Mathematics · Quantitative Aptitude

Geometry and Coordinate Geometry

84 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 geometrical construction constructing a perpendicular bisector construction of a perpendicular bisector construction of penpendicual bisector set squares

For drawing the perpendicular bisector of $PQ$, which of the following radii can be taken to draw arcs from $P$ and $Q$?

  1. $\dfrac{PQ}2$
  2. $\dfrac{PQ}3$
  3. $\dfrac{2PQ}3$
  4. $\dfrac{PQ}4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To draw a perpendicular bisector of a given side, take any length that is greater than half the length of the side. Draw the arcs from the edges of the base. The point where arcs meet is on the perpendicular bisector.


From the given options,

$\dfrac{2PQ}{3}$ can be considered to draw to draw arcs from edges $P, \ Q$

Remaining options has the value $\leq \dfrac{PQ}{2}$

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

$P, Q, R$ and $S$ are the mid-points of sides $AB. BC, CD$ and $DA$ respectively of rhombus $ABCD$. Show that $PQRS$ is a rectangle.
Under what condition will $PQRS$ be a square ?

  1. When $ABCD $ is a square.
  2. When $ABCD$ is a parallelogram
  3. When $ABCD$ is a rectangle
  4. When $ABCD$ is a square or a rectangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $ABCD$ is a rhombus. $P, Q, R, S$ are mid points of $AB, BC, CD, DA$ respectively.
Join: $AC$ and $BD$

In $\triangle ABC$
$P$ is mid point of $AB$ and $Q$ is mid point of $AC$.
Thus, by mid point theorem, $PQ \parallel AC$ and $PQ = \dfrac{1}{2} AC$
Similarly, In $\triangle ACD$,
$S$ is mid point of $AD$ and $R$ is mid point of $CD$.
Thus, by mid point theorem, $SR \parallel AC$ and $SR = \dfrac{1}{2} AC$
Hence, $PQ \parallel SR$ and $PQ = SR$

Similarly, $PS = QR$ and $PS \parallel QR$
Thus, the opposite sides of $PQRS$ are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since $AC \perp BD$ thus, $PS \perp PQ$ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of $PQRS$ are equal and parallel and the sides meet each other at right angles. Hence, $PQRS$ is a rectangle.

If $PQRS$ had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if $ABCD$ is a square.

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 third 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. $3$
  2. $4$
  3. $2$
  4. $5$
  5. $1$
Reveal answer Fill a bubble to check yourself
C 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 third step is $2$
Option $C$ is correct.

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}$
$\angle L+\angle M+\angle N=180^{\circ}$
$60^{\circ}+50^{\circ}+\angle N=180^{\circ}$
$\angle N=180^{\circ}-110^{\circ}$
$\angle N=70^{\circ}$
$\therefore \angle R=70^{\circ}$