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 theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

There are m points on a straight line AB & n points on the line AC none of them being the point A. Triangles are formed with these points as vertices, when (i) A is excluded (ii) A is included.

The ratio of number of triangles in the two cases is?

  1. $\dfrac{m+n-2}{m+n}$
  2. $\dfrac{m+n-2}{m+n-1}$
  3. $\dfrac{m+n-2}{m+n+2}$
  4. $\dfrac{m(n-1)}{(m+1)(n+1)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider triangle without vertex
we can choose $2$ vertices from line $AB$ and one vertex from $A$ the possibilities are 
$\ ^{m}C _{2}\times n$
We can choose $2$ vertices from line $AC$ and one vertex from $AB$ the possibilities are:
$\ ^{n}C _{2}\times m$
As anyone of the above can be done so number of possibilities is 
$\ ^{m}C _{2}\times n+\ ^{n}C _{2}\times m$
Solving 
$\ ^{m}C _{2}\times \ ^{n}C _{2}\times m$
$=\dfrac{m!}{2!(m-2)!}\times n+\dfrac{n!}{2!(n-2)!}\times m$
$=\dfrac{m(m-1)}{2}\times n+\dfrac{n(n-1)}{2}\times m$
$=\dfrac{mn(m+n-2)}{2}$
Consider triangles with vertex $A$
As one vertex is $A$, we can choose one vertex from $AC$ and one from $AB$ the possibilities are 
$l\times m\times n$
$=mn$
Number of triangle is mn(m+n)/2$
Taking the ratio of $1$ and $2$
$\dfrac{mn(n+m-2)}{2}/\dfrac{mn(m+n)}{2}$
$\dfrac{m+n-2}{m+n}$



Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

The position vectors of vertices of $\Delta ABC$ are $(1, -2), (-7, 6)$ and $\left(\dfrac{11}{5}, \dfrac{2}{5}\right)$ respectively. The measure of the interior angle $A$ of the $\Delta ABC$, is

  1. acute and lies in $(75^o, 90^o)$
  2. acute and lies in $(60^o, 75^o)$
  3. acute and lies in $(45^o, 60^o)$
  4. obtuse and lies in $(120^o, 150^o)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A=(1, -2), B=(-7, 6), C=(11/5, 2/5). Vector AB = (-8, 8), vector AC = (11/5 - 1, 2/5 + 2) = (6/5, 12/5). The dot product AB dot AC = (-8)(6/5) + (8)(12/5) = -48/5 + 96/5 = 48/5. The magnitudes are |AB| = sqrt(64+64) = 8*sqrt(2) and |AC| = sqrt(36/25 + 144/25) = sqrt(180/25) = (6/5)*sqrt(5). Cos(A) = (48/5) / (8*sqrt(2) * (6/5)*sqrt(5)) = 48 / (48*sqrt(10)) = 1/sqrt(10). Since cos(A) is positive and approximately 0.316, A is acute and arccos(0.316) is approximately 71.5 degrees.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle
Let $ A(1,2,3), B(0,0,1), C(-1,1,1)$ are the vertices of a $\triangle ABC$. Then, the equation of internal angle bisector through A to side BC is 
  1. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+2\widehat{j}+3\widehat{k})$
  2. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+4\widehat{j}+3\widehat{k})$
  3. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+3\widehat{j}+2\widehat{k})$
  4. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+3\widehat{j}+4\widehat{k})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The internal angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides. The lengths of sides AB and AC can be found to determine the ratio, and then the coordinates of the dividing point on BC are used to find the direction vector of the angle bisector.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In a  $\triangle A B C,$  side  $A B$  has the equation  $2 x + 3 y = 29$  and the side  $A C$  has the equation  $x + 2 y = 16.$  If the mid point of  $B C$  is  $( 5,6 ) ,$  then the equation of  $B C$  is

  1. $2 x + y = 16$
  2. $x + y = 11$
  3. $2 x - y = 4$
  4. $x + y = - 11$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\cfrac { x _{ 1 }+x _{ 2 } }{ 2 } =5\Rightarrow x _{ 1 }+x _{ 2 }=10.....(1)\quad and\quad y _{ 1 }+y _{ 2 }=12.....(2)$

$Point(x _1,y _1)$ lie on line AC
then
$x _1+2y _1=16...(3)$
Similarly $2x _2+3y _2=29....(4)$
$\Rightarrow 2(x _1+x _2)+4y _1+3y _2=32+29\2\times 10+4y _1+36-3y _1=61\y _1=5\Rightarrow x _1=6\Rightarrow x _2=4\ \Rightarrow y _2=7$
now,
we take these two points and make equation,
$AC= x+y=11$

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In triangle, three angles are  $x , x + 10 ^ { \circ } + x + 20 ^ { \circ }$  then the biggest is

  1. $70 ^ { \circ }$
  2. $80 ^ { \circ }$
  3. $90 ^ { \circ }$

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

The sum of angles in a triangle is 180 degrees. x + (x + 10) + (x + 20) = 180. 3x + 30 = 180, so 3x = 150, x = 50. The angles are 50, 60, and 70 degrees. The biggest is 70 degrees.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In. triangle ABC,$\angle A$ + $\angle B$ = 144 and$\angle A$ + $\angle C$ = 124.
Calculate smallest angle of the triangle.

  1. $36^o$
  2. $56^o$
  3. $46^o$

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

$\angle A + \angle B = 144$...(I)
$\angle A + \angle C = 124$...(II)
In triangle ABC,
$\angle A  + \angle B + \angle C = 180 $
Add, I and II,
$\angle A + \angle B + \angle A + \angle C = 144+ 124$
$180 + \angle A = 268 $
$\angle A = 268 - 180 $
$\angle A = 88$
Put this value in (I)
$\angle A + \angle B = 144$
$88 + \angle B = 144$
$\angle B = 56$
Put this value in (II)
$\angle A + \angle C = 124$
$88 + \angle C = 124$
$\angle C = 36$

Multiple choice mathematical modelling proof by contradiction similar triangles

If a triangle is equiangular, then it is an obtuse angled triangle. Which of the following statements doesn't convey the same meaning as of this mentioned sentence.

  1. A triangle is equiangular only if it is an obtuse angled triangle

  2. If a triangle is not obtuse angled triangle then it is not an equiangular triangle.

  3. Equiangularity is a sufficient condition for triangle to be obtuse angled.

  4. A triangle is only obtuse is obtuse angled if it is equiangular
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the given statements to be in the form of $p\rightarrow{q}$

Options A, B and C represents $q\rightarrow{p}, \sim{p}\rightarrow\sim{q}$ and $\sim{q}\rightarrow\sim{p}$ respectively.

Hence, option C, which is in the form of $p\rightarrow{q}$ is the correct answer.




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 $w=9k=9(15^{\circ })=135^{\circ }$