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 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 ABC ; M is mid-point of AB, N mid-point of AC and D is any point in base BC. Then:

  1. MN bisects AD

  2. MN divides AD in the ratio 1:3

  3. MN divides AD in the ratio 1:2

  4. MN divides AD in the ratio 1:4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

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 $DEF$ and $PQR$, $\angle D=\angle Q$ and $\angle R=\angle E$, then which of the following is not true?

  1. $\cfrac{EF}{PR}=\cfrac{DF}{PQ}$
  2. $\cfrac{DE}{PQ}=\cfrac{EF}{RP}$
  3. $\cfrac{DE}{QR}=\cfrac{DF}{PQ}$
  4. $\cfrac{EF}{RP}=\cfrac{DE}{QR}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given triangles DEF and PQR, with angle D = angle Q and angle R = angle E. By AA similarity, triangle DEF is similar to triangle QRP. The corresponding sides are proportional: DE/QR = EF/RP = DF/QP. Option B claims DE/PQ = EF/RP, which is not necessarily true because PQ is not the corresponding side to DE.

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 the triangles $ABC$ and $DEF$, angle $A$ is equal to angle $E$, both are equal to ${40}^{o}$, $AB:ED=AC:EF$ and angle $F$ is ${65}^{o}$, then angle $B$ is:

  1. ${35}^{o}$
  2. ${65}^{o}$
  3. ${75}^{o}$
  4. ${85}^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given triangle ABC and DEF. Angle A = Angle E = 40 degrees. AB/ED = AC/EF. This satisfies the SAS similarity criterion, so triangle ABC is similar to triangle EDF. Therefore, Angle B = Angle D and Angle C = Angle F. We are given Angle F = 65 degrees, so Angle C = 65 degrees. In triangle ABC, Angle A + Angle B + Angle C = 180. 40 + Angle B + 65 = 180 => Angle B = 180 - 105 = 75 degrees.

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 a cyclic quadrilateral ABCD, $\displaystyle \angle ABC=60^{\circ}$ and if O be the centre of the circle then the measure of $\displaystyle \angle OAC$ is

  1. $\displaystyle 20^{\circ}$
  2. $\displaystyle 30^{\circ}$
  3. $\displaystyle 40^{\circ}$
  4. $\displaystyle 50^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a cyclic quadrilateral, the angle subtended by a chord at the center is twice the angle at the circumference. If angle ABC = 60, the central angle AOC = 120. In triangle OAC, OA = OC (radii), so angle OAC = (180 - 120)/2 = 30 degrees.