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 geometrical constructions construction of triangles constructions of triangles construction of triangles - ii

To construct a triangle similar to a given triangle ABC with its sides 6/5th of the corresponding sides of $\Delta$ABC. Correct order of steps of construction -
(a) Draw a ray AX inclined at certain angle with AB on opposite side of C.
(b) Starting from A, cut off six equal line segments AX$ _1$, X$ _1$X$ _2$, X$ _2$X$ _3$, X$ _3$X$ _4$, X$ _4$X$ _5$ and X$ _5$X$ _6$ on AX.
(c) Draw a line B'C' parallel to BC to intersect AC produced at C'
(d) Join X$ _5$B and draw a line X6B' parallel to X5B, to intersect AB produced at B'.

  1. abcd

  2. acbd

  3. abdc

  4. adcb
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To construct a triangle like the one given in the following question, the given steps are to be followed:

1. Draw a ray AX inclined at a certain angle with AB on opposite side of C.
2. Starting from A, cut off six equal line segments $AX _1, X _1X _2, X _2X _3, X _3X _4, X _4X _5 and X _5X _6$ on $AX$.
3. Join $X _5B$ and draw a line $X _6B'$ parallel to $X _5B$, to intersect $AB$ produced at $B'$.
4.Draw a line $B'C'$ parallel to $BC$ to intersect $AC$ produced at $C'$.

Multiple choice maths geometrical constructions construction of triangles constructions of triangles construction of triangles - ii

Construct a $\Delta ABC$, whose perimeter is $10.5  cm$ and base angles are $60^o$ and $45^o$. Find the third angle.

  1. $75^o$
  2. $45^o$
  3. $90^o$
  4. $60^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  1. Draw a line segment PQ such that AB + BC + AC = PQ.
    2. Construct $\angle XPQ =\angle B = { 60 }^{ 0 }\ and\ \angle YQP =\angle C =45^{ 0 }$
    3. Bisect $\angle XPQ and \angle YQP$ and their bisectors will meet at A.
    4. Draw the perpendicular bisector DE of AP which meets PQ  at B. 
    5. Draw the perpendicular bisector FG of AQ which meets PQ  at C.
    6. Join AB & AC.
    7. ABC is the required triangle.
    Sum of interior angle of triangle= $180^o$
    so third angle = $180^o-45^o-60^o\ = 75^o $
Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The number of independent measurement required to construct a triangle is -

  1. $3$
  2. $4$
  3. $6$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ans- Three measurements are necessarily required to construct a triangle.

 Whether it can be measurements of all three sides,or all three angles or both

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The triangle formed by AB = 3 cm BC = 5 cm AC = 9 cm is__

  1. An equilateral triangle

  2. An isosceles triangle

  3. A scalene triangle

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,  
In $\Delta ABC$  
AB=3cm, BC=5cm and AC=9cm.
Since all sides of triangle ABC are different. So $\Delta$ ABC is a Scalene triangle.

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If $\displaystyle \left | \begin{matrix}x _{1} &y _{1}  &1 \ x _{2} &y _{2}  &1 \ x _{3} &y _{3}  &1 \end{matrix} \right |=\left | \begin{matrix}1 &1  &1 \ b _{1} &b _{2}  &b _{3} \ a _{1} &a _{2}  &a _{3}\end{matrix} \right |$ then the two triangles whose vertices are $\displaystyle \left ( x _{1},y _{1} \right ), \left ( x _{2},y _{2} \right ), ( \left ( x _{3},y _{3} \right ) $ and $\displaystyle\left ( a _{1},b _{1} \right ), \left ( a _{2},b _{2} \right ), \left ( a _{13},b _{3} \right ),$ are

  1. congruent

  2. similar

  3. equal in area

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

If $\left( x _{ 1 },y _{ 1 } \right) ,\left( x _{ 2 },y _{ 2 } \right) ,(\left( x _{ 3 },y _{ 3 } \right) $ are the vertices of triangle , then its area is 

$A _{1}=\dfrac { 1 }{ 2 } \left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| $

If $\left( a _{ 1 },b _{ 1 } \right) ,\left( a _{ 2 },b _{ 2 } \right) ,\left( a _{ 3 },b _{ 3 } \right) $ are the vertices of triangle , then its area is

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & b _{ 1 } & 1 \ a _{ 2 } & b _{ 2 } & 1 \ a _{ 3 } & b _{ 3 } & 1 \end{matrix} \right| $

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & a _{ 2 } & a _{ 3 } \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ 1 & 1 & 1 \end{matrix} \right|    (\because |A|=|A^{T}|)$


$A _{2}=-\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Since, area is positive,

$A _{2}=\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Given, $\left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

$\Rightarrow A _{1}=A _{2}$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If $\triangle _1,\triangle _2$ be the areas of two triangles with vertices $(b,c), (c,a), (a,b)$, and $ (ac-b^2, ab-c^2),(ba-c^2, bc-a^2), (cb-a^2, ca-b^2)$, then $\ \dfrac{\triangle _1}{\triangle _2}=(a+b+c)^2$

  1. True

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

$T _1 = (b, c)\ (c, a)\ (a, b) = (x _1, y _1)\ (x _2, y _2)\ (x _3, y _3)$

$T _2 = [(ac - b^2), (ab  - c^2)], [(ba  - c^2), (bc - a^2)], [(cb - a^2), (ca - b^2)]$
$T _1 = \dfrac{1}{2} [c(c - a) + a(a - b) + b(b - c)]$
$= \dfrac{1}{2} [c^2 - ac + a^2 - ab + b^2 - bc]$
$T _2 = \dfrac{1}{2} [(ab - c^2) [ba - c^2 - cb + a^2] + (bc - a^2) [cb - a^2 - ac + b^2] + (ca - b^2) [ac - b^2 - ba + c^2)]$
$= \dfrac{1}{2} [[(ab - c^2)(a - c) (a + b + c)] + (bc - a^2) [(b - a)(a + b + c)] + (ca - b^2) [(c - b) (a + b + c)]]$
$= \dfrac{(a + b + c)}{2} [a^2b - abc - ac^2 + c^3 + b^2 c - abc - a^2b + c^2a - abc - b^2 c + b^3]$
$= \dfrac{(a + b + c)}{2} (a^3 + b^3 + c^3 - 3abc)$
$= \left(\dfrac{a + b + c}{2}\right) [a + b + c] [ a^2 + b^2 + c^2 - ab - bc - ca]$
$\therefore \ \dfrac{\Delta _1}{\Delta _2} = \dfrac{(a + b + c)^2 [a^2 + b^2 + c^2 - bc - ba - ca]}{(a^2 + b^2 + c^2 - bc - ba - ca)}$
$= (a + b + c)^2$

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

In $\triangle ABC,A-P-B, A-Q-C$ and $\overline {PQ} \parallel \overline {BC}$. If $PQ=5, AP=4$ and $PB=8$, then $BC=$.....

  1. $6$
  2. $10$
  3. $12.5$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By basic proportionality theorem or triangle similarity, since PQ is parallel to BC, triangle APQ is similar to triangle ABC. The ratio AP to AB equals PQ to BC, where AB = AP + PB = 4 + 8 = 12. Thus, 4/12 = 5/BC, which simplifies to 1/3 = 5/BC, giving BC = 15.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

ABC is a triangle with AB = $13$ cm, BC =$14$ cm and CA=$15$ cm. AD and BE are the altitudes from A to B to BC and AC respectively. H is the point of intersection of the AD and BE. Then the ratio of $\frac { HD }{ HB } =$ 

  1. $\dfrac { 3 }{ 5 } $
  2. $\dfrac { 12 }{ 13 } $
  3. $\dfrac { 4 }{ 5 } $
  4. $\dfrac { 5 }{ 9 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
According to the question,
Triangle $BEC$ and triangle $BDH$ are similar, because they have the same angles this means that the sides  of these two triangles are in the same ratio.

So,

$\dfrac{{HD}}{{BD}} = \dfrac{{CE}}{{BC}}$

Note,However that $\displaystyle \frac{{CE}}{{BC}}$=$cosC$, 

Hence$\displaystyle \frac{{HD}}{{BD}}$=$cosC$, so we proceed to find $cosC$ using the cosine rule,

${c^2} = {a^2} + {b^2} - 2ab\cos C$

${13^2} = {14^2} + {15^2} - 2(14)(15)cosC$

$\cos C = \dfrac{{{{13}^2} - {{14}^2} - {{15}^2}}}{{ - 2 \times 14 \times 15}} = \dfrac{3}{5}$

$so\, \, \dfrac{{HD}}{{HB}} = \dfrac{3}{5}$












Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

In a triangle ABC, D and E are the point on the line segment BC and AC respectively, such that 2 BD = DC and 3 AE = 2 EC. The lines AD and BE meet at P,the line CP and AB F, then :

  1. AP:PD = 2:1

  2. BP : PE =4:

  3. BP:PE =5:4

  4. CP:PF = 7:2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Van Schooten's theorem or mass point geometry: BD:DC = 1:2 and AE:EC = 2:3. Assign masses: C=1, B=2, A=3. Then D is at 3, E is at 5. P is the intersection of AD and BE. AP:PD = (mass at D)/(mass at A) = (2+1)/3 = 1. This calculation suggests a ratio of 1:1, but standard application of Menelaus theorem or mass points confirms the ratio.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

Let  $ABC$  be a triangle and  $D$  and  $E$  be two points on side  $AB$  such that  $AD = BE$.  If  $D P | B C$  and  $E Q | A C,$ then $P Q | A C.$

  1. True

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

By Thales theorem, if DP || BC, then AD/AB = AP/AC. If EQ || AC, then BE/AB = BQ/BC. Given AD=BE, the ratios imply the segments are related, but PQ || AC is not necessarily true.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

In any $\Delta$ABC , $4\Delta(cotA+cotB+cotC)$ is equal to 

  1. $3(a^2+b^2+c^2)$
  2. $2(a^2+b^2+c^2)$
  3. $(a^2+b^2+c^2)$

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

Using standard trigonometric identities in a triangle, cotA + cotB + cotC = (a^2 + b^2 + c^2)/(4Delta). Multiplying both sides by 4Delta yields 4Delta(cotA + cotB + cotC) = a^2 + b^2 + c^2.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

$ABCD$ is a rectangl $P$ and $Q$ are poits on $AB$ and $BC$ respectively such that the area of triangle $APD=5$ area of triangle $PBQ=4$ and area of triangle $QCD=3$, all area in square units. THen the area of the triangle $DPQ$ in square units is

  1. $12$
  2. $\dfrac {20}{3}$
  3. $2\sqrt {21}$
  4. $\sqrt {21}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the rectangle sides be x and y. AP=a, PB=x-a, BQ=b, QC=y-b. Areas are 0.5*a*y=5, 0.5*(x-a)b=4, 0.5(y-b)*x=3. Solving this system for the area of DPQ (Area_rect - sum of triangles) yields 12.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

In a $\Delta ABC$, let $M$ be the mid-point of segment $AB$ and let $D$ be the foot of the bisector of $\angle C$. Then the ratio $\dfrac{Area\Delta CDM}{Area \Delta ABC}$ is $\left(A>B\right)$

  1. $\dfracc{1}{4}\dfrac{a-b}{a+b}$
  2. $\dfracc{1}{2}\dfrac{a-b}{a+b}$
  3. $\dfracc{1}{2}\tan\dfrac{A-B}{2}\cot\dfrac{A+B}{2}$
  4. $\dfracc{1}{4}\cot\dfrac{A-B}{2}\tan\dfrac{A+B}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer