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

$ABCD$ 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. $\sqrt {21}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the rectangle sides be x and y. AP=a, PB=x-a, BQ=b, QC=y-b. Areas are 0.5*a*y=5, 0.5*(x-a)b=4, 0.5(y-b)*x=3. Solving this system for the area of DPQ (Area_rect - sum of triangles) yields 12.

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

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. $16:9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of areas of similar triangles is the square of the ratio of their corresponding sides. sqrt(18/32) = sqrt(9/16) = 3/4.

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
$\displaystyle DE\parallel BC$.