Tag: basic proportionality theorem and its converse

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

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. $\dfracc{1}{4}\cot\dfrac{A-B}{2}\tan\dfrac{A+B}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

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. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of areas is 9/4, so the ratio of corresponding sides is sqrt(9/4) = 3/2. Since ABC ~ QPR, AB/QP = BC/PR = 3/2. 15/PR = 3/2, so 3*PR = 30, PR = 10.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$\displaystyle DE\parallel BC$.

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

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
  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. $\displaystyle 1\rightarrow c,2\rightarrow b,3\rightarrow d,4\rightarrow a$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse 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 $\Delta A P C \sim \Delta B C Q$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the geometric properties of the isosceles triangle and the condition AP * BQ = AC^2, the triangles APC and BCQ satisfy the criteria for similarity (SAS similarity).

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

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
$\triangle DEF=\dfrac {1}{2}\triangle ABC$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If D, E, F divide the sides in 1:2 ratio, the area of triangle DEF is (1 - 3*(1/3)*(2/3)) = 1/3 of the area of triangle ABC. The statement that it is 1/2 is false.