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 mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

The sides $AB, BC$ and $CA$ of a triangle $ABC$ have $3, 4$ and $5$ interior points respectively on them.The number of triangles that can be constructed using these interior points as vertices is 

  1. 60

  2. 205

  3. 115

  4. 405
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

No. of ways $={^{ 3 }{ { C } _{ 1 } }}\times {^{ 4 }{ { C } _{ 1 }} }\times {^{ 5 }{ { C } _{ 1 } }}+{^{ 3 }{ { C } _{ 2 } }}\left( ^{ 4 }{ { C } _{ 1 } }+{^{ 5 }{ { C } _{ 1 } }}\right)+{^{ 4 }{ { C } _{ 2 }} } \left(^{ 3 }{ { C } _{ 1 } }+{^{ 5 }{ { C } _{ 1 }} }\right) +{^{ 5 }{ { C } _{ 2 }} }\left( ^{ 3 }{ { C } _{ 1 } }+{^{ 4 }{ { C } _{ 1 } }}\right)$

$\Rightarrow$ No of ways $=60+3\left( 4+5\right) +6\left( 3+5\right) +10 \left( 3+4\right)$
$\Rightarrow$ No of ways $=60+27+48+70=205.$
Hence, the answer is $205.$

Multiple choice maths plane geometry line and angle set squares constructing a perpendicular bisector construction of a perpendicular bisector

In the sides a,b,c of a triangle ABC are in A.P then $\dfrac{b}{c}$ belong to

  1. $(0, \dfrac{2}{3})$
  2. $(1,2)$
  3. $(\dfrac{2}{3}, 2)$
  4. $(\dfrac{2}{3}, \dfrac{7}{3})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a, b, c are in A.P., then 2b = a + c. By triangle inequality, a + c > b, so 2b > b (always true). Also, a + b > c and b + c > a. Substituting a = 2b - c into the inequalities leads to the range for b/c.

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If the vector $a, b$ and $c$ form the sides $BC, CA $ and $AB $ and equal magnitute respectively of a triangle $ABC,$ then

  1. $ a \cdot b + b\cdot c + c \cdot a = 0$
  2. $a \times b = b \times c = c \times a$
  3. $a \cdot b = b\cdot c = c \cdot a$
  4. $a \times b + b \times c + c \times a = O$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By triangle law, $\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { 0 } $
Taking cross product by $\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ respectively 
$\overrightarrow { a } \times \left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }  \right) =\overrightarrow { a } \times \overrightarrow { 0 } =\overrightarrow { 0 } $
$\Rightarrow \overrightarrow { a } \times \overrightarrow { a } \times \overrightarrow { a } \times \overrightarrow { b } +\overrightarrow { a } \times \overrightarrow { c } =\overrightarrow { a } $
$\Rightarrow \overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { c } \times \overrightarrow { a } \quad \left[ \because \overrightarrow { a } \times \overrightarrow { a } =\overrightarrow { 0 }  \right] $
Similarly, $\overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { b } \times \overrightarrow { c. } $
$\therefore \overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { b } \times \overrightarrow { c } =\overrightarrow { c } \times \overrightarrow { a. } $

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

The angles of a pentagon in degrees are $y^\circ$, $(y+20^\circ)$, $(y+40^\circ)-(y+60^\circ)$ and $(y+80^\circ)$. The smallest angle of the pentagon is

  1. $88^\circ$
  2. $78^\circ$
  3. $68^\circ$
  4. $58^\circ$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the given angles.

${{y}^{\circ }},\left( {{y}^{\circ }}+{{20}^{\circ }} \right),\left( {{y}^{\circ }}+{{40}^{\circ }} \right),\left( {{y}^{\circ }}+{{60}^{\circ }} \right),\left( {{y}^{\circ }}+{{80}^{\circ }} \right)$

 

We know that the sum of all angles of pentagon

$ {{y}^{\circ }}+\left( {{y}^{\circ }}+{{20}^{\circ }} \right)+\left( {{y}^{\circ }}+{{40}^{\circ }} \right)+\left( {{y}^{\circ }}+{{60}^{\circ }} \right)+\left( {{y}^{\circ }}+{{80}^{\circ }} \right)={{540}^{\circ }} $

$ 5{{y}^{\circ }}+{{200}^{\circ }}={{540}^{\circ }} $

$ 5{{y}^{\circ }}={{340}^{\circ }} $

 

Hence, the smallest angle of the pentagon is ${{68}^{\circ }}$.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

The sum of all the angles of a pentagon are

  1. $360^\circ$
  2. $540^\circ$
  3. $720^\circ$

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

Pentagon is a five sided polygon.

The sum of the interior angles of the pentagon is the sum of interior angles of the three triangles.The sum of interior angles of the three triangles is 180 degree.so the sum of interior angles of the pentagon is 3 times 180 degree which is 540 degree.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

State true or false:

$ D, E $ and $ F $ are the mid-points of the sides $ AB, BC $ and  $ CA $ of an isosceles $ \bigtriangleup ABC $  in which $ AB= BC $. then
 $ \bigtriangleup DEF $  is also isosceles.

  1. True

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

AB = AC
Hence, $\angle ABC = \angle ACB$ (Isosceles triangle property)
Now, since, D and F are mid point of AB and AC respectively, thus DF II BC (Mid point theorem)
Hence, 
$\angle ADF = \angle ABC$ and $\angle AFD = \angle ACB$ (Corresponding angles)
Thus, $\angle ADF = \angle ABC = \angle AFD = \angle ACB$ 
Now, In $\triangle$ ADF and FEC
$\angle ADF = \angle FEC$ (Corresponding angles of parallel lines EF and AB)
$\angle AFD = \angle ACB $(Corresponding angles of parallel lines DF and BC)
AF = FC (F is the mid point of AC)
Thus $\triangle ADF \cong \triangle FEC$ (AAS rule)
Hence, AD = FE (corresponding sides of congruent triangles)
Similarly, we can prove, AF = DE
Since, AD = AF (half lengths of equal sides, AB and AC)
Thus, EF = DE or $\triangle$ DEF is an isosceles triangle.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

In $\Delta ABC$, D and E are mid points of AB and BC respectively and $\angle ABC=90^o$, then

  1. $AE^2+CD^2=AC^2$
  2. $AE^2+CD^2=\frac {5}{4}AC^2$
  3. $AE^2+CD^2=\frac {3}{4}AC^2$
  4. $AE^2+CD^2=\frac {4}{5}AC^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a right triangle ABC, using the Pythagorean theorem for triangles ABE and BCD, AE^2 = AB^2 + BE^2 and CD^2 = BC^2 + BD^2. Substituting BE = BC/2 and BD = AB/2, we get AE^2 + CD^2 = 5/4(AB^2 + BC^2), which equals 5/4(AC^2).

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

If a line cuts sides $BC, CA$ and $AB$ of $\triangle ABC$ at $P, Q, R$ respectively then " $\dfrac {BP}{PC}\cdot \dfrac {CQ}{QA}\cdot \dfrac {AR}{RB} = 1$. "  that statement is ?

  1. True

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

This is Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

In a triangle $ABC,D$ and  $E$ are the mid-points of $BC,CA$ respectively. If $AD=5,BC=BE=4$, then $CA=$

  1. $5$
  2. $\sqrt{7}$
  3. $2\sqrt{7}$
  4. $5\sqrt{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Apollonius theorem on triangle ABC with median AD, 2(AD^2 + BD^2) = AB^2 + AC^2. With BE as a median to AC, 2(BE^2 + AE^2) = AB^2 + BC^2. Solving these equations with given values leads to AC = 2*sqrt(7).

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

Consider $\Delta$ABC and $\Delta A {1}B _{1}C _{1}$ in such a way that $\bar { AB } =\bar { { A } _{ 1 }{ B } _{ 1 } } $ and M,N,$M _{1}N _{1}$ be that mid points of AB,BC, $A _{1}B _{1}$ and $B _{1}C _{1}$ respectively, then ____________.

  1. $\bar { M{ M } _{ 1 } } =\bar { NN _{ 1 } } $
  2. $\bar { { CC } _{ 1 } } =\bar { MM _{ 1 } } $
  3. $\bar { { CC } _{ 1 } } =\bar { NN _{ 1 } } $
  4. $\bar { { MM } _{ 1 } } =\bar { BB _{ 1 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given AB = A1B1 and M, N, M1, N1 are midpoints, the vectors MM1 and NN1 represent the displacement between the midpoints of the sides of two congruent triangles, which must be equal.