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 general knowledge math & puzzles

According to him, sum of the angles of a triangle is greater than 180 degree...

  1. Euclid

  2. Riemann

  3. Lobachevsky

  4. It is impossible!
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In non-Euclidean geometries, the sum of angles in a triangle differs from 180 degrees. In Riemannian geometry (spherical/elliptic geometry), the angle sum is always greater than 180 degrees - think of a triangle drawn on a sphere where the sides are great circle arcs. Lobachevsky developed hyperbolic geometry where the sum is less than 180 degrees.

Multiple choice

A symmetric frame PQR consists of two inclined members PQ and QR, connected at 'Q' connected with a rigid joint and hinged at 'P' and 'R'. The horizontal length PR is l. If a weight W is suspended at 'Q' the bending moment of 'Q' is
  1. $\frac{WI}{2}$
  2. $\frac{WI}{4}$
  3. $\frac{WI}{8}$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a symmetric frame hinged at the ends and connected by a rigid joint at the apex, a point load at the apex results in axial forces in the members but no bending moment at the joint Q if it is a pin-jointed truss behavior.

Multiple choice

The length and bearings of a closed traverse PQRSP are given below |||| |---|---|---| | Line| Length (m)| Bearing (WCB)| | PQ| 200| 0o| | QR| 1000| 45o| | RS| 907| 180o| | SP| ?| ?|

The missing length and bearing, respectively of the line SP are

  1. 207 m and 270o

  2. 707 m and 270o

  3. 707 m and 180o

  4. 907 m and 270o
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Summing the latitudes and departures of the closed traverse: PQ (200, 0), QR (1000*cos45, 1000*sin45) = (707.1, 707.1), RS (907*cos180, 907*sin180) = (-907, 0). Sum of latitudes = 200 + 707.1 - 907 = 0.1. Sum of departures = 0 + 707.1 + 0 = 707.1. To close, SP must have latitude -0.1 and departure -707.1. This corresponds to a length of 707 m and bearing of 270 degrees.

Multiple choice

For machining a rectangular island represented by coordinates P(0, 0), Q(100, 0), R(100, 50) and S(0, 50) on a casting using CNC milling machine, an end mill with a diameter of 16 mm is used. The trajectory of the cutter centre to machine the island PQRS is

  1. (- 8, - 8), (108, - 8), (108, 58), (- 8, 58), (- 8, - 8)

  2. (8, 8), (94, 8), (94, 44), (8, 44), (8, 8)

  3. (- 8, 8), (94, 8), (94, 44), (8, 44), (- 8, 8)

  4. (0, 0), (100, 0), (100, 50), (50, 0), (0, 0)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Correct option is (1).

Multiple choice

Two tools P and Q have signatures 50 - 50 - 60 - 80 - 300 - 0 and 50 - 50 - 70 -70 - 80 - 150 - 0 (both ASA) respectively. They are used to turn components under the same machining conditions. If hP and hQ denote the peak to-valley heights of surfaces produced by the tools P and Q, the ratio $\frac{h_p}{h_Q}$ will be
  1. $\frac{tan8^0+cot15^0}{tan8^0+cot30^0}$
  2. $\frac{tan15^0+cot8^0}{tan30^0+cot8^0}$
  3. $\frac{tan15^0+cot7^0}{tan30^0+cot7^0}$
  4. $\frac{tan7^0+cot15^0}{tan7^0+cot30^0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

In $\Delta$PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find the respective values of sin P, cos P and tan P.

  1. 12/13, 5/13 and 12/5

  2. 12/13, 12/5 and 5/13

  3. 5/13, 12/13 and 12/5

  4. 12/5, 12/13 and 5/13
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 $\Delta$PQR is right-angled at Q.

Let                   QR =   x cm                         PR  =  (25 - x) cm                      PR2  =    PQ2  + QR2                                    (25 - x)2 =    (5)2 + x2 $\Rightarrow$       625 + x2 - 50x   = 25 + x2                                             $\Rightarrow$                       50x    =   600 $\Rightarrow$                           x       =  12 Thus, in  $\Delta$PQR, PQ = 5 cm, QR  = 12 cm and PR  = 13 cm. Therefore, sin P = QR/ PR  = 12 / 13, cos P = PQ / PR = 5 / 13 and tan P = QR/ PQ = 12/ 5.

Multiple choice maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

$VPQRS$ is rectangle based pyramid where $PQ = 30\ cm, QR = 20\ cm$ and volume is $2000\ {cm}^3$, then height (in cm) is

  1. $20$
  2. $40$
  3. $10$
  4. $30$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given : Length of base$(l)=30\ cm$, width of base$(h)=20\ cm$, Volume of pyramid$=2000\ cm^3$

Let $h$ be the height of the pyramid
We know that, volume of pyramid $=\dfrac{l\times w\times h}{3}$
$\implies 2000\ cm^3 = \dfrac{30 cm\times 20 cm\times h}{3}$
$\implies h=\dfrac{2000\times 3}{30\times 20} cm=10 cm$
Hence, height is $10 cm$.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

Which is the smallest side in the following triangle?
$\displaystyle \angle P:\angle Q:\angle R=1:2:3$

  1. $PQ$
  2. $QR$
  3. $PR$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\angle P: \angle Q: \angle R=1:2:3$

By applying relationship between sides and angles of a triangle, if two sides of a triangle are unequal, the side opposite to smaller angle is smaller.
Since, $\angle P$ is smallest, so $QR$ is smallest.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

In $\Delta PQR, \angle P = 60^{\circ}$ and $\angle Q = 50^{\circ}$. Which side of the triangle is the longest ?

  1. PQ

  2. QR

  3. PR

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

Using angle sum property of triangle 

$\angle P+\angle Q+\angle R={ 180 }^{ \circ  }\ \Rightarrow { 60 }^{ \circ  }+{ 50 }^{ \circ  }+\angle R={ 180 }^{ \circ  }\ \Rightarrow \angle R={ 180 }^{ \circ  }-{ 110 }^{ \circ  }={ 70 }^{ \circ  }$

So $\angle R$ is the largest angle and side opposite to largest angle is the longest side.
$\therefore PQ$ is the longest side.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

If a $\triangle PQR$ is constructed taking QR = $5$ cm, PQ = $3$ cm and PR = $4$ cm, then the correct order of the angles of the triangle is:

  1. $\displaystyle \angle P$ < $\displaystyle \angle Q$ < $\displaystyle \angle R$
  2. $\displaystyle \angle P$ > $\displaystyle \angle Q$ < $\displaystyle \angle R$
  3. $\displaystyle \angle P$ > $\displaystyle \angle Q$ >$\displaystyle \angle R$
  4. $\displaystyle \angle P$ < $\displaystyle \angle Q$>$\displaystyle \angle R$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a triangle, the angle is determined by their sides if it is given. The largest side will have the largest angle opposite it. The smallest side will have the smallest angle opposite to it.


So, $QR=5\ cm$. It is the largest side. Hence the angle opposite to it will also be largest that is$\angle P.$


Then the side$PR=4\ cm$, smaller than $QR$. Hence the $\angle Q$ will be smaller than $\angle P$

Finally, the smallest side $PQ=3\ cm$ with its corresponding angle $\angle R$ is smallest.

Hence the option C is right.