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

Consider $\triangle ABC$ and $\triangle { A } _{ 1 }{ B } _{ 1 }{ C } _{ 1 }$ in such a way that $\overline { AB } =\overline { { A } _{ 1 }{ B } _{ 1 } } $ and M, N, ${ M } _{ 1 }$, ${ N } _{ 1 }$ be the mid points of AB, BC, ${ A } _{ 1 }{ B } _{ 1 }$ and ${ B } _{ 1 }{ C } _{ 1 }$ respectively, then

  1. $\overline { M{ M } _{ 1 } } =\overline { N{ N } _{ 1 } } $
  2. $\overline { C{ C } _{ 1 } } =\overline { M{ M } _{ 1 } } $
  3. $\overline { C{ C } _{ 1 } } =\overline { N{ N } _{ 1 } } $
  4. $\overline { M{ M } _{ 1 } } =\overline { B{ B } _{ 1 } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given AB = A1B1 and M, N, M1, N1 are midpoints, the triangles are congruent. The vector relationship between the midpoints of sides in congruent triangles preserves the distance between corresponding points. Specifically, M and M1 are midpoints of AB and A1B1, and N and N1 are midpoints of BC and B1C1; the displacement MM1 is equal to BB1.

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

The sides  $A B , B C , C A$  of a triangle  $A B C$  have  $3,4$  and  $5$  interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

  1. $205$
  2. $210$
  3. $315$
  4. $216$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total number of points $12$. If no three points are co-linear then total number of the   triangles would be  $^{12}{C _3}$.

But $3$ points on AB , $4$ points on BC and $5$ points on CA are co-linear.

So, total number of triangles formed should be 

$\begin{array}{l} { =^{ 12 } }{ C _{ 3 } }-\left( { ^{ 3 }{ C _{ 3 } }{ +^{ 4 } }{ C _{ 3 } }{ +^{ 5 } }{ C _{ 3 } } } \right)  \\ =220-\left( { 1+4+10 } \right)  \\ =205 \end{array}$



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

Mark the correct alternative of the following.
In which of the following cases can a right triangle ABC be constructed?

  1. $AB=5$cm, $BC=7$cm, $AC=10$cm
  2. $AB=7$cm, $BC=8$cm, $AC=12$cm
  3. $AB=8$cm, $BC=17$cm, $AC=15$cm

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

A right triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For option C, 8^2 + 15^2 = 64 + 225 = 289, which is 17^2. Thus, 8, 15, 17 forms a right triangle.

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

Mark the correct alternative of the following.
In which of the following cases, a right triangle cannot be constructed?

  1. $12$cm, $5$cm, $13$cm
  2. $8$cm, $6$cm, $10$cm
  3. $5$cm, $9$cm, $11$cm

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

Any triangle is right angled if it satisfies $a^2 + b^2 = c^2$, where c is the largest side from option (A)  we take 

$a = 12 cm$
$b = 5 cm$
$c = 13 cm$
we know $(12)^2 + (5)^2 = 169 = (13)^2$
So we can construct a right angled triangle.

Similarly in option (B) we can take 
$a = 8 cm$
$b = 6 cm$
$c = 10 cm$
with these sides also we can construct a right angled triangle.

Consider option (C) then 
$a = 5 cm$
$b = 9 cm$
$c = 11 cm$
$a^2 + b^2 = 25 + 81$   But  $c^2 = 121$
             $= 106$

So we cannot construct right angled triangle from $5 cm, 9 cm, 11 cm$.

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

Construct a right angled $\triangle ABC$ with $\angle B = 90^\circ, BC = 5\ cm$ and $AC = 10\ cm$ and find the the length of side $AB$

  1. $6.2\ cm$
  2. $5\ cm$
  3. $8.7\ cm$
  4. $7.2\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Step 1. Draw a line segment $BC=5\ \ cm$

Step 2. At $B$ draw an angle of $90^{\circ}$ and extend the ray.
Step 3. Now taking $C$ and centre draw an arc of radius $10$ cm intersecting the previous ray at $A$.
Step 4. Join $C$ to $A$.
Now using a ruler measure $AB$
$AB=8.7$ cm

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

Construct a triangle $ABC$ in which $AB = 5 cm$ and $BC = 4.6 cm$ and $AC = 3.7 cm$
Steps for the construction is given in jumbled form.Choose the appropriate sequence for the above
1) With radius as $5\ cm$ from $C$, cut an arc.
2)They arcs will intersect at point $A$. Join $AB$ and $AC$. $ABC$ is the required triangle.
3)Draw a line segment $BC = 4\ cm.$
4)With radius as $3$ cm from $B$, cut the arc. 

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

Correct sequence is:

1. Draw a line segment $AB=4$ cm.
2. With radius $5$ cm from $C$ , cut an arc.
3. With radius $3$ cm from $B$ , cut an arc.
4. The arc will intersect at point $A$, Join $AB$ and $AC$ .$ABC$ is required triangle.
So correct sequence is $3,1,4,2$
So option $C$ is correct.

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

Construct an isosceles $\triangle  XYZ,$ where $YZ=5$ units and $\angle XYZ=35^{o}$. Also, find the measure of $\angle YXZ$.

  1. $35^{o}$
  2. $70^{o}$
  3. $110^{o}$
  4. $140^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$YZ=5$ CM $,\angle XYZ=35^{\circ}$

As the triangle is isosceles therefore $\angle XZY=50^{\circ}$
Steps of construction:
1. Draw a line segment $XY=5$ cm.
2. At $Y$ draw an angle of $35^{\circ}$ and extend the arm.
3. At $Z$ draw an angle of $35^{\circ}$ and extend the ray such that it intersect the previous ray at $X$
4. Join $Y$ to $X$ and $Z$ to $X$
Now measure $\angle YXZ$
$\angle YXZ=110^{\circ}$

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

The side of a triangle are a. b and $\sqrt{a^2+ab+b^2}$. The greatest angle is 

  1. $90^0$
  2. $135^0$
  3. $120^0$

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

The greatest angle is opposite the longest side, which is sqrt(a^2 + ab + b^2). Applying the law of cosines with sides a, b, and c = sqrt(a^2 + ab + b^2) gives cos(theta) = (a^2 + b^2 - c^2) / (2ab) = -1/2, meaning theta = 120 degrees.

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.
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

Sum of the length of any two sides of a triangle is always greater than the length of third side.

  1. True

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

The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.