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 circle and its elements angle subtended by arc sector of a circle arcs and sectors

Write True or False:

The tangent to the circumcircle of an isosceles $\triangle ABC$ at A, in which $AB = AC$, is parallel to BC.

  1. True

  2. False

  3. Ambiguous

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

Given-

PQ is a tangent to a circle at a point A when the circle is a circumcircle of the isosceles $\Delta ABC$.
$ AB=AC.$ 
To find out -
The statement, $PQ\parallel BC$, is true or not.
Justification-
In $\Delta ABC$ we have,
$ AB=AC.$ 
$\therefore  \angle ABC=\angle ACB$    ...(base angles of an isosceles triangle)    ........(i)
Again, PQ is the tangent to the circle at A & AB is a chord drawn from A. 
And AB subtends \angle ACB to the corresponding alternate segment of the circle.
$ \therefore  \angle PAB=$ corresponding alt. segment $\angle ACB$ ......(ii)
So, from (i) & (ii), 
$\angle PAB=\angle ABC.$ 
But they are alternate angles $\Longrightarrow  PQ\parallel BC$ 
$\therefore$ The statement, $PQ\parallel BC$, is true.

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

If the position vectors of the vertices of atriangle are $2 \overline { i } - \overline { j } + \overline { k } , \overline { i } - 3 \vec { j } - 5 \overline { k }$ and $3 \vec { i } - 4 \overline { j } - 4 \overline { k }$ then the triangle is

  1. Equilateral triangle

  2. Isosceles triangle

  3. Right angled isosceles triangle

  4. Right angled triangle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\\Let\>A\>(2\hat{i}-\hat{j}+\hat{k}),\>B(\hat{i}-3\hat{j}-5\hat{k})\>and\\C(3\hat{i}-4\hat{j}-4\hat{k})\\then\\\overrightarrow{AB}=-\hat{i}-2\hat{j}-6\hat{k}\\\therefore\>|\overrightarrow{AB}|=\sqrt{1+4+36}=\sqrt{41}\\\overrightarrow{BC}=2\hat{i}-\hat{j}+\hat{k}\\\therefore\>|\overrightarrow{BC}|=\sqrt{4+1+1}=\sqrt{6}\\\overrightarrow{CA}=-\hat{i}+3\hat{j}+5\hat{k}\\\therefore\>|\overrightarrow{CA}|=\sqrt{1+9+25}=\sqrt{35}\\clearly\>\>\>|\overrightarrow{BC}|^2+|\overrightarrow{CA}|^2=|\overrightarrow{AB}|^2\\\therefore\>Triangle\>is\>a\>right\>angled\>triangle$

Multiple choice maths direct proportion and inverse proportion inverse proportion rule of three types of proportions

Four angles of a quadrilateral are in the ratio $3:5:7:9$. The greatest angle is _________.

  1. $125^o$
  2. $75^o$
  3. $135^o$
  4. $120^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the angles of quadrilateral be $x,y,z,w.$
 
Given ,$x:y:z:w=3:5:7:9$

$x=3k$
$y=5k$
$z=7k$
$w=9k$

we k.n.t sum of angles of a quadrilateral is $360^{\circ }$
$x+y+z+w=3k+5k+7k+9k=360^{\circ }$
$\Rightarrow 24k=360^{\circ }$
$\Rightarrow k=\dfrac{360^{\circ }}{24}=15^{\circ }$

Greatest angle is $w=9k=9(15^{\circ })=135^{\circ }$
Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In $\Delta ABC$, AB$ =5$cm, $BC=8$cm and $CA=7$cm. If D and E are respectively, the mid-points of AB and BC, then determine the length of DE.

  1. $3.5$ cm
  2. $2.5$ cm
  3. $2.8$ cm
  4. $2.0$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mid-Point Theorem : 
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

On applying the midpoint theorem, we get
 $DE = \cfrac {AC}{2} = \cfrac 72 = 3.5$cm

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 3, 0)$.

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

using formula incenter $x= \dfrac{ax _{1}+bx _{2}+cx _{3}}{a+b+c}$ 


    $y=\dfrac{ay _{1}+by _{2}+cy _{3}}{a+b+c}$ 
    $z=\dfrac{az _{1}+bz _{2}+cx _{3}}{a+b+c}$
    $a=5$    $x=\dfrac{5(0)+4(3)+3(0)}{5+4+3}$       ;     $x=1$


    $b=4$      $y=\dfrac{5(0)+4(0)+3(3)}{5+4+3}$       ; $y=\dfrac{3}{4}$


    $c=3$      $z=\dfrac{5(0)+4(0)+3(0)}{5+4+3}$        ; $z=0$

 incentre =$(1,\dfrac{3}{4},0)$ 

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

The mid-points of the sides of a triangle are $D(6,1),E(3,5)$ and $F(-1,-2)$ then vertex opposite to D is 

  1. $(-4,2)$
  2. $(-4,5)$
  3. $(2,2)$
  4. $(10,8)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the vertices be A, B, C. Midpoints are D(6,1), E(3,5), F(-1,-2). The vertex opposite to D is A. Using the property that the quadrilateral formed by midpoints is a parallelogram, A = E + F - D = (3-1-6, 5-2-1) = (-4, 2).

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

ABC is an isosceles triangle with AB=AC. D,E, F are mid point of sides BC,AB and AC respectively then line segment $A D \perp E F$ and is bisected by it.

  1. True

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

In an isosceles triangle with AB=AC, the median AD to the base BC is also the altitude. EF, connecting midpoints of AB and AC, is parallel to BC. Thus, AD is perpendicular to EF and bisected by it.

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

State true or false:

In triangle $ ABC $, $ P $ is the mid-point of side $ BC $. A line through $ P $ and Parallel to $ CA $ meets $ AB $ at  point  $ Q $; and a line through $ Q $ and parallel to $ BC $ meets median $ AP $ at point $ R $. Can it be concluded that, $ BC= 4QP $

  1. True

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

Given: $\triangle ABC$, P is mid point of BC, $QR \parallel BC$ and $PQ \parallel AC$

Since, $ PQ \parallel AC$ and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.

Now, In $\triangle ABP$
Since, $QR \parallel BP$ and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence, $QR = \frac{1}{2} BP$ (Mid point theorem)
$QR = \frac{1}{2} (\frac{1}{2} BC)$ (P is midpoint of BC)
$BC = 4 QR$

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In triangle $ ABC $; $ M $ is mid-point of $ AB $, $ N $ is mid-point of $ AC $ and $ D $ is any point in base $ BC $. Then:

  1. MN bisects AD

  2. MN divides AD in the ratio 1:3

  3. MN divides AD in the ratio 1:2

  4. MN divides AD in the ratio 1:4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In $\triangle ABC, D$ is a point on AB and E is a point on BC such that DE || AC and $ar (DBE) = \dfrac {1}{2} ar (ABC)$. Find $\dfrac{AD}{AB}$

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

In $\triangle ABC$

$DE \parallel BC$
$ \cfrac{AD}{BD} = \cfrac {AE}{EC} $ Basic proportionality theorm
$ AD = \cfrac{BD}{2}$
$ \cfrac {AD}{BD} = \cfrac{1}{3}$
$ \cfrac {area(DBE)}{area(ABC)} = \left (\cfrac {AD}{AB} \right)^{2} $

$ \sqrt {\cfrac {area(DBE)}{area(ABC)}} = \left (\cfrac {AD}{AB} \right) $
$ 1 - \cfrac{AD}{AB}$
$ = 1- \cfrac{1}{\sqrt {2}}$
$ = \cfrac {\sqrt{2} - 1}{\sqrt{2}}$

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In any triangle ABC state whether following statements are true or false:
(1) the bisectors of the angles A, B, and C meet in a point,
(2) the medians, i.e. the lines joining each vertex to the middle point of the opposite side, meet in a point, and
(3) the straight lines through the middle points of the sides perpendicular to the sides meet in a point.

  1. True

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

The angle bisectors meet at the incenter, the medians meet at the centroid, and the perpendicular bisectors of the sides meet at the circumcenter. All three statements are standard geometric properties.

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

D,E,F are midpoints of sides BC, CA and AB of $\Delta ABC$. If perimeter of $\Delta ABC$ is 12.8 cm, then perimeter of $\Delta DEF$ is :

  1. $17 cm$
  2. $38.4 cm$
  3. $25.6 cm$
  4. $6.4 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given in $\triangle ABC$, $D,E,F$ are the mid points of sides $AB,BC$ and $CA$ respectively

Now using mid point theorem line segment joining the mid points of two sides is parallel to third side and also half of it.
$\therefore DF=\dfrac{1}{2}BC$
$\Rightarrow \dfrac{DF}{BC}=\dfrac{1}{2}.......(i)$
Similarly $\dfrac{DE}{AC}=\dfrac{1}{2}.........(ii)$
$\dfrac{EF}{AB}=\dfrac{1}{2}...........(iii)$
Using $(i),(ii)$ and $(iii)$
$\dfrac { DF }{ BC } =\dfrac { DE }{ AC } \dfrac { EF }{ AB } =\dfrac { 1 }{ 2 } $
$\dfrac { DF }{ BC } =\dfrac { DE }{ AC } \dfrac { EF }{ AB } =\dfrac { 1 }{ 2 } \ \therefore \triangle ABC\sim \triangle DEF$
Now if triangles are similar then ratio of their perimeter is equal to ratio of  of their corresponding sides.
$\ \dfrac { Perimeter(\triangle DEF) }{ Perimeeter(\triangle ABC) } =\dfrac { 1 }{ 2 } $
$\Rightarrow $ Perimeter of $\triangle ABC=\dfrac{1}{2}\times 12.8=6.4$ cm