Free Triangles 01 Practice Test - 10th Grade 

Let $AB$ and $CD$ be two trees such that $AB=20m,CD=28m\&BD=17m$

Draw $BE$  parallel to $AC$. Then,
$ED=8m$.
By applying Pythagoras' theorem,
$B{E}^2+D{E}^2=B{D}^2$.

$\therefore BE=\sqrt{{BD}^2-{DE}^2}=\sqrt{{\left(17\right)}^2-8^2}$
           $=\sqrt{289-64}=\sqrt{225}=15m$

$\therefore AC=BE=15m$

Consider $△AEF$
AE = AF (As $△ABC$ is equilateral and E and F are midpoints of sides AC and AB respectively)
$\Rightarrow \angle AFE=\angle AEF=\angle A={60}^o$
$\Rightarrow △AEF$ is equilateral.

Similarly we can prove that $△DEC$  and $△BFD$ are equilateral.
Since $△DEF$ share a common side with each of the other three triangles, $△DEF$ is also equilateral.

$\Rightarrow △DEF\cong △AEF$
$\Rightarrow △DEF\cong △DEC$
$\Rightarrow △DEF\cong △BFD$ by using SSS postulate.

If two triangles are similar, then the ratio of the corresponding sides are equal.
Hence,
the ratio of heights = the ratio of sides = 1 : 3.

The ratio of areas of two similar triangles is equal to ratio of squares of corresponding sides.
$\frac{49}{64}=\frac{(\text{Side of first 1st triangle}{)}^2}{(\text{Corresponding side of 2nd triangle}{)}^2}$
 
$\frac{7}{8}=\frac{\text{Side of first 1st triangle}}{\text{Corresponding side of 2nd triangle}}$ 

$\Rightarrow$ Ratio of corresponding sides $=7:8$

In $△$ABC,

It's given LM $\parallel$ AB,
So, $\frac{AL}{CA}=\frac{BM}{CB}$      [By BPT]
$\frac{(x-3)}{2x}=\frac{(x-2)}{(2x+3)}$

$(x-3)\times (2x+3)$ = $(x-2)\times 2x$

$2x^2$ - $3x$ - 9 = $2x^2$ - $4x$

$x=9$
$SoAC=2x=18$

By applying Pythagoras theorem:

$A{B}^2$+$B{C}^2$=$A{C}^2$

$6^2$+${2.5}^2$=$A{C}^2$

We get, AC = 6.5m

Consider  $△ADB$ and  $△ABC$

∠BAD = ∠BAC                [common angle]

∠BDA = ∠ABC                [ ${90}^{\circ }$]

Therefore by AA similarity criterion, $△ADB$ and  $△ABC$ are similar.

So,  $\frac{AB}{AC}=\frac{AD}{AB}$  ⇒  ${AB}^2$ = AC.AD   ---(I)

Similarly, $△BDC$ and  $△ABC$ are similar.

So, $\frac{BC}{AC}$ =  $\frac{DC}{BC}$  ⇒  ${BC}^2$ = AC.DC   ---(II)

Dividing (I) and (II) and cancelling out AC, we get,   ${\left(\frac{AB}{BC}\right)}^2$ =   $\frac{AD}{DC}$  ----(III)

Also, In $△ADB$, ${AB}^2$ = ${AD}^2$ + ${DB}^2$ and in  $△BDC$, ${CB}^2$ = ${CD}^2$ + ${DB}^2$ [Pythagoras theorem]

Subtracting these two equations above and cancelling off ${DB}^2$ on both sides, we get 

${AB}^2$ - ${BC}^2$ = ${AD}^2$ - ${CD}^2$ ⇒    ${AB}^2$ + ${CD}^2$ = ${AD}^2$  + ${BC}^2$  --------------(IV)

Dividing this equation with ${BC}^2$ on both sides,  $\frac{{AB}^2}{{BC}^2}$ +  $\frac{{CD}^2}{{BC}^2}$ =  $\frac{{AD}^2}{{BC}^2}$ +  $\frac{{BC}^2}{{BC}^2}$   ⇒  $\frac{{AB}^2}{{BC}^2}$ + $\frac{{CD}^2}{{BC}^2}$ = $\frac{{AD}^2}{{BC}^2}$ + 1

$⟹$  $\frac{{AB}^2}{{BC}^2}$ - 1 = $\frac{{AD}^2}{{BC}^2}$ - $\frac{{CD}^2}{{BC}^2}$

$⟹$ $\frac{AD}{DC}$ - 1 =  $\frac{{AD}^2-{CD}^2}{{BC}^2}$    [Using (III)]

$⟹$$\frac{AD-DC}{DC}$ = $\frac{(AD-CD)(AD+CD)}{{BC}^2}$

$⟹$$\frac{1}{DC}$ = $\frac{AC}{{BC}^2}$ 

$⟹$ ${BC}^2$ = AC.DC

Alternatively,

Consider $△ABC$ and $△BDC$,

$\angle ABC=\angle BDC={90}^{\circ }$

$\angle C=\angle C$    [Common angle]

Therefore by AA similarity criterion, $△ABC$ and  $△BDC$ are similar.

$\frac{AC}{BC}$ = $\frac{BC}{DC}$

$B{C}^2$ = $AC\times DC$

Since ΔADB and ΔDBC are right triangles, using Pythagoras theorem, we can write:

${AB}^2={AD}^2+{BD}^2...\left(1\right)$

${BC}^2={CD}^2+{BD}^2...\left(2\right)$

Subracting (2) from (1), we get,

${AB}^2={AD}^2+{BD}^2\underset{–}{{BC}^2={CD}^2+{BD}^2(-)(-)(-)}{AB}^2-{BC}^2={AD}^2-{CD}^2$

Rearranging, we get,
$\Rightarrow {AB}^2+{CD}^2={AD}^2+{BC}^2$

Let the other two sides of the triangle be a and b. Since the largest side is the hypotenuse, we have, by Pythagoras theorem,

${Hypotenuse}^2$ = $a^2$ +  $b^2$

${13}^2$ = $a^2$ + $b^2$ = 169           ------------------- (I)

Now,

Sum of the areas of the three squares = ${\left(sideofyellowsquare\right)}^2$ + ${\left(sideofbrownishsquare\right)}^2$  +

${\left(sideofbluesquare\right)}^2$

Sum of the areas of the three squares =$a^2$  + $b^2$ + ${13}^2$ = ${13}^2$ + ${13}^2$ (from (I))

Sum of the areas of the three squares = 169 + 169 = 338.

Let O be the centre of the circle. The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle. (here that point is P). 

$\angle AOB={180}^{\circ }$
$\Rightarrow \angle APB=\frac{\angle AOB}{2}={90}^{\circ }$

$\therefore △APB$ is a triangle, right angled at $P$.

$\Rightarrow$ By pythagoras theorem,
${AB}^2={AP}^2+{PB}^2={\left(2PB\right)}^2+{PB}^2=5{PB}^2$

$\Rightarrow AB=\sqrt{5}PB$

$\Rightarrow PB=\frac{AB}{\sqrt{5}}=\frac{d}{\sqrt{5}}$