Answer
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Hint: We have to find the additional value of the two sides in a triangle. In order to solve this question we need to prove that the sides of the two triangles $\vartriangle ACB$ and $\vartriangle ECD$ are in proportion.
Complete step-by-step answer:
It is given in the question that \[\angle A = \angle CED\]
Therefore from $\vartriangle ACB$ and $\vartriangle ECD$ we get,
\[\angle A = \angle CED\] already given in the question
$\angle ACB = \angle ECD$ common angle
Thus $\vartriangle ACB$ and $\vartriangle ECD$are similar triangles as they satisfy the condition of similarity of triangles by AA criterion.
So it can be said that their sides are in proportion.
So, $\dfrac{{AC}}{{EC}} = \dfrac{{CB}}{{CD}} = \dfrac{{BA}}{{DE}}$
Putting the values of $CD = 8cm$, $CE = 10cm$, $BE = 2cm$,$AB = 9cm$,$AD = b$ and $DE = a$ in the above equation we get-
$\Rightarrow$$\dfrac{{8 + b}}{{10}} = \dfrac{{12}}{8} = \dfrac{9}{a}$ since $AC = AD + DC$
So from the above equation we get-
$\Rightarrow$$\dfrac{{12}}{8} = \dfrac{9}{a}$
By doing cross-multiplication we get-
$\Rightarrow$$12a = 72$
So, $a = 6$
Again we can write-
$\Rightarrow$$\dfrac{{8 + b}}{{10}} = \dfrac{{12}}{8}$
By doing cross multiplication we get-
$\Rightarrow$$64 + 8b = 120$
By moving $64$ on the right hand side we get-
$\Rightarrow$$8b = 120 - 64$
$\Rightarrow$$8b = 56$
By division we get-
So, $b = 7$
Therefore the value of $a + b = 7 + 6 = 13$
Option A is the correct answer
Note: There are three conditions of similarity of triangles which are given below
If two angles of a triangle are equal to the other two angles of another triangle, then the two triangles are similar to each other.
If the two sides of a triangle are in the same proportion of the two sides of another triangle, and the angle inscribed by the two sides in both the triangles are equal, then two triangles are considered to be similar.
If all the three sides of a triangle are in proportion to the three sides of another triangle, then the two triangles will be treated as similar.
Complete step-by-step answer:
It is given in the question that \[\angle A = \angle CED\]
Therefore from $\vartriangle ACB$ and $\vartriangle ECD$ we get,
\[\angle A = \angle CED\] already given in the question
$\angle ACB = \angle ECD$ common angle
Thus $\vartriangle ACB$ and $\vartriangle ECD$are similar triangles as they satisfy the condition of similarity of triangles by AA criterion.
So it can be said that their sides are in proportion.
So, $\dfrac{{AC}}{{EC}} = \dfrac{{CB}}{{CD}} = \dfrac{{BA}}{{DE}}$
Putting the values of $CD = 8cm$, $CE = 10cm$, $BE = 2cm$,$AB = 9cm$,$AD = b$ and $DE = a$ in the above equation we get-
$\Rightarrow$$\dfrac{{8 + b}}{{10}} = \dfrac{{12}}{8} = \dfrac{9}{a}$ since $AC = AD + DC$
So from the above equation we get-
$\Rightarrow$$\dfrac{{12}}{8} = \dfrac{9}{a}$
By doing cross-multiplication we get-
$\Rightarrow$$12a = 72$
So, $a = 6$
Again we can write-
$\Rightarrow$$\dfrac{{8 + b}}{{10}} = \dfrac{{12}}{8}$
By doing cross multiplication we get-
$\Rightarrow$$64 + 8b = 120$
By moving $64$ on the right hand side we get-
$\Rightarrow$$8b = 120 - 64$
$\Rightarrow$$8b = 56$
By division we get-
So, $b = 7$
Therefore the value of $a + b = 7 + 6 = 13$
Option A is the correct answer
Note: There are three conditions of similarity of triangles which are given below
If two angles of a triangle are equal to the other two angles of another triangle, then the two triangles are similar to each other.
If the two sides of a triangle are in the same proportion of the two sides of another triangle, and the angle inscribed by the two sides in both the triangles are equal, then two triangles are considered to be similar.
If all the three sides of a triangle are in proportion to the three sides of another triangle, then the two triangles will be treated as similar.
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