Tag: angle theorems for a right angled triangle

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

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side is called ______ theorem.

  1. Pythagoras

  2. Apollinius

  3. Stewart

  4. Ceva's
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side is called Apollinius theorem.

Option $B$ is correct.

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

In a $\Delta$ $ABC, AC = 6, BC = 2, AB = 8$, find the value of $AD$. (Use Apollonius theorem).

  1. 5

  2. 6

  3. 7

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left[AD^{2}+\dfrac{BC}{2}^{2}\right]$
$8^{2}+6^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$64+36= 2[AD^{2} + 1]$
$2AD^{2}=100 -2$
$2AD^{2}= 98$
$AD^{2} = \dfrac{98}{2}$
$AD^{2} = 49$
$AD = 7$

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

Which one of the following formula is used to find apollinius theorem for isosceles triangle?

  1. $a^{2}+b^{2}=2m^{2}+\dfrac{c}{2}^{2}$
  2. $b^{2}=m^{2}+\dfrac{c}{4}^{2}$
  3. $b^{2}+b^{2}=2m^{2}+\dfrac{c}{2}^{2}$
  4. $a^{2}+b^{2}=2m^{2}+\dfrac{b}{2}^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side.
Apollinius theorem formula,
$a^{2}+b^{2}= 2[m^{2}+\dfrac{c}{2}^{2}]$
When the given triangle is isosceles, than $b = a$.
So, $b^{2}=m^{2}+\dfrac{c}{4}^{2}$ is the formula used for isosceles triangle.

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

In a $\triangle ABC$, $AB= 4$ cm and $AC = 8$ cm. If M is the midpoint of BC and $AM = 3$ cm, then the length of $BC$ in cm is:

  1. ${2\sqrt{26}}$
  2. ${2\sqrt{31}}$
  3. ${\sqrt{31}}$
  4. ${\sqrt{26}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given : In $\triangle ABC$, $AB=4$ cm and $AC=8$ cm.

M is the midpoint of BC and $AM=3$ cm
Using Apollonius theorem,
$AB^2+AC^2=2(AM^2+BM^2)$
$\implies$ $4^2+8^2=2(3^2+BM^2)$
$\implies$ $16+64=2(9+BM^2)$
$\implies$ $BM^2=31$
$\implies$ $BM=\sqrt{31}$.
$\because$ $BC=2BM$
$\therefore$ $BC=2\sqrt{31}$.

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

In $\triangle PQR$, $\angle P=30^o$, $\angle Q=60^0$, $\angle R= 90^o$ and $PQ=10 $ units. 

Find $PR$ and $QR$.

  1. $PR =$ $10$ units, $QR =$ $5\sqrt 3$ units
  2. $PR =$ $5$ units, $QR =$ $5\sqrt 3$ units
  3. $PR =$ $5\sqrt 3$ units, $QR =$ $5$ units
  4. $PR =$ $5$ units, $QR =$ $10\sqrt 3$ units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\triangle PQR$ is a $30^o-60^o-90^o$ triangle        ....given

Since, $\angle R = 90^o$, side $PQ$ is hypotenuse.

$\Rightarrow $ By $30^o-60^o-90^o$ theorem,

$\Rightarrow PR = \dfrac {\sqrt 3}{2} \times PQ$ and $QR = \dfrac 12 \times PQ$

$\Rightarrow  PR = \dfrac {\sqrt 3}{2} \times 10$ and $QR = \dfrac 12 \times 10$ 

$\Rightarrow  PR = 5 \sqrt 3$ units and $QR = 5$ units
So, option C is correct.

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

According to Apolloneous Theorem, if $\overline AD$ is a median of $\triangle ABC$, then $AB^{2}+AC^{2}=$

  1. $AD+BD$
  2. $AD-BD$
  3. $2(AD^{2}+BD^{2})$
  4. $2(AD^{2}-BD^{2})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Then,\, \, A{ B^{ 2 } }+A{ C^{ 2 } }=? &  \ AD\, \, is\, \, median &  \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2\left| { \frac { { B{ C^{ 2 } } } }{ 4 }  } \right| +2A{ D^{ 2 } } & \left[ \begin{array}{l} BC=2BD \ or\, \, BC=DC \end{array} \right]  \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2{ \left| { \frac { { B{ C^{  } } } }{ 2 }  } \right| ^{ 2 } }+2A{ D^{ 2 } } &  \ A{ B^{ 2 } }+A{ C^{ 2 } }=2B{ D^{ 2 } }+2A{ D^{ 2 } }=2\left( { A{ D^{ 2 } }+B{ D^{ 2 } } } \right)  &  \end{array}$