Free Triangles 02 Practice Test - 10th Grade 

$\frac{AB}{DE}$ =$\frac{AC}{DF}$ = $\frac{BC}{EF}$ = $\frac{2}{4}$ = $\frac{1}{2}$

$DE=2\times AB=6cm,DF=2\times AC=5cm$.     

$\therefore$ $\text{Perimeter of}\Delta DEF=(DE+EF+DF)$
                                  $=6+4+5=15cm$.

In $△CDB$,

${BC}^2={CD}^2+{BD}^2\text{[Pythagoras theorem]}$

${BC}^2={CD}^2+{(DA+AB)}^2$ 

${BC}^2={CD}^2+{DA}^2+{AB}^2+(2\times DA\times AB)...\left(i\right)$

In $△ADC$,

${CD}^2+{DA}^2={AC}^2...\left(ii\right)\text{[Pythagoras Theorem]}$

Here, $\angle CAB={120}^{\circ }$ (given)

$\Rightarrow \angle CAD={60}^{\circ }$ (since  $\angle CAD$ and $\angle CAB$ form a linear pair of angles)

Also, $\cos {60}^{\circ }=\frac{AD}{AC}$

$AC=2AD...\left(iii\right)$

Substituting  the values from $\left(ii\right)\&\left(iii\right)\text{in}\left(i\right)$ we get,

${BC}^2={AC}^2+{AB}^2+(AC\times AB)$

$a^2=b^2+c^2+bc$

Alternatively,

 Since  $\angle A$ is an obtuse angle in  $△ABC$ so,

${BC}^2={AB}^2+{AC}^2+2AB.AD={AB}^2+{AC}^2+2\times AB\times \frac{1}{2}\times AC[\because AD=AC\cos {60}^{\circ }=\frac{1}{2}AC]={AB}^2+{AC}^2+AB\times AC$
 

  $\Rightarrow a^2=b^2+c^2+bc$

$\angle S={180}^o-(\angle A+\angle C)$

The angles of the traingle are known. With the given measurement of the angles, we can draw infinite number of traingles (Similar triangles).

In $△$ABC,

DE $\parallel$ BC      (Given)

Therefore, $\frac{AD}{AB}$ = $\frac{AE}{AC}$    [By Basic Proportionality Theorem]

Comparing the above with $\frac{AD}{AB}$ = $\frac{AE}{x}$

$x$ = AC

It is given that

$\frac{PS}{SQ}$ = $\frac{PT}{TR}$

So, ST||QR     [By converse of Basic Proportionality Theorem]

$\therefore$, $\angle$PST=$\angle$PQR (Corresponding Angles)

Also, it is given that

$\angle$PST = $\angle$PRQ

So,$\angle$PRQ =  $\angle$PQR

Therefore, PQ = PR (Sides opposite the equal angles)

i.e., $\Delta$PQR is an isosceles triangle.

Let AB be a units long.

We know that diagonal of a square = $\sqrt{2}\times$ side length
$\Rightarrow AC=\sqrt{2}\times AB$

Now, $△$ ABE and $△$ ACF are equilateral triangles.

Each angle of both $△$ ABE and $△$ ACF is ${60}^{\circ }$.

$\therefore △ABE\sim △$ ACF (by AAA similarity criterion)

So, $\frac{ar\left(ACF\right)}{ar\left(ABE\right)}=\frac{A{C}^2}{A{B}^2}=\frac{2a^2}{a^2}=\frac{2}{1}$

Let the triangles be $△ABC$ and $△DEF$

Then  $AB=3$ units and $DE=4$ units.

Each angle of both $△ABC$ and $△DEF$ is ${60}^{\circ }$.

$△ABE\sim △$ ACF
( AAA similarity)

So,
$\frac{ar\left(ABC\right)}{ar\left(DEF\right)}=\frac{B{C}^2}{E{F}^2}=\frac{3^2}{4^2}=\frac{9}{16}$

Consider  ΔABD and ΔACB:

∠BAD =  ∠BAC    [common angle]

∠BDA =  ∠ABC     [ ${90}^{\circ }$]

By AA similarity criterion,
 $△ABD\text{~}△$ACB

Hence,

$\frac{ar\left(\Delta ABD\right)}{ar\left(\Delta ACB\right)}=({\frac{AB}{AC})}^2=({\frac{AD}{AB})}^2=({\frac{BD}{CB})}^2$ 

Consider the above figure:

AD is the part of the ladder in the first case and AC is the ladder in the second case

AB = BD = 10 $\sqrt{7}$m          [given]

2AD = AC        [since AD is the half of the ladder AC]

$\therefore$ By Pythagoras Theorem in  $△$ABD,

${AD}^2$ =  ${AB}^2$ +  ${BD}^2$

⇒ ${AD}^2$ = $({10\sqrt{7})}^2$   +$({10\sqrt{7})}^2$ 
             = 2 $({10\sqrt{7})}^2$  

⇒$AD=10\sqrt{7}$$\sqrt{2}$

Since AD is half of the ladder, the complete length of the ladder

AC = 2AD = 20 $\sqrt{2}$$\sqrt{7}$

Now,
In  ΔABC,  
${AC}^2$ =  ${AB}^2$ +  ${BC}^2$

⇒ ${BC}^2={AC}^2-{AB}^2$
             $={\left(20\sqrt{7}\sqrt{2}\right)}^2$ -  $({10\sqrt{7})}^2$
             $=\left({10\sqrt{7})}^2\right[{\left(2\sqrt{2}\right)}^2-1]$
             = $({10\sqrt{7})}^2\times 7$

$\therefore BC=10\sqrt{7}\times \sqrt{7}=70m$