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 tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

In a right-angled triangle ABC, $\angle B=90^{o}, BC = 12 cm $ and $AB = 5 cm$.The radius of the circle inscribed in the triangle (in cm) is

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

We know in $\triangle ABC, AB=5cm, BC=12cm$.
So, by pythagoras theorem we can find the length of side $AC$
$AC^2= AB^2 +BC^2=5^2+ 12^2$
$\therefore AC=13cm$
Circle is inscribed in a triangle. This type of circle is called as Incircle.
So, radius of incircle $=\displaystyle \frac {2 \triangle }{a+b+c}$
where $\triangle$ is the area of $\triangle ABC$ and $a,b,c$ are the sides of the triangle.
Area of $\triangle ABC= \displaystyle \frac {1}{2} AB \times BC= \frac {1}{2} \times 5 \times 12= 30sq.cm$
$\therefore$ radius of incircle $= \displaystyle \frac {2 \times 30}{5+12+13}=\frac {60}{30}=2cm$

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

ABC is a right angled triangle right angled at B such that $BC = 6$ cm and $AB = 8$ cm. A circle with center O is inscribed in $\displaystyle \Delta ABC$. The radius of the circle is

  1. 1 cm

  2. 2 cm

  3. 3 cm

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

Given, $BC = 6$ and $AB = 8$
using Pythagoras Theorem,
$AC^2 = AB^2 + BC^2$
$AC^2 = 6^2 + 8^2$
$AC = 10$
Radius = $\cfrac{2\times Area}{Perimeter}$
Radius = $\cfrac{2 \times (\dfrac{1}{2} \times 6 \times 8)}{10+8+6}$
Radius = $2$ cm

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

$ABC$ is a right triangle with $\angle A = 90^{\circ}$. Let a circle touch tangent $\overline {AB}$ at A and tangent $\overline {BC}$ at some point D. Suppose the circle intersects $\overline {AC}$ again at E and $CE = 3 cm, CD = 6 cm$, find the measure of BD

  1. $9 cm$
  2. $3\sqrt {5} cm$
  3. $3 cm$
  4. $2 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $\angle BAC = 90^\circ$

$AB$ is tangent at $A$

$BDC$ is tangent at $D$

$CE = 3,CD =  6$

According to the tangent-secant theorem the length of tangent segment squared equals the product of secant segment and its external segment.

$\implies AC \times CE = CD^2$

$AC = \dfrac{6^2}{3} = 12$

$AE = 12 – CE = 9 = d = 2r$

Let $O$ be the center of the circle

$OA = OD = r$

$AB = BD = l$, tangents drawn from $B$

Since $\angle A = 90$

$BC^2 = AC^2 + AB^2$

$\implies (l + CD)^2 = AC^2 + l^2$

$\implies l^2 + 6^2 + 12l = 12^2 + l^2$

$\implies 12l = 108$

$BD = l = 9 \, cm = $ length of tangent

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The base at a triangle passes through a fixed point $(a, b)$ and its sides are respectively bisected at right angles by the lines $y^{2} - 4xy - 5x^{2} = 0$. Find locus of its vertex.

  1. $2 \, (x^2 \, + \, y^2) + (3a + 2b) x + (2a - 3b) y = 0$
  2. $2 \, (x^2 \, + \, y^2) - (3a + 2b) x + (2a - 3b) y = 0$
  3. $2 \, (x^2 \, + \, y^2) + (3a + 2b) x - (2a - 3b) y = 0$

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

The lines are y^2 - 4xy - 5x^2 = 0, which factor into (y - 5x)(y + x) = 0. Using the properties of the orthocenter and the given fixed point, the locus of the vertex is derived as 2(x^2 + y^2) + (3a + 2b)x + (2a - 3b)y = 0.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

A triangle ${ABC}$ is formed by the lines $2x-3y-6=0$; $3x-y+3=0$ and $3x+4y-12=0$. If the points $P(\alpha,0)$ and $Q(0,\beta)$ always lie on or inside the $\triangle {ABC}$, then

  1. $\alpha \in [-1,2]$ and $\beta\in [-2,3]$
  2. $\alpha \in [-1,3]$ and $\beta\in [-2,4]$
  3. $\alpha \in [-2,4]$ and $\beta\in [-3,4]$
  4. $\alpha \in [-1,3]$ and $\beta\in [-2,3]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The points P and Q lie inside the triangle if they satisfy the inequalities defined by the three lines forming the triangle. Testing the bounds for alpha and beta confirms the range.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The combined equation of two sides of an equilateral tringle is $x^{2}-3y^{2}-2x+1=0$. If the length of a side of the triangle is $4$ then the equation of the third side is

  1. $x=2\sqrt{3}+1$
  2. $y=2\sqrt{3}+1$
  3. $x+2\sqrt{3}=1$
  4. $x=2\sqrt{3}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$x^{2}-3y^{2}-2x+1=0$

$(x-1)^{2}=3y^{2}$

$x-1=\pm\sqrt{3}y$

Hence the equation of the sides are 

$x-\sqrt{3}y=1$ and $x+\sqrt{3}y=1$

They intersect at $(1,0)$. Hence one of the vertex will be $(1,0)$.

Now we can clearly observe that the equation of the third side will be perpendicular to x-axis and parallel to y-axis since

$x-\sqrt{3}y=1$ and $x+\sqrt{3}y=1$ are equally inclined to positive x axis- one in clockwise sense and another in anticlockwise sense, and both have the same x-intercept while equal and opposite y intercept. In other words we can imagine $x-\sqrt{3}y=1$ as the image of the line $x+\sqrt{3}y=1$ with respect to x axis.

Hence the third line will be of the form $x=c$.

Now distance of the vertex $(1,0)$ from the above line will be 

$=asin60^{0}$

$=4sin60^{0}$

$=2\sqrt{3}$.

$=\dfrac{|1-c|}{1}$

Or 

$|1-c|=2\sqrt{3}$

Hence

$c-1=2\sqrt{3}$

$c=2\sqrt{3}+1$ and $c=-2\sqrt{3}+1$

Hence the corresponding equations are 

$x=2\sqrt{3}+1$ and $x+2\sqrt{3}=1$.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If G is the centroid and O is the circumcentre of the triangle with vertices (1, 2, 0), (0, 0, 2) and (2, 1, 1), then equation/s of line OG is/are

  1. x = y = z

  2. y = 1, z = 1
  3. $\frac{x-2}{1}=\frac{y-2}{1}=\frac{z-2}{1}$
  4. $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A(1, 2, 0), B (0, 0, 2) and C(2, 1, 1)
$\therefore$ G(1, 1, 1)
$AB^{2} = 1 + 4 + 4 = 9$, $AC^{2} = 1 + 1 + 1 = 3$ and $BC^{2} = 4 + 1 + 1 = 6$
$\therefore$ $AB^{2} =AC^{2} + BC^{2}$
$\therefore$ $\Delta $ ABC is right angled at C
$\therefore$ O is the mid point of AB
$\therefore$ coordinates of O are $\left ( \frac{1}{2},1,1 \right )$
$\therefore$ equation of OG are $\frac{x-1}{\frac{1}{2}}=\frac{y-1}{0}=\frac{z-1}{0}$
$\Rightarrow y=1,z=1$

Multiple choice physics simple machine common machines terms related to machines introduction to simple machines

The mechanical advantage of inclined plane of angle of inclination $60^{\circ}$ is equal to :

  1. $\dfrac{2}{\sqrt{3}}$
  2. $cosec 30^o$

  3. both (1) and (2)

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mechanical advantage of an inclined plane at an angle $\theta$ is  $ \dfrac{Length }{height}$$ = \dfrac{l}{h} = \dfrac{1}{sin \theta}$


$ = cosec \theta = cosec 60^o = \dfrac{1}{sin 60^o} = \dfrac{2}{\sqrt{3}}$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

If an triangle ABC, A = {1, 10}, circumference = $\left( -\dfrac { 1 }{ 3 } ,\dfrac { 2 }{ 3 }  \right) $ and orthocenter = $\left( \dfrac { 11 }{ 3 } ,\dfrac { 4 }{ 3 }  \right) $ then the co-ordinate of mid-point of side opposite to A is ________.

  1. (1, 11/3)

  2. (1, 5)

  3. (1, -3)

  4. (1, 6)
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths constructions mid-point formula midpoints division of a line segment

If $(-2,3), (4,-3), (4,5)$ are mid-points of the sides of a triangle, find the coordinates of the centroid of the triangle formed by these mid-points.

  1. $\left (3,\dfrac43 \right )$
  2. $\left (2,\dfrac43 \right )$
  3. $\left (2,\dfrac53 \right )$
  4. $\left (3,\dfrac53\right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given vertices of a triangle be A$(-2,3)$ and B $(4,-3)$ let the third vertex be C $(4,5)$.


Let Centroid be $G= \left [\dfrac {x _1 + x _2 + x _3}{3} , \dfrac{y _1 +y _2 + y _3}{3}\right ]$


$\Rightarrow G$ = $\left (\dfrac{-2 +4+4 }{3} ,\dfrac{3-3+5}{3}\right )$

$\Rightarrow G$ = $\left (\dfrac{6 }{3} , \dfrac{5}{3}\right )$

$\Rightarrow G$ = $\left (2 , \dfrac {5}{3}\right )$.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

Find the third vertex of a triangle, if two of its vertices are $(-3,1), (0,-2)$ and centroid is at the origin.

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

Let the third vertex be $(a,b)\equiv(x _1,y _1),(-3,1)\equiv(x _2,y _2), (0,-2)\equiv(x _3,y _3)$


Centroid of triangle is $(0,0)$

$G =\left [\dfrac {x _1 + x _2 + x _3}{3} , \dfrac{y _1 +y _2 + y _3}{3}\right ]$

$\Rightarrow \left( \dfrac { a-3+0 }{ 3 } ,\dfrac { b+1-2 }{ 3 }  \right) =(0,0)$

$\Rightarrow\dfrac{a-3}3=0$ and $\dfrac{b-1}3=0$

$\Rightarrow a=3,b=1$

So the third vertex is $(3,1)$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

$A\equiv(0, b), B\equiv(0, 0) $ and $C\equiv(a, 0)$ are the vertices of $\triangle ABC. D, E, F$ are the mid-points of the sides $BC, CA $ and $AB $ respectively. If  $a^{2}+ b^{2} = 20$ then

  1. $(AD)^{2}=9$
  2. $(BE)^{2}=4$
  3. $(AD)^{2}+(CF)^{2}=25$
  4. $(AD)^{2}+(CF)^{2}=(BE)^{2}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

As $A\equiv\left( 0,b \right) ,B\equiv\left( 0,0 \right) $ and $C\equiv\left( a,0 \right) $ 
Then $D\equiv\left( \cfrac { a }{ 2 } ,0 \right) ,E\equiv\left( \cfrac { a }{ 2 } ,\cfrac { b }{ 2 }  \right) $ and $F\equiv\left( 0,\cfrac { b }{ 2 }  \right) $
Using $a^{ 2 }+b^{ 2 }=20$, we have
${ \left( AD \right)  }^{ 2 }={ \left( \cfrac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( 0-b \right)  }^{ 2 }=\cfrac { { a }^{ 2 } }{ 4 } +{ b }^{ 2 }\ Similary\,\, {(CF)}^2\ { \left( AD \right)  }^{ 2 }+{ \left( CF \right)  }^{ 2 }={ \left( \cfrac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( 0-b \right)  }^{ 2 }+{ \left( 0-a \right)  }^{ 2 }+{ \left( \cfrac { b }{ 2 } -0 \right)  }^{ 2 }\ =\cfrac { 5\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ 4 } =25={ \left( BE \right)  }^{ 2 }$