Mathematics · Quantitative Aptitude

Geometry and Coordinate Geometry

72 Questions

Geometry questions test your knowledge of triangle properties, congruence, similarity, and quadrilaterals. They form a significant portion of the mathematics section in competitive exams. Practicing these builds spatial reasoning and theorem application skills.

triangle similaritycongruence rulesquadrilateral propertiesright angle properties

Geometry and Coordinate Geometry Questions

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Construct a right angled triangle $PQR$, in which $\angle Q = 90^\circ $, hypotenuse $PR=8\,cm$ and $QR=4.5\,cm$. Draw bisector of angle $PQR$ and let it meet $PR$ at point $T$ then $T$ is equidistant from$PQ$ and $QR$.

  1. True

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

The angle bisector theorem states that any point on the angle bisector of an angle is equidistant from the two sides forming the angle. Since T lies on the bisector of angle PQR, it is by definition equidistant from PQ and QR.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Construct a $\triangle PQR$ in which $QR= 4.6\ cm., {\angle Q}={\angle R=50 ^{0}}$. Then the perimeter of the triangle is:

  1. $10.2\ cm$
  2. $13.2\ cm$
  3. $11.8\ cm$
  4. $12.4\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In triangle PQR, if angle Q = angle R = 50 degrees, then angle P = 180 - 100 = 80 degrees. Using the sine rule, PQ/sin(50) = QR/sin(80) = PR/sin(50). PQ = PR = QR * sin(50) / sin(80) = 4.6 * 0.766 / 0.985 approx 3.57. Perimeter = 4.6 + 3.57 + 3.57 = 11.74, which rounds to 11.8.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles
Construct a $\triangle PQR$ such that $\angle P = 30^\circ, \angle Q = 60^\circ$ and $PQ = 10\ cm$.Find the measure of $\angle R$
  1. $30^\circ$
  2. $45^\circ$
  3. $60^\circ$
  4. $90^\circ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sum of angles in any triangle is 180 degrees. Given angle P = 30 and angle Q = 60, angle R = 180 - (30 + 60) = 180 - 90 = 90 degrees.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

For construction of a $\triangle PQR$, where $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the fourth step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass .
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

  1. $5$
  2. $1$
  3. $2$
  4. $3$
  5. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $P$ to $Q$. $PQR$ is required triangle.
So the fourth step is $3$
Option $D$ is correct.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

For construction of a $\triangle PQR$, where $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the second step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass.
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

  1. $2$
  2. $1$
  3. $4$
  4. $5$
  5. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $P$ to $Q$. $PQR$ is required triangle.
So the second step is $1$
Option $B$ is correct.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

For construction of a $\triangle PQR$, when $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the fifth step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass.
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

  1. $2$
  2. $3$
  3. $1$
  4. $5$
  5. $4$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $P$ to $Q$. $PQR$ is required triangle.
So the fifth step is $4$
Option $E$ is correct.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

For construction of a $\triangle PQR$, where $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the first step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass .
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

  1. $2$
  2. $1$
  3. $3$
  4. $5$
  5. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $P$ to $Q$. $PQR$ is required triangle.
So the first step is $5$
Option $D$ 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

If  in $\Delta PQR$,M and N are  points on PQ and PR and $PQ=1.28 ,PR=2.56,PM=0.18,PN=0.36$ cm. 
then $MN||QR.$

  1. True

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

According to the Converse of the Basic Proportionality Theorem, if PM/PQ = PN/PR, then MN || QR. Here, PM/PQ = 0.18/1.28 = 18/128 = 9/64. PN/PR = 0.36/2.56 = 36/256 = 9/64. Since the ratios are equal, the lines are parallel.

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

$\displaystyle \triangle ABC\sim \triangle PQR$ If ar(ABC)=2.25$\displaystyle m^{2}$ ar(PQR)=6.25$\displaystyle m^{2}$, PQ=0.5 m, then length of AB is

  1. $30 cm$
  2. $1.5 cm$
  3. $50 cm$
  4. $2 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The area of two similar triangles are proportional to the square of their corresponding sides

$\therefore \dfrac{Ar(ABC)}{ar(PQR)}=\dfrac{(AB)^2}{(PQ)^2}$
$\Rightarrow \dfrac{2.25}{6.25}=\dfrac{(AB)^2}{(.5)^2}$
$\Rightarrow AB^2=\dfrac{2.25\times .5}{6.25}$
$\Rightarrow AB^2=.09$
$\Rightarrow AB=.3m=30cm$

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 $\displaystyle \triangle ABC\cong \triangle RQP,\angle A=80^{\circ},\angle B=60^{\circ}$, then the value of $\displaystyle \angle P$ is

  1. $\displaystyle 60^{\circ}$
  2. $\displaystyle 50^{\circ}$
  3. $\displaystyle 40^{\circ}$
  4. $\displaystyle 80^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If two triangles are equal then their corresponding angle are also equal

$\therefore \angle A=\angle R,$$\angle B=\angle Q$ and$ \angle C=\angle P$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^{\circ}$
$80^{\circ}+60^{\circ}+\angle C=180^{\circ}$
$\angle C=180^{\circ}-140^{\circ}=40^{\circ}$
$\therefore \angle P=40^{\circ}$

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

If $\displaystyle \triangle ABC$ and $\displaystyle \triangle PQR$ are similar triangles such that $\displaystyle \angle A=32^{\circ}$ and $\displaystyle \angle R=65^{\circ}$ then $\displaystyle \angle B$ is

  1. $\displaystyle 83^{\circ}$
  2. $\displaystyle 42^{\circ}$
  3. $\displaystyle 65^{\circ}$
  4. $\displaystyle 97^{\circ}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since triangle ABC anD PQR are similar 

$\therefore \angle A=\angle P,\angle B=\angle Q   and  \angle  C=\angle R=65^\circ$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^\circ$
$32^\circ+\angle B+65=180^\circ$
$\angle B=180^\circ-97^\circ$
$\angle B=83^\circ$

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

In $\displaystyle \triangle LMN,\triangle L=60^{\circ},\angle M=50^{\circ}$ If $\displaystyle \angle LMN\sim \triangle PQR$ then the value of $\displaystyle \angle R$ is

  1. $\displaystyle 40^{\circ}$
  2. $\displaystyle 60^{\circ}$
  3. $\displaystyle 70^{\circ}$
  4. $\displaystyle 110^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\because \triangle LMN=\triangle PQR$ then

$\angle L=\angle P$,$\angle M=Q$,$\angle N=\angle R$
In $\triangle LMN$$\angle L=60^{\circ},\angle M=50^{\circ}$
$\angle L+\angle M+\angle N=180^{\circ}$
$60^{\circ}+50^{\circ}+\angle N=180^{\circ}$
$\angle N=180^{\circ}-110^{\circ}$
$\angle N=70^{\circ}$
$\therefore \angle R=70^{\circ}$

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

If $\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be ________ similarity of postulate.

  1. SAS similarity postulate

  2. AAA similarity postulate

  3. SSS similarity postulate

  4. AAS similarity postulate
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be SAS similarity of postulate.
Because, SAS Similarity Postulate states, "If an angle of one triangle is congruent to the corresponding angle of another triangle and the sides that include this angle are proportional, then the two triangles are similar."

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

For $\triangle ABC$ and $\triangle PQR$, if $m\angle A=m\angle R $ and $m\angle C=m\angle Q$, then $ABC \longleftrightarrow $_________ is a similarity.

  1. $RQP$
  2. $PQR$
  3. $RPQ$
  4. $QPR$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For $\triangle ABC $ and $\triangle PQR$,
$m\angle A = m\angle R$
$m\angle C= m\angle Q$
$\therefore $ by AA criteria for similarity 

$ABC \longleftrightarrow RPQ $ is a similarity.