Mathematics · Quantitative Aptitude

Geometry of Triangles and Angles

758 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 axioms, postulates and theorems euclid's fifth postulate conditional statements and converse euclid's postulates

Select the correct statement for the following:

$A$: The angles of an equilateral triangle are equal.
$B$: Angles opposite to two congruent sides of a triangle are congruent.

  1. $A$ is a theorem and $B$ is its corollary
  2. $B$ is a theorem and $A$ is its corollary
  3. $A$ and $B$ are both theorems
  4. $A$ and $B$ are both corollaries
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$B$ is called the  Triangle theorem.  $A$ can be proved using this theorem, as in an equilateral triangle, all sides are equal and the angles opposite to these congruent sides are Equal. So, $A$ is the corollary of $B$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The points $(10,7,0)$, $(6,6-1)$ and $(6,9,-4)$ form a 

  1. Right -angled triangle

  2. Isosceles triangle
  3. Both $(1)$ & $(2)$

  4. Equilateral triangle
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P(10, 7, 0),\ Q(6, 6, -1),\ R(6, 9, -4)$
$PQ=\sqrt {4^2+(-1)^2 +(-1)^2}$
$=\sqrt {16+1+1}=3\sqrt 3$
$QR=\sqrt {0^2+3^2+-3^2}$
$PR=\sqrt {4^2+2^2+(-4)^2}$
$=\sqrt {16+4+16}=6$
$PQ^2+QR^2=(3\sqrt 2)^2+(3\sqrt 2)^2=6^2=PR^2$
$\therefore \ \triangle PQR$ is a right angle $D$ of $Q,\ PQ=QR$
$\triangle PQR$ is a isoscels traingle
$(C)$ Both $(1)$ & $(2)$


Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If $(1, 1, a)$ is the centroid of the triangle formed by the points $(1, 2, -3)$ , $(\mathrm{b}, 0, 1)$ and $(-1, 1, -4)$ then $a-b$ $=$

  1. $-5$
  2. $-7$
  3. $5$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The coordinates of the vertices of the triangle are given by $(1,2,-3) , (b,0,1), (-1,1,-4)$

Accordingly the coordinates of the centroid of this triangle will be given by 

($ \dfrac{b}{3}, 1 , -2 $)

Hence, $ \dfrac{b}{3} $ $= 1$ Or, $b= 3$

and $a = -2$

So $a- b = -5$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The circum centre of the triangle formed by the points $(2, 5, 1), (1, 4, -3)$ and $(-2, 7, -3)$ is

  1. $(6,0,1)$
  2. $(0,6,-1)$
  3. $(-1,6,2)$
  4. $(6,1,-2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the points $A(2,5,1) B(1,4,-3)$ and $C(-2,7,-3)$
Now using distance formula in $3D$, we have
$AB {=}$$\sqrt{18}$, $BC {=}$$\sqrt{18}$, and $AC {=}$$\sqrt{36}$
Since, ${AB}^{2}+{BC}^{2}$${=}$${AC}^{2}$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e $AC$.
Therefore by using section formula (1:1), circumcentre ${=}(0,6,-1)$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Assertion (A): The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus
Reason (R): $AB = BC = CD = DA$ and $AC = BD$

  1. Both A and R are individually true and R is the correct explanation of A

  2. Both A and R individually true but R is not the correct explanation of A

  3. A is true but R is false

  4. Both A and R false
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given: The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus. 
So using distance formula Reason is not true. 
Thus both A and R false. 
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

In a $\triangle {ABC}$, side $AB$ has the equation $2x+3y=29$ and the side $AC$ has the equation $x+2y=16$. If the mid point of $BC$ is $(5,6)$, then the equation of $BC$ is

  1. $2x+y=7$
  2. $x+y=1$
  3. $2x-y=17$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let co-ordinates of $B$ be $(x _1, y _1)$ & $C$ be $(x _2, y _2)$

$\therefore$ $(5,6)$ is the mid point,

so, $\dfrac{x _1 + x _2}{2} = 5, \dfrac{y _1 + y _2}{2} = 6$

$\Rightarrow x _1 + x _2 = 10, y _1 + y _2 = 12$

$B(x _1, y _1)$ lies on the line $2x + 3y = 29$

$\therefore 2x _1 + 3y _1 = 29$  ----(1)

$C(x _2, y _2)$ lies on the line $x + 2y = 16,$

$\therefore x _2 + 2y _2 = 16$  ----(2)

$\therefore$ putting $x _1, y _1$ in the form of $x _2, y _2$ in (1)

$2(10 - x _2) + 3(12 - y _2) = 29$  {$x _1 = 10 - x _2, y _1 = 12 - y _2$}

$\Rightarrow 20 - 2x _2 + 36 - 3y _2 = 29$

$\Rightarrow 2x _2 + 3y _2 = 27$  ----(3)

on subtracting $(3)$ and $(2)$ $\times$ $2$

$-y _2 = -5$

$y _2 = 5$

Putting $y _2 \,  in (2)$

$x _2 + 2(5) = 16$

$x _2 = 6$

$x _1 = 10 - x _2$

      $= 4$

$y _1 = 12 - 5 = 7$

Equation :

$\dfrac{x - x _1}{x _2 - x _1} = \dfrac{y - y _1}{y _2 - y _1}$

$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$

$\Rightarrow -x + 4 = y - 7$

$\Rightarrow x + y = 11$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If $A= \left ( 5,-1,1 \right ),B= \left ( 7,-4,7 \right ),C= \left ( 1,-6,10 \right ),D= \left ( -1,-3,4 \right )$. Then $ABCD$ is a

  1. square

  2. rectangle

  3. rhombus

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

AB${=}$ $\sqrt{{(7-5)}^{2}+{(-4+1)}^{2}+{(7-1)}^{2}}$
AB${=}$ $\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$
AB${=}$ $\sqrt{49}$
AB${=}$ 7
Similarly you find that BC${=}$ $\sqrt{49}$  CD${=}$ 7  and DA${=}$7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$
AC${=}$ $\sqrt{122}$
similarly BD${=}$ $\sqrt{74}$ 
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through  BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus

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 $\frac{x-1}{\frac{1}{2}}=\frac{y-1}{0}=\frac{z-1}{0}$
$\Rightarrow y=1,z=1$