Tag: similarity and right angled triangle

Questions Related to similarity and right angled triangle

In $\Delta ABC, \, m \angle B = 90$ and $\overline{BM}$ is an altitude. If AB = 2 AM, then AC = ......

  1. 2 AM

  2. 4 AM

  3. 6 AM

  4. 8 AM


Correct Option: A

P, Q, R are the points of intersection of a line 1 with sides BC, CA, AB of a $\Delta$ ABC 
respectively, then $\dfrac{BP}{PC} \dfrac{CQ}{QA} \dfrac{AR}{RB}$

  1. 1

  2. 2

  3. -1

  4. -2


Correct Option: A

Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

  1. 7 cm, 24 cm, 25 cmj

  2. 3 cm, 8 cm, 6 cm

  3. 50 cm, 80 cm, 100 cm

  4. 13 cm, 12 cm, 5 cm


Correct Option: A

The hypotenuse and the semi-perimeter of right triangle are 20 cm and 24 cm, respectively. The other two sides of the triangle are :

  1. 16 cm, 15 cm

  2. 14 cm, 16 cm

  3. 20 cm, 16 cm

  4. None of these


Correct Option: A

In $\Delta PQR,\angle P$ is right angle and $\bar{PM}$ is an altitude. $PQ=\sqrt{20}$ and QM=4 then RM=____.

  1. 8

  2. 5

  3. 9

  4. 1


Correct Option: A

The lengths of the medians through acute angles of a right-angled triangle are 3 and 4. Find the area of the triangle:

  1. $\displaystyle \frac{4}{3}\sqrt{11}$

  2. $\displaystyle \frac{2}{3}\sqrt{11}$

  3. $\displaystyle \frac{1}{3}\sqrt{11}$

  4. None of these


Correct Option: A
Explanation:

Given, $AD=3,CE=4$
Using Appaloneaus theorem for median $AD$
We have $\displaystyle{ c }^{ 2 }+{ b }^{ 2 }=2\left( \frac { { a }^{ 2 } }{ 4 } +9 \right) $   ...(1)
Using Appaloneaus theorem for median $CE$
We have $\displaystyle{ b }^{ 2 }+{ a }^{ 2 }=2\left( \frac { { c }^{ 2 } }{ 4 } +10 \right) $   ...(2)
Also, ${ a }^{ 2 }+{ c }^{ 2 }={ b }^{ 2 }$
Adding (1) and (2)
$\displaystyle 3{ b }^{ 2 }=2\left( \frac { { b }^{ 2 } }{ 4 } +25 \right) \Rightarrow { b }^{ 2 }=20$
Solving (1) and (2) we get,
$\displaystyle c=\frac { 4 }{ \sqrt { 3 }  }$ and $\displaystyle a=2\frac { 4 }{ \sqrt { 3 }  } $
Hence, area of triangle
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.

The length of the hypotenuse of an isosceles right triangle whose one side is $4\surd {2}\ cm$  is___________$cm$

  1. $12$

  2. $8$

  3. $8\surd {2}$

  4. $12\surd {2}$


Correct Option: A

Let $ABC$ be a fixed triangle and $P$ be variable point in the plane of a triangle $ABC$. Suppose $a, b, c$ are lengths of sides $BC,  CA,AB$ opposite to angles $A, B, C $ respectively. If $a(PA)^{2} + b(PB)^{2} + c(PC)^{2}$ is minimum, then the point $P$ with respect to $\triangle{ABC}$ is

  1. Centroid

  2. Circumcenter

  3. Orthocenter

  4. Incenter


Correct Option: A

If $AD,BE$ and $CF$ are the medians of a $\Delta ABC,$ then evaluate  $\displaystyle \left ( AD^{2}+BE^{2}+CF^{2} \right ):\left ( BC^{2}+CA^{2}+AB^{2} \right )=$

  1. $3:4$

  2. $4:3$

  3. $5:3$

  4. $4:1$


Correct Option: A
Explanation:

Given, $AD,BE$ and $CF$ are the medians of a $\Delta ABC$.
$\Rightarrow AB^2+AC^2=2(AD^2+BD^2)$
$\Rightarrow AB^2+AC^2=2AD^2+\displaystyle\frac{BC^2}{2}$
$\Rightarrow 2AD^2=AB^2+AC^2-\displaystyle\frac{BC^2}{2}$ -----(1)
Similarly,
$2BE^2=BC^2+BA^2-\displaystyle\frac{AC^2}{2}$ -----(2)
$2CF^2=CA^2+CB^2-\displaystyle\frac{AB^2}{2}$ -----(3)
Adding equation 1,2 and 3, we get
$2(AD^2+BE^2+CF^2)=\displaystyle\frac{3}{2}(AB^2+BC^2+CA^2)$
$\therefore (AD^2+BE^2+CF^2):(AB^2+BC^2+CA^2)=3:4$

The distances of the circumcentre of the acute-angled $  \Delta \mathrm{ABC}  $ from the sides $  \mathrm{BC},  $ CA and AB are in the ratio

  1. asinA: bsinB:csinC

  2. $

    \cos A : \cos B : \cos C

    $

  3. $

    \operatorname{acot} A : \operatorname{bcot} B : \operatorname{ccot} C

    $

  4. none of these


Correct Option: A