Tag: maths

Questions Related to maths

$ABCD$ is a rectangl $P$ and $Q$ are poits on $AB$ and $BC$ respectively such that the area of triangle $APD=5$ area of triangle $PBQ=4$ and area of triangle $QCD=3$, all area in square units. THen the area of the triangle $DPQ$ in square units is

  1. $12$

  2. $\dfrac {20}{3}$

  3. $2\sqrt {21}$

  4. $\sqrt {21}$


Correct Option: A

If G is the centroid of $\Delta ABC$ and if area of $\Delta AGB$ is 5 sq. nits then the area of $\Delta ABC$ is 

  1. 20 sq. unit

  2. 10 sq.unit

  3. 15 sq. unit

  4. 25 sq. unit


Correct Option: A

The areas of two similar triangle are $18\ cm^{2}$ and $32\ cm^{2}$ respectively. What is the ratio of their corresponding sides?

  1. $3:4$

  2. $4:3$

  3. $9:16$

  4. $16:9$


Correct Option: A

In a triangle PQR, S and T are points on QR and PR respectively, such that QS = 3SR and PT = 4TR Let M be the point of intersection of PS and QT.  FInd the ration QM : MT 

  1. 15: 8

  2. 16 : 5

  3. 15 : 4

  4. 5 : n2


Correct Option: A

In a triangle PQR, S and T are points on QR and PR respectively, such that QS = 3SR and PT = 4TR Let M be the point of intersection of PS and QT.  FInd the ration QM : MT 

  1. 15 : 8

  2. 16 : 5

  3. 15 : 4

  4. 5 :2


Correct Option: A

If the areas of two similar triangles are equal,  then they are congruent.

  1. True

  2. False


Correct Option: A

In a $\Delta ABC$, let $M$ be the mid-point of segment $AB$ and let $D$ be the foot of the bisector of $\angle C$. Then the ratio $\dfrac{Area\Delta CDM}{Area \Delta ABC}$ is $\left(A>B\right)$

  1. $\dfracc{1}{4}\dfrac{a-b}{a+b}$

  2. $\dfracc{1}{2}\dfrac{a-b}{a+b}$

  3. $\dfracc{1}{2}\tan\dfrac{A-B}{2}\cot\dfrac{A+B}{2}$

  4. $\dfracc{1}{4}\cot\dfrac{A-B}{2}\tan\dfrac{A+B}{2}$


Correct Option: A

If $\triangle ABC \cong \triangle QPR$ and $\dfrac {ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac {9}{4}$, $AB=18\ cm$ and $BC=15\ cm$, then $PR$ is equal to________ $cm$

  1. $10$

  2. $12$

  3. $20/3$

  4. $8$


Correct Option: A

The sides of a triangle are $3x+4y,\,4x+3y$ and $5x+5y$ units, where $x,y>0$.The triangle is ______________.

  1. right  angled 

  2. equilateral

  3. obtuse angled

  4. none of these


Correct Option: C
Explanation:
Let $a=3x+4y,\,b=4x+3y$ and $c=5x+5y$ be the largest side
$\Rightarrow \cos{C}=\dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$
$=\dfrac{{\left(3x+4y\right)}^{2}+{\left(4x+3y\right)}^{2}-{\left(5x+5y\right)}^{2}}{2\left(3x+4y\right)\left(4x+3y\right)}$
$\Rightarrow \cos{C}=\dfrac{9{x}^{2}+16{y}^{2}+24xy+16{x}^{2}+9{y}^{2}+24xy-25{x}^{2}-25{y}^{2}-50xy}{2\left(3x+4y\right)\left(4x+3y\right)}<0,\,\,\,x,y>0$
$\Rightarrow \cos{C}=\dfrac{-2xy}{2\left(3x+4y\right)\left(4x+3y\right)}<0,\,\,x,y>0$
$\Rightarrow \theta>{90}^{\circ}$
$\therefore,\, $ the triangle is obtuse angled.

D and E are respectively the points on the sides AB and AC of a $\displaystyle \Delta ABC$ such that $AB = 12 cm$, $AD = 8 cm$, $AE = 12 cm$ and $AC = 18 cm$, then

  1. DE $\parallel$ BD is true

  2. DE $\parallel$ BC is true

  3. AD $\parallel$ BD is true

  4. AD $\parallel$ CD is true


Correct Option: B
Explanation:

We have,
AB = 12 cm, AC = 18 cm, AD = 8 cm and AE = 12 cm.
$\displaystyle \therefore \quad BD=AB-AD=\left( 12-8 \right) cm=4cm$
$\displaystyle CE=AC-AE=\left( 18-12 \right) cm=6cm$
Now, $\displaystyle \frac { AD }{ BD } =\frac { 8 }{ 4 } =\frac { 2 }{ 1 } $
And, $\displaystyle \frac { AE }{ CE } =\frac { 12 }{ 6 } =\frac { 2 }{ 1 } $
$\displaystyle \Rightarrow \quad \frac { AD }{ BD } =\frac { AE }{ CE } $
Thus, DE divides sides AB and AC of $\displaystyle \Delta ABC$ in the same ratio. Therefore, by the converse of basic proportionality theorem, we have
$\displaystyle DE\parallel BC$.