Tag: maths

Questions Related to maths

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

Mark the correct alternative of the following.
In a right triangle, one of the acute angles is four times the other. Its measure is?

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

  1. $68^o$

  2. $84^o$

  3. $80^o$

  4. $72^o$

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Correct Option: D
Explanation:

In the right-angled triangle the sum of the other two angles is $90^o$.

Let one acute angle is $x$, then the other angle is $4x$. [ Given]
Then we get,
$4x+x=90^o$
or, $5x=90^o$
or, $x=18^o$.
So the measure of that angle is $4\times 18^o=72^o$.

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

Show answer
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.

If the sides of a triangle are in the ratio $1\, :\, \sqrt2\, :\, 1$, then the triangle is:

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

  1. an equilateral triangle

  2. an isosceles triangle

  3. a right angled triangle

  4. a right angled isosceles triangle

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Correct Option: D
Explanation:

Given ratio of sides of the triangle, $1 : \sqrt{2} : 1$
Let the triangle be $ABC$ and sides be
$AB = x$
$BC = x $
$AC = \sqrt{2}x$

Clearly, $AC^2 = BC^2 + AB^2$
Hence, by converse of Pythagoras theorem, $ABC$ is a right-angled isosceles triangle.

In $\triangle ABC$, AP is the median. If $AP=7$ and $AB^2+AC^2=260$, then find BC.

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

  1. $14$ cm

  2. $18$ cm

  3. $15$ cm

  4. $12$ cm

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Correct Option: B
Explanation:

By Apollonius Theorem (states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side".) , we have


$BC^2=AB^2+AC^2+2AP^2$

$BC^2=260+2(7)^2$

$BC^2=260+98=358$

$BC=18.92$

Find the length of median. If the sides of triangle are:
$a = 5, b = 6, c = 8$. and $m = 3, n = 2$.

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

  1. $\sqrt{\dfrac{206}{5}}$

  2. $\sqrt{206}$

  3. $\dfrac{\sqrt{206}}{5}$

  4. None

Show answer
Correct Option: A
Explanation:
We have from Appollonius theorem,

$a(mn+p^2)=b^2m+c^2n$

$5(3\times2+p^2)=6^2\times3+8^2\times2$

$5(6+p^2)=36\times3+64\times2$

$30+5p^2=108+128$

$5p^2=236−30 \ \implies 5p^2=206$

$p^2=\dfrac{206}{5}$

$p=\sqrt{\dfrac{206}{5}}$

Option A.

In a $\Delta$ $ABC, AD = 3, BC = 2, AB = 1$, find the value of $AC$. (Use Apollonius theorem).

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

  1. $2.35$

  2. $3.42$

  3. $4.35$

  4. $5.61$

Show answer
Correct Option: C
Explanation:

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2[AD^{2}+\dfrac {BC}{2}^{2}]$
$1^{2}+AC^{2}= 2[3^{2}+\dfrac{2}{2}^{2}]$
$1+AC^{2}= 2[9 + 1]$
$AC^{2}=20 -1$
$AC^{2}= 19$
$AC = \sqrt{19}$
$AC = 4.35$