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 the plane angle between planes angle between two planes problems involving equation of plane

The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where $\displaystyle a\equiv (3, -2, 1); B\equiv (3, 1, 5); C\equiv (4, 0, 3)and D\equiv (1, 0, 0)is$

  1. $\displaystyle \frac{2}{\sqrt{29}}$
  2. $\displaystyle \frac{5}{\sqrt{29}}$
  3. $\displaystyle \frac{3\sqrt3}{\sqrt{29}}$
  4. $\displaystyle \frac{-2}{\sqrt{29}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\overrightarrow { AD } =-2\hat { i } +2\hat { j } -\hat { k } ,\overrightarrow { Ac } =\hat { i } +2\hat { j } +2\hat { k } ,\overrightarrow { AB } =3\hat { j } +4\hat { k } : \ \overrightarrow { n _{ 1 } } =\overrightarrow { AD } \times \overrightarrow { AC } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ -2 & 2 & -1 \ 1 & 2 & 2 \end{vmatrix}=6\hat { i } +3\hat { j } -6\hat { k } =3\left( 2\hat { i } +\hat { j } -2\hat { k }  \right) \ \overrightarrow { n _{ 2 } } =\overrightarrow { AC } \times \overrightarrow { AB } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 1 & 2 & 2 \ 0 & 3 & 4 \end{vmatrix}=2\hat { i } -4\hat { j } +3\hat { k } : \ \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right| =3\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 1 & -2 \ 2 & -4 & 3 \end{vmatrix}=3\left( 5\hat { i } -10\hat { j } -10\hat { k }  \right) \ \sin  \theta =\dfrac { 5 }{ \sqrt { 29 }  } \left( \because \sin  \theta =\dfrac { \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right|  }{ \left| \overrightarrow { n _{ 1 } }  \right| \left| \overrightarrow { n _{ 2 } }  \right|  }  \right) $

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Tetrahedron has Vertices at $O(0,0,0)$ , $A(1,2, 1)$ , $B(2,1,3)$ , $C(-1,1,2)$ . Then the angle between the faces $OAB$ and $ABC$ will be

  1. $\cos^{-1} (\dfrac{19}{35})$
  2. $\cos^{-1} (\dfrac{17}{31})$
  3. $30^{0}$
  4. $90^{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n _{1}= \overrightarrow{OA}\times \overrightarrow{OB}= \begin{vmatrix}\hat{i} &\hat{j}  &\hat{k} \1  &2  &1 \2  &1  &3 \end{vmatrix}= 5\hat{i}-\hat{j}-3\hat{k}$


$n _{2}=\overrightarrow{AB}\times \overrightarrow{AC}= \begin{vmatrix}\hat{i} &\hat{j}  &\hat{k} \1  &-1  &2 \-2  &-1  &1 \end{vmatrix}= \hat{i}-5\hat{j}-3\hat{k}$

$\cos \theta = \dfrac{\vec n _{1}-\vec n _{2}}{\left | n _{1} \right |\left | n _{2} \right |}$

$\theta = \cos^{-1}\left ( \dfrac{19}{35} \right )$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If the angles  $A,B,C$ of a $\triangle ABC$ are in $A.P.$, then:-

  1. ${c}^{2}={a}^{2}+{b}^{2}-ab$
  2. ${b}^{2}={a}^{2}+{c}^{2}-ac$
  3. ${c}^{2}={a}^{2}+{b}^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If A, B, C are in AP, then 2B = A+C. Since A+B+C = 180, B=60. Using the cosine rule b^2 = a^2 + c^2 - 2ac cos(B), and cos(60)=1/2, we get b^2 = a^2 + c^2 - ac.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

The sides of triangle are in A.P. and the greatest angle exceeds the least by 90. The sides are in the ratio _____________.

  1. $1 : 2 : \sqrt { 2 }$
  2. $1 : \sqrt { 3 } : 2$
  3. $\sqrt { 7 } + 1 : \sqrt { 7 } : \sqrt { 7 } - 1$
  4. $\sqrt { 3 } + 1 : 1 : \sqrt { 3 } - 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the sides be a-d, a, a+d. Using the law of cosines and the condition that the largest angle exceeds the smallest by 90 degrees, one can derive the ratio of the sides as sqrt(7)+1 : sqrt(7) : sqrt(7)-1.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

ABC is a triangle right angle at B. D is a point on AC such that $\angle ABD = 45^0$. If AC =$6$ and AD =$2$ , then AB is 

  1. $\dfrac{6}{\sqrt{5}}$
  2. ${3}{\sqrt{2}}$
  3. $\dfrac{12}{\sqrt{5}}$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the area of triangle ABC as the sum of areas of ABD and BCD, or using trigonometry in right triangles, we find AB = 6/sqrt(5).

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

If in a $\Delta ABC,\sin A=\sin^{2} B$ and $2\cos^{2}A=3\cos^{2}B$, then the $\Delta ABC$ is 

  1. Right angled

  2. Obtuse angled

  3. Isosceles

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

Given sin A = sin^2 B and 2 cos^2 A = 3 cos^2 B, substituting cos^2 A = 1 - sin^2 A = 1 - sin^4 B into the second equation allows solving for sin^2 B, which leads to A = 90 degrees.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Which of the following  can be the sides of a right-angled triangle?

  1. $0.5cm, 1.2 cm, 1.3cm$
  2. $2.4cm, 3.2 cm, 7.9cm$
  3. $5.0cm, 5.25 cm, 7.25cm$
  4. $1.6cm, 3.0 cm, 3.4cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A right-angled triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For 0.5, 1.2, 1.3: 0.25 + 1.44 = 1.69, which is 1.3^2.

Multiple choice maths pythagoras theorem similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

  1. 7 cm, 24 cm, 25 cmj

  2. 3 cm, 8 cm, 6 cm

  3. 50 cm, 80 cm, 100 cm

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

For option A: 7, 24, 25. Check if right triangle: 7² + 24² = 49 + 576 = 625 = 25². Also forms valid triangle (7+24 > 25). Option A is a right triangle with hypotenuse 25 cm. The question has typo ('25 cmj') but this doesn't affect the answer.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Let $ABC$ be a fixed triangle and $P$ be variable point in the plane of a triangle $ABC$. Suppose $a, b, c$ are lengths of sides $BC,  CA,AB$ opposite to angles $A, B, C $ respectively. If $a(PA)^{2} + b(PB)^{2} + c(PC)^{2}$ is minimum, then the point $P$ with respect to $\triangle{ABC}$ is

  1. Centroid

  2. Circumcenter

  3. Orthocenter

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

The point P that minimizes the weighted sum of squared distances a(PA)^2 + b(PB)^2 + c(PC)^2 is the centroid of the triangle.