# Find the value of x in the figure?

Aug 6, 2016

In Fig.(a), $x = 8.75$

In Fig.(b), $x \cong 8.57$

#### Explanation:

In Similar Triangles, the corresponding sides are in proportion.

In Fig.(a), the small triangle ling inside the big one is similar to each other.

Hence, $\frac{x}{7} = \frac{8 + 2}{8} \Rightarrow x = \frac{70}{8} = 8.75$

In Fig.(b), x/15=12/21 rArr x=180/21~=8.57

Aug 6, 2016

In Figure (a): $\textcolor{g r e e n}{x = 8 \frac{3}{4}}$

In Figure (b): $\textcolor{g r e e n}{x = 8 \frac{4}{7}}$

#### Explanation:

Figure (a)
I have assumed that the lines labelled with $7$ and $x$ are parallel (otherwise this question can not be solved).
Reproducing the figure (a) with labelled vertices for reference purposes:

Notice the similar triangles:
color(white)("XXX")triangleABC~triangleADE

rarrcolor(white)("XXX")abs(BC)/abs(AB)=abs(DE)/(abs(AD)

$\rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{7}{8} = \frac{x}{8 + 2}$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} 8 x = 70$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} x = 8 \frac{6}{8} = 8 \frac{3}{4}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Figure (b)
Similarly in figure (b) I have had to assume that sides with lengths $15$ and $x$ are parallel.
Again, reproducing the image with labelled vertices:

trianglePQR ~ trianglePST#

$\rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{\left\mid Q R \right\mid}{\left\mid P Q \right\mid} = \frac{\left\mid S T \right\mid}{\left\mid P S \right\mid}$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{x}{12} = \frac{15}{9 + 12}$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} 21 x = 15 \times 12$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} x = \frac{180}{21} = 8 \frac{4}{7}$