Tag: maths

Questions Related to maths

In a $\Delta$ $ABC, AC = 6, BC = 2, AB = 4$, find the value of $AD$. (Use Apollonius theorem).

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

  1. 5

  2. 4

  3. 3

  4. 2

Show answer
Correct Option: A
Explanation:

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$4^{2}+6^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$16+36= 2[AD^{2} + 1]$
$2AD^{2}=52 -2$
$2AD^{2}= 50$
$AD^{2} = \dfrac{50}{2}$
$AD^{2} = 25$
$AD = 5$

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

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

  1. 3

  2. 5

  3. 7

  4. 9

Show answer
Correct Option: C
Explanation:

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

In a $\Delta$ $ABC, AC = 4, BC = 2, AB = 6$, find the value of $AD$. (Use Apollonius theorem).

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

  1. 2

  2. 3

  3. 4

  4. 5

Show answer
Correct Option: D
Explanation:

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$6^{2}+4^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$36+16= 2[AD^{2} + 1]$
$2AD^{2}=52 -2$
$2AD^{2}= 50$
$AD^{2} = \dfrac{50}{2}$
$AD^{2} = 25$
$AD = 5$

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.

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

  1. Pythagoras

  2. Apollinius

  3. Stewart

  4. Ceva's

Show answer
Correct Option: B
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.

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

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

  1. 5

  2. 6

  3. 7

  4. 8

Show answer
Correct Option: C
Explanation:

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

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

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

  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. $a^{2}+b^{2}=2m^{2}+\dfrac{b}{2}^{2}$

Show answer
Correct Option: B
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.

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:

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

  1. ${2\sqrt{26}}$

  2. ${2\sqrt{31}}$

  3. ${\sqrt{31}}$

  4. ${\sqrt{26}}$

Show answer
Correct Option: B
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,
$AB^2+AC^2=2(AM^2+BM^2)$
$\implies$ $4^2+8^2=2(3^2+BM^2)$
$\implies$ $16+64=2(9+BM^2)$
$\implies$ $BM^2=31$
$\implies$ $BM=\sqrt{31}$.
$\because$ $BC=2BM$
$\therefore$ $BC=2\sqrt{31}$.

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

Find $PR$ and $QR$.

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

  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

Show answer
Correct Option: C
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.

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

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

  1. $AD+BD$

  2. $AD-BD$

  3. $2(AD^{2}+BD^{2})$

  4. $2(AD^{2}-BD^{2})$

Show answer
Correct Option: C
Explanation:

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

Diagonals of trapezium $ABCD$ with $AB\parallel DC$ intersect each other at the point $O$. If $AB=2CD$, find the ratio of the areas of triangles $AOB$ and $COD$.

maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

  1. $4:1$

  2. $4:3$

  3. $3:1$

  4. $4:7$

Show answer
Correct Option: A
Explanation:

Given: $AB \parallel CD$ and $AB = 2 CD$
In $\triangle OAB$ and $\triangle OCD$
$\angle AOB = \angle COD$ (Vertically opposite angles)
$\angle ODC = \angle OBA$ (Alternate angles)
$\angle OCD = \angle OAB$ (Alternate angles)
thus, $\triangle OAB \cong \triangle OCD$ (AAA rule)
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = \dfrac{AB^2}{CD^2}$ (Similar triangle Property)
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = \dfrac{(2 CD)^2}{CD^2}$
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = 4 : 1$