Mathematics · Quantitative Aptitude

Triangle Properties

243 Questions

Triangle properties encompass the rules governing the sides, angles, and area of different types of triangles. Key areas include the Pythagorean theorem, similar triangles, and centroid calculations. These concepts form a foundational part of geometry in various competitive examinations.

Similar trianglesArea and perimeterPythagorean theoremTriangle inequalityEquilateral properties

Triangle Properties Questions

Multiple choice maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

The base of a right pyramid is an equilateral triangle of perimeter $8$ dm and the height of the pyramid is $30$$\sqrt{3}$ cm. The volume of the pyramid is

  1. $1600$ cm$^{3}$
  2. $16000$ cm$^3$
  3. $\displaystyle \frac{16000}{3} cm^3$
  4. $\displaystyle \frac{5}{4} cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Base of pyramid is an equilateral triangle of parameter
$8dm=80cm$
Let the side of the equilateral triangle be $'a'cm$
$\therefore $Parameter of equilateral triangle$=3a$
$\Rightarrow 3a=80\Rightarrow a=\cfrac { 80 }{ 3 } cm$
Height of pyramid$=30\sqrt { 3 } cm$
Volume of pyramid=Area of base $\times $ height
$=\cfrac { \sqrt { 3 }  }{ 4 } { a }^{ 2 }\times 30\sqrt { 3 } $

$=\cfrac { \sqrt { 3 }  }{ 4 } \times \cfrac { 80 }{ 3 } \times \cfrac { 80 }{ 3 } \times 30\sqrt { 3 } $

$=\cfrac { 3 }{ 4 } \times \cfrac { 80 }{ 3 } \times 80 \times 10 $

$=\cfrac { 1 }{ 4 } \times 80 \times 80 \times 10 $

$=20 \times 80 \times 10 $

$=16000cm^3$
Multiple choice maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

The base of a right pyramid is an equilateral triangle of perimeter 8 cm and the height of the pyramid is $30\sqrt 3$ cm. The volume of the pyramid is

  1. $160 cm^3$
  2. $1600 cm^3$
  3. $\dfrac {160}{3} cm^3$
  4. $\dfrac {5}{4} cm^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of right pyramid $=$ $\dfrac { 1 }{ 3 } \times area\quad of\quad base\times height\quad of\quad pyramid$

Now, base is equilateral $\triangle $, therefore,
area $=\dfrac { \sqrt { 3 }  }{ 4 } \times { \left( side \right)  }^{ 2 }$
Perimeter of triangle $=8cm$
$\therefore \quad \quad 3a=8\Rightarrow a=\dfrac { 8 }{ 3 } cm$
$\therefore \quad \quad area=\dfrac { \sqrt { 3 }  }{ 4 } \times \dfrac { 8 }{ 3 } \times \dfrac { 8 }{ 3 } =\dfrac { 16\sqrt { 3 }  }{ 9 } { cm }^{ 2 }$
Now,  Volume $=\dfrac { 1 }{ 3 } \times \dfrac { 16\sqrt { 3 }  }{ 9 } \times 30\sqrt { 3 } $
                        $=\dfrac { 160\times 3 }{ 9 } =\dfrac { 160 }{ 3 } { cm }^{ 3 }$

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

If every side of a triangle is doubled, then the area of the new triangle is 'K' times the area of the old one. The value of K is

  1. 2

  2. 3
  3. $\sqrt 2$

  4. 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the area of the triangle be $x$.

We know that the area of the triangle
$=\dfrac{1}{2}\times Height \times Base$
$x=\dfrac{1}{2}\times Height \times Base$              $........ (1)$

According to the question,
$Kx=\dfrac{1}{2}\times 2 \times Height \times 2 \times Base$
$Kx=4\times x$
$K=4$

Hence, this is the answer.
Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

Each side of $\triangle ABC$ is 12 units. D is the foot of the perpendicular dropped from A on BC and E is the mid point of AD. The length of BE in the same units is: 

  1. $\sqrt{18}$
  2. $\sqrt{28}$

  3. 6

  4. 7.93
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In an equilateral triangle with side 12, the altitude AD = 12 * sqrt(3)/2 = 6 * sqrt(3) approx 10.39. E is the midpoint of AD, so AE = 3 * sqrt(3). In right triangle ABE, BE^2 = AE^2 + AB^2 is not correct; rather, use triangle BDE where BD=6 and DE=3*sqrt(3). BE^2 = 6^2 + (3*sqrt(3))^2 = 36 + 27 = 63. BE = sqrt(63) approx 7.937.

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

If A, B and C are the midpoint of the sides PQ, QR and PR of $\triangle $PQR respectively, then the area of $\triangle $ABC equals if area of $\triangle PQR$ is $4$ units

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

The triangle formed by joining the midpoints of the sides of a triangle has an area equal to 1/4 of the area of the original triangle. (1/4) * 4 = 1.

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

The minimum number of dimensions needed to construct an equilateral triangle is:

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

As we know that all angles in an equilateral triangle measures $60^o$. Hence we need only the length of the side to construct an equilateral triangle.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

In a $\triangle DEF$; $A,B$ and $C$ are the mid-points of $EF,FD$ and $DE$ respectively. If the area of $\triangle DEF$ is $14.4{ cm }^{ 2 }$, then find the area of $\triangle {ABC}$.

  1. $1.75$cm
  2. $2.54$cm
  3. $3.2$cm
  4. $3.6$cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Fact: Using mid-point theorem $\dfrac{\Delta{DEF}}{\Delta{ABC}}=4$


Here $\Delta{DEF}=14.4$ cm$^2$ is given 
Hence are of triangle $ABC $ is given by $ \dfrac{14.4}{4}=3.6$ cm$^2$ 

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles
Which of the following pair of sides can form triangle?
  1. $5\ cm, 6\ cm , 4\ cm$
  2. $13\ cm , 12\ cm , 24\ cm$
  3. $2\ cm , 7\ cm , 9\ cm$
  4. $5.6\ cm , 6.5\ cm , 12\ cm$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

As in a triangle, the sum of any two sides should be strictly greater than the third side.

 
In $(A), 5+6 > 4$   $\therefore$ This pair will form triangle.


In $(B), 13+12 > 24$ This pair will form triangle.

In $(C), 2+7\ \ngtr 9$  This pair will not form triangle.

In $(D), 5.6+6.5 > 12$  This pair will form triangle.

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

Length of two sides of a $\triangle ABC$ is $AB=6\ cm$ and $BC=7\ cm$. Then, which of the following can represent the third side of the triangle ? Also, construct the triangle formed by these three sides.

  1. $8\ cm$
  2. $13\ cm$
  3. $14\ cm$
  4. $15\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A triangle can be formed if sum of any two sides is greater then the third side.

Here $AB=6$ cm and $BC=7$ cm
Now $AB+BC>AC$
$6+7>AC$
$AC<13$ cm
So only option $A$ is possibel.
Steps of construction

Step 1. Draw a line segment $AB=6\ \ cm$
Step 2. Assuming $A$ a centre draw an arc of radius $8 \ \ cm$
Step 3. Now assuming $B$ as centre draw an arc of $7 \ \ cm$ intersecting the previous arc at $C$.
Step 4. Now join $A$ to $C$ and $B$ to $C$.

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

The perimeter of a triangle is $45\ cm$. Length of the second side is twice the length of first side. The third side is $5$ more than the first side. Find the length of each sides and construct the triangle made by these three sides.

  1. $11,19,15$
  2. $10,20,15$
  3. $10,16,19$
  4. $13,15,17$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the length of first side $=x$

Length of second side $=2x$
Length of third side $=x+5$
Perimeter $=45cm$
$x+2x+x+5=45\ 4x=45-5\ 4x=40\ x=10$
So the sides are
$x=10$
$2x=2\times 10=20\ x+5=10+5=15$
Option $B$ is correct.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

The perimeters of two similar triangles ABC and LMN are 60 cm and 48 cm respectively If LM=8 cm, the length of AB is

  1. $10\ cm$
  2. $8\ cm$
  3. $6\ cm$
  4. $4\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If two triangles are similar then the ratio of their perimeter is equal to the ratio of their corresponding sides.

$\therefore \dfrac{perimeter\ ABC}{perimeter\ LMN}=\dfrac{AB}{LM}$
$\Rightarrow \dfrac{60}{48}=\dfrac{AB}{8}$
$\Rightarrow AB=\dfrac{60\times 8}{48}=10 cm$

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

The area of two similar triangles ABC and PQR are 25 $\displaystyle cm^{2}$ and $\displaystyle 49cm^{2}$ If QR=9.8 cm then BC is

  1. 9.0 cm

  2. 7 cm

  3. 49 cm

  4. 41 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

$\therefore \dfrac{arc(\triangle ABC)}{arc(\triangle PQR)}=\dfrac{BC^2}{QR^2}$
$\Rightarrow \dfrac{25}{49}=\dfrac{BC^2}{(9.8)^2}$
$\Rightarrow \dfrac{5}{7}=\dfrac{BC}{9.8}$
$\Rightarrow BC=\dfrac{9.8\times 5}{7}=7.0 cm^2$

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

The perimeters of two similar triangles ABC and PQR are 60 cm and 48 cm respectively If PQ=8 cm length of AB is

  1. $10\ cm$
  2. $8\ cm$
  3. $6\ cm$
  4. $4\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P _1=AB+BC+AC=60  cm$

$P _2=PQ+QR+RP=48  cm$
PQ=8 cm
$\dfrac{P _1}{P _2}=\dfrac{AB}{PQ}$
$\Rightarrow \dfrac{60}{48}=\dfrac{AB}{8}$
$\Rightarrow AB=\dfrac{60\times 8}{48}=10 cm$

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

Triangle $ABC$ is such that $AB=3cm, BC=2cm$ and $CA=2.5cm$. Triangle $DEF$ is similar to $\triangle ABC$. If $EF=4cm$, then the perimeter of $\triangle DEF$ is:

  1. $7.5cm$
  2. $15cm$
  3. $22.5cm$
  4. $30cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: $AB = 3 cm$, $BC = 2 cm$ and $CA = 2.5 cm$ and $EF = 4 cm$ 

Also, $\triangle ABC \sim \triangle DEF$

Thus, $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

$\dfrac{3}{DE} = \dfrac{2}{4} = \dfrac{2.5}{DF}$

Hence, $DE = 6 cm$ and $DF = 5 cm$

Perimeter of $\Delta$ DEF = $DE + EF + EF$

Perimeter of $\Delta$ DEF = $4 + 5 + 6$

Perimeter of $\Delta$ DEF = $15$ cm

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

$\triangle ABC\sim \triangle DEF$. IF $BC=4cm$, $EF=5cm$ and area $(\triangle ABC)=32{cm}^{2}$, determine the area of $\triangle DEF$.

  1. $50{m}^{2}$
  2. $40{m}^{2}$
  3. $40{cm}^{2}$

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

Ar. ($\triangle ABC$) = $32 cm^2$
$BC = 4 cm$
$EF = 5 cm$
For similar triangles the ratio of areas is equal to the ratio of square of its sides.
Thus, $\frac{A(\triangle ABC)}{A(\triangle DEF)} = \frac{BC^2}{EF^2}$
$\frac{32}{A(\triangle DEF)} = \frac{4^2}{5^2}$
$A(\triangle DEF) = \frac{32 \times 25}{16}$
$A(\triangle DEF) = 50 cm^2$