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 construction of triangle construction of triangles - i construction of triangles constructions of triangles

Construct an isosceles $\triangle  ABC,$ where base $AB=7\ cm$ and $\angle ABC=50^{o}$. Also, find the measure of $\angle ACB$.

  1. $50^{0}$
  2. $80^{o}$
  3. $100^{o}$
  4. $120^{o}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$AB=7$ cm $,\angle ABC=50^{\circ}$

As the triangle is isosceles therefore $\angle CAB=50^{\circ}$
Steps of construction:
1. Draw a line segment $AB=7$ cm.
2. At $A$ draw an angle of $50^{\circ}$ and extend the arm.
3. At $B$ draw an angle of $50^{\circ}$ and extend the ray such that it intersect the previous ray at $C$
4. Join $A$ to $C$ and $B$ to $C$
Now measure $\angle ACB$
$\angle ACB=80^{\circ}$

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

State the following statement is True or False
In a right angle triangle $ABC$ such as $AC=5 cm ,BC=2 cm$ , $\angle B=90^o$
Then the length of $AB$ after construction is $7$cm

  1. True

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

In the given triangle $\Delta ABC$:


$AC=5$ and $BC=2$.

So by the property of triangle(sum of two sides are always greater than the third side):

$AB<(AC+BC)\implies AB<7$.

But in the given question it is given that $AB=7$, which is not possible.
So given statement is incorrect.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Construct a triangle $ABC$, in which $AB = 5.5 cm, AC = 6.5 cm$ and $\angle BAC = 70^{\circ}$.
Steps for its construction is given in a jumbled form.Identify its correct sequence.
1) At $A$, construct a line segment $AE$, sufficiently large, such that $\angle BAC$ at $70^\circ$, use protractor to measure $70^\circ$
2) Draw a line segment which is sufficiently long using ruler.
3) With $A$ as centre and radius $6.5cm$, draw the line cutting $AE$ at C, join $BC$, then $ABC$ is the required triangle.
4) Locate points $A$ and $B$ on it such that $AB = 5.5cm$.

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

Below are the correct steps.

i) Draw a line segment which is sufficiently long using ruler.
ii) Locate points $A$ and $B$ on it such that $AB=5.5 \ cm$
iii) At $A$ construct a line segment $AE$ , sufficiently large, such that $\angle BAC=70^\circ$, use protractor to measure
iv) With $A$ as centre and radius $6.5 \ cm$ draw the line cutting $AE$ at $C$, join $BC$ then $ABC$ is the required triangle.

So, the correct sequence of given steps is $2,4,1,3$.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

Which of the following steps is INCORRECT, while constructing $\triangle$LMN, right angled at M, given that LN = 5 cm and MN = 3 cm?
Step 1. Draw MN of length 3 cm.
Step 2. At M, draw MX $\perp$ MN. (L should be some where on this perpendicular).
Step 3. With N as centre, draw an arc of radius 5 cm. (L must be on this arc, since it is at a distance of 5 cm from N).
Step 4. L has to be on the perpendicular line MX as well as on the arc drawn with centre N. Therefore, L is the meeting point of these two and $\triangle$LMN is obtained.

  1. Only Step 4

  2. Both Step 2 and Step 3

  3. Only Step 2

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

In a right-angled triangle, the square of the hypotenuse is equal to twice the product of the other two sides. One of the acute angles of the triangle is

  1. $40^{\circ}$
  2. $42^{\circ}$
  3. $44^{\circ}$
  4. $45^{\circ}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a right angled triangle, if the square of the hypotenuse is  equal to twice the product of  the other two sides, then the two angles are equal.
Since, one of the angle is 90, the sum of other two will be 90. 
Thus, each angle should be $45^{\circ}$

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

A line $OP$ through origin $O$ is inclined at $30^{o}$ and $45^{o}$ to $OX$ and $OY$ respectively. The angle at which it is inclined to $OZ$ is-

  1. $\cos^{-1} \sqrt{\dfrac{4}{6}}$
  2. $\cos^{-1} \left(\dfrac{2}{6}\right)$
  3. $\cos^{-1} \left(\dfrac{1}{2}\right)$
  4. $Not\ defined$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

The area of $\displaystyle \Delta $ whose vertices are (a,b+c),(b,c+a),(c,a+b) will be -

  1. 0

  2. a+b+c

  3. ab+bc+ac

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

Area of $\displaystyle \Delta $
$\displaystyle =\frac { 1 }{ 2 } \left[ a\left( c+a \right) -b\left( b+c \right) +b\left( a+b \right) -c\left( c+a \right) +c\left( b+c \right) -a\left( a+b \right)  \right] $
$\displaystyle =\frac { 1 }{ 2 } \left[ ac+{ a }^{ 2 }-{ b }^{ 2 }-bc+ab+{ b }^{ 2 }-{ c }^{ 2 }-ac+bc+{ c }^{ 2 }-{ a }^{ 2 }-ab \right] $
$\displaystyle =\frac { 1 }{ 2 } \times 0$
$\displaystyle =0$

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

If $A,B,C$ are the vertices of a triangle whose position vectors are $\vec { a } ,\vec { b } ,\vec { c } $ and $G$ is the centroid of the $\triangle ABC$, then $\overrightarrow { GA } +\overrightarrow { GB } +\overrightarrow { GC } $ is

  1. $\vec { 0 } $
  2. $\vec { A } +\vec { B } +\vec { C } $
  3. $\cfrac { a+b+c }{ 3 } $
  4. $\cfrac { a-b-c }{ 3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the position vectors of vertices $A,B$ and $C$ of the triangle $ABC$ are $\vec { a } ,\vec { b } $ and $\vec { c } $


ie $\overrightarrow { OA } =\vec { a } \quad \overrightarrow { OB } =\vec { b } \quad \overrightarrow { OC } =\vec { c } $

$\therefore$ Centroid of triangle $(G)=\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 } $

Now $\overrightarrow { GA } +\overrightarrow { GB } +\overrightarrow { GC } $

$=\left( \overrightarrow { OA } -\overrightarrow { OG }  \right) +\left( \overrightarrow { OB } -\overrightarrow { OG }  \right) +\left( \overrightarrow { OC } -\overrightarrow { OG }  \right) $

$=\left( \vec { a } -\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 }  \right) +\left( \vec { b } -\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 }  \right) +\left( \vec { c } -\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 }  \right) \quad $

$=\cfrac { 1 }{ 3 } \left( 3\vec { a } -\vec { a } -\vec { b } -\vec { c } +3\vec { b } -\vec { a } -\vec { b } -\vec { c } +3\vec { c } -\vec { a } -\vec { b } -\vec { c }  \right) $

$=\cfrac { 1 }{ 3 } \left[ \vec { 0 }  \right] =\vec { 0 } $

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

If the coordinates of vertices of a triangle is always rational then the triangle cannot be

  1. Scalene

  2. Isosceles

  3. Rightangle

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

The triangle cannot be equilateral if coordinates of vertices of the triangle is always rational.

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The points (1, -1), $\displaystyle \left ( -\frac{1}{2},\frac{1}{2} \right )$ and (1, 2) are the vertices of an isosceles triangle

Say yes or no.

  1. Yes

  2. No

  3. Ambiguous

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

Let the point (1, -1), $\displaystyle \left ( -\frac{1}{2},\frac{1}{2} \right )$ and (1, 2) be denoted by P, Q and R respectively
Now PQ = $\displaystyle \sqrt{\left ( -\frac{1}{2}-1 \right )^{2}+\left ( \frac{1}{2}+1 \right )^{2}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$
QR = $\displaystyle \sqrt{\left ( 1+\frac{1}{2} \right )^{2}+\left ( 2-\frac{1}{2} \right )^{2}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$
PR = $\displaystyle \sqrt{\left ( 1-1 \right )^{2}+\left ( 2+1 \right )^{2}}=\sqrt{9}=3 $
From the above we see that PQ = QR
$\displaystyle \therefore $ The triangle is isosceles

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The points $A\left( {2a,\,4a} \right),\,B\left( {2a,\,6a} \right)\,$ and $C\left( {2a + \sqrt 3 a,\,5a} \right)$ (when $a>0$) are vertices of 

  1. an obtuse angled triangle

  2. an equilateral triangle

  3. an isosceles obtuse angled triangle

  4. a right angled triangle
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the given point

$A\left( 2a,4a \right),\,B\left( 2a,6a \right)\,and\,C\left( 2a+\sqrt{3}a,5a \right)$

Distance between AB,BC and CA  respectively.

$ AB=\sqrt{{{\left( 2a-2a \right)}^{2}}+{{\left( 4a-6a \right)}^{2}}} $

$ =\sqrt{0+{{(-2a)}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

$ BC=\sqrt{{{\left( 2a-(2a+\sqrt{3}a) \right)}^{2}}+{{\left( 6a-5a \right)}^{2}}} $

$ =\sqrt{3{{a}^{2}}+{{a}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

$ CA=\sqrt{{{\left( 2a+\sqrt{3}a-2a \right)}^{2}}+{{\left( 5a-4a \right)}^{2}}} $

$ =\sqrt{3{{a}^{2}}+{{a}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

Hence, $AB=BC=CA$

It is an equilateral triangle.

Option (B) is correct answer.

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

O is a point that lies in the interior of $\Delta ABC$. Then $2(OA - OB -OC) > \text{Perimeter}\ of\ \Delta ABC$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
From the $\triangle ABC,$ by triangle inequality,
$ OA+OB>AB$ ....... $(i)$
$ OB+OC>BC$ ........ $(ii)$
$ OA+OC>AC$ ........ $(iii)$
By adding $(i),(ii)$ and $(iii)$
$ 2(OA+OB+OC)>AB+BC+AC$
$ \therefore 2(OA+OB+OC)>\text{Perimeter of triangle } ABC$
Hence, the statement is false.