Tag: relation between perimeters of similar shapes

Questions Related to relation between perimeters of similar shapes

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 

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. 15: 8

  2. 16 : 5

  3. 15 : 4

  4. 5 : n2

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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 

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. 15 : 8

  2. 16 : 5

  3. 15 : 4

  4. 5 :2

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Correct Option: A

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

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. True

  2. False

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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 $\left(A>B\right)$

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

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

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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________ $cm$

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. $10$

  2. $12$

  3. $20/3$

  4. $8$

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Correct Option: A

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

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. right  angled 

  2. equilateral

  3. obtuse angled

  4. none of these

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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

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. DE $\parallel$ BD is true

  2. DE $\parallel$ BC is true

  3. AD $\parallel$ BD is true

  4. AD $\parallel$ CD is true

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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
$\displaystyle DE\parallel BC$.

Match the column.

1. In $\displaystyle \Delta ABC$ and $\displaystyle \Delta PQR$,$\displaystyle \frac{AB}{PQ}=\frac{AC}{PR},\angle A=\angle P$ (a) AA similarity criterion 
2. In $\displaystyle \Delta ABC$ and $\displaystyle \Delta PQR$,$\displaystyle \angle A=\angle P,\angle B=\angle Q$ (b) SAS similarity criterion 
3. In $\displaystyle \Delta ABC$ and $\displaystyle \Delta PQR$,$\displaystyle \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$$\angle A=\angle P$ (c) SSS similarity criterion 
4. In $\displaystyle \Delta ACB,DE
maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. $\displaystyle 1\rightarrow a,2\rightarrow b,3\rightarrow c,4\rightarrow d$

  2. $\displaystyle a\rightarrow d,2\rightarrow a,3\rightarrow c,4\rightarrow b$

  3. $\displaystyle 1\rightarrow b,2\rightarrow a,3\rightarrow c,4\rightarrow d$

  4. $\displaystyle 1\rightarrow c,2\rightarrow b,3\rightarrow d,4\rightarrow a$

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Correct Option: C
Explanation:

In $\triangle ABC$ and $\triangle PQR$

Option A:

If $\angle A = \angle P$      ....Given

And, $\dfrac {AB}{PQ} = \dfrac {AC}{PR}$    ...Given

$\triangle ABC \sim \triangle PQR$        ...SAS test of similarity


Option B:

If $\angle A = \angle P$      ....Given

And $\angle B = \angle Q$      ....Given

$\triangle ABC \sim \triangle PQR$        ...AA test of similarity


Option C:

If $\angle A = \angle P$      ....Given

And $\dfrac {AB}{PQ} = \dfrac {AC}{PR} = \dfrac {BC}{QR}$      ....Given

$\triangle ABC \sim \triangle PQR$        ...SS S test of similarity


Option D:

In $\triangle ACB, DE \parallel BC$

$\dfrac {AD}{BD} = \dfrac {AE}{CE} $      ....Given

This is known as basic proportionality theorem.

In an isosceles $\Delta A B C$ the base $A B$ is produced both the ways to $P$ and $Q$ such that $A P \times BO = A C ^ { 2 }$ then $\Delta A P C \sim \Delta B C Q$

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. True

  2. False

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Correct Option: A

In the sides $BC,CA,AB$ of a triangle $ABC$, three points $D,E,F$ are taken such that each of $BD,CE,AE$ is equal to one-third of the corresponding side, then
$\triangle DEF=\dfrac {1}{2}\triangle ABC$.

maths triangles relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. True

  2. False

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Correct Option: B