle B=\angle E   and  \angle  C=\angle F$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^\circ$
$47^\circ+83^\circ+\angle C=180^\circ$
$\angle C=180^\circ-130^\circ$
$\angle C=50^\circ$
$\therefore \angle F=50^\circ$

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

SAS criterion is true when two sides  and the included angle is congruent with the when two sides  and the included angle of the other triangle are equal. The included angle means

  1. The side between two sides

  2. The angle not between two sides

  3. The line between two sides

  4. The angle between two sides
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.
Therefore, D is the correct answer.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

  1. AAA similarity criterion

  2. SAS similarity criterion

  3. SSS similarity criterion

  4. All of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the corresponding angles are equal then the triangles are similar by $AAA$ similarity criteria.

Option $A$ is correct.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

When we construct a triangle similar to a given triangle as per given scale factor, we construct on the basis of ...........

  1. SSS Similarity

  2. AAA similarity

  3. Basic proportionality theorem
  4. $A$ and $C$ are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As we consider only sides, therefore, SSS similarity is used.
Option A is correct.  

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid point of $BC$. Ratio of the areas of triangle $ABC$ and $BDE$ is

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

$\triangle ABC \sim \triangle BDE$                            (both are equilateral triangles)


$\Rightarrow \triangle ABC : \triangle BDE = AB^2 : BD^2$

                                          $= AB^2 :  (\dfrac{1}{2} BC)^{2} $
                                          
                                          $ = AB^2 : \dfrac{1}{4} BC^2 $

                                          $= 4 : 1$           $(\because AB = BC)$
Hence proof.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

In $\triangle A B C$, D is a point on AB such that $A D = \frac { 1 } { 4 } A B$ and E is a point on AC such that $A E = \frac { 1 } { 4 } A C$ then $D E = \frac { 1 } { 8 } B C$

  1. True

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

By the Basic Proportionality Theorem converse, if AD/AB = AE/AC = 1/4, then DE is parallel to BC and triangle ADE is similar to triangle ABC with a scale factor of 1/4. Therefore, DE = (1/4)BC, not 1/8.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

State true or false:

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P, then
$\displaystyle \Delta APB$ is similar to $\displaystyle \Delta CPD.$

  1. True

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

In $\triangle$ APB and $\triangle$ CPD,
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle ABP = \angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP = \angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $\triangle APB \sim \triangle CPD$ (AAA rule)

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

In quadrilateral ABCD, the diagonals AC and BD intersect each at point O. If $AO=2CO$ and $BO=2DO$; Then,

$\displaystyle \Delta AOB$ is similar to $\displaystyle \Delta COD$

  1. True

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

Given: $AO = 2 CO$ or $\dfrac{AO}{CO} = 2$
Also given, $BO = 2 DO$ or $\dfrac{BO}{DO} = 2$
In $\triangle AOB$ and $\triangle COD$, we know 
$\angle AOB = \angle COD$
$\dfrac{AO}{CO} = \dfrac{BO}{DO}$
Thus, $\triangle AOB \sim \triangle COD$ (SAS rule)

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

$\angle BAC$ of triangle $ABC$ is obtuse and $AB=AC$. $P$ is a point in $BC$ such that $PC= 12$ cm. $ PQ $ and $PR$ are perpendiculars to sides $AB$ and $AC$ respectively. If $PQ= 15$ cm and $=9$ cm; find the length of $PB$.

  1. $20$
  2. $24$
  3. $36$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $AB = AC$, $PQ \perp AB$ and $PR \perp AC$
Since, $AB = AC$
$\angle ABC = \angle ACB$...(I) (Isosceles triangle property)

Now, In $\triangle PBQ$ and $\triangle PRC$
$\angle PBQ = \angle PCR$ (From I)
$\angle PQB = \angle PRC$ (Each $90^{\circ}$)
$\angle QPB = \angle RPC$ (Third angle)
Thus, $\triangle QPB \sim \triangle RPC$ (AAA rule)
Hence, $\dfrac{PQ}{PR} = \dfrac{PB}{PC}$
$\dfrac{15}{9} = \dfrac{PB}{12}$
$PB = \dfrac{15 \times 12}{9}$
$PB = 20$ cm

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, $\triangle DEF \cong \triangle ADF$

$\triangle DEF \cong \triangle DBE$

$\triangle DEF \cong \triangle FEC$

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