Tag: geometry

Questions Related to geometry

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

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  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

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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
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.

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

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. Centroid

  2. Circumcenter

  3. Orthocenter

  4. Incenter

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Correct Option: A

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

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $3:4$

  2. $4:3$

  3. $5:3$

  4. $4:1$

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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

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  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

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Correct Option: A

Let ABC be a triangle having its centroid at G. If S is any point in the plane of the triangle, then $S\vec { A } +S\vec { B } +S\vec { C } =$

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $S\vec { G } $

  2. $2S\vec { G } $

  3. $3S\vec { G } $

  4. $\vec { 0 } $

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Correct Option: A

Find the perimeter of an isosceles right triangle with each of its congruent as 7cm.

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $7\sqrt 2$ cm

  2. $14$ cm

  3. $(2+ \sqrt 2)$ cm

  4. $7(2+ \sqrt 2)$ cm

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Correct Option: D
Explanation:

Let the other side of triangle is x cm

Then in isosceles right angle triangle two congruent sides are 7 cm
$x^{2}=(7)^{2}+(7)^{2}$
$\Rightarrow x^{2}=49+49$
$\Rightarrow x^{2}=198$
$\Rightarrow x=7\sqrt{2}$
Then perimeter of right angle isosceles triangle =$7+7+7\sqrt{2}=14+7\sqrt{2}=7(2+\sqrt{2})$ 

So, option D is correct.