Question

ABC is a right triangle. \[AD\] is drawn perpendicular to \[BC\]. If \[BC = 9{\text{ }}cm\], \[BD = 4{\text{ }}cm\] then find \[AB\].

Answer 1

Hint: To find \[AB\], we will consider two triangles, \[\vartriangle BDA\] and \[\vartriangle CDA\]. We will assume \[AD = x{\text{ }}cm\]. Then we will find the value of \[\tan B\] and \[\tan C\] in \[\vartriangle BDA\] and \[\vartriangle CDA\] respectively. Then using \[\tan B\] and \[\tan C\] we will find a relation between the ratio of sides of these two triangles and last, we will use the Pythagoras theorem in \[\vartriangle BDA\] to find \[AB\].

Complete step by step answer:

Given, a right-angle triangle \[ABC\] and \[AD\] is drawn perpendicular to \[BC\].
Let \[AD = x{\text{ }}cm\].
In \[\vartriangle BDA\],
\[ \Rightarrow \tan B = \dfrac{{AD}}{{BD}}\]
On putting the values, we get
\[ \Rightarrow \tan B = \dfrac{x}{4} - - - (1)\]
Similarly, in \[\vartriangle CDA\],
\[ \Rightarrow \tan C = \dfrac{{AD}}{{CD}}\]
As \[BC = 9{\text{ }}cm\] and \[BD = 4{\text{ }}cm\], we get \[CD = \left( {9 - 4} \right){\text{ }}cm\] i.e., \[5{\text{ }}cm\].
On putting the values, we get
\[ \Rightarrow \tan C = \dfrac{x}{5} - - - (2)\]
As \[ABC\] is a right-angle triangle, right-angled at \[A\]. So, we can write
\[ \Rightarrow B + C = {90^ \circ }\]
\[ \Rightarrow C = {90^ \circ } - B\]
Taking \[\tan \] both the sides, we get
\[ \Rightarrow \tan C = \tan \left( {{{90}^ \circ } - B} \right)\]
As, \[\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \], we get
\[ \Rightarrow \tan C = \cot B\]
As, \[\cot \theta = \dfrac{1}{{\tan \theta }}\], we get
\[ \Rightarrow \tan C = \dfrac{1}{{\tan B}}\]
From \[(1)\] and \[(2)\], we get
\[ \Rightarrow \dfrac{x}{5} = \dfrac{1}{{\left( {\dfrac{x}{4}} \right)}}\]
On rewriting we get
\[ \Rightarrow \dfrac{x}{5} = \dfrac{4}{x}\]
\[ \Rightarrow {x^2} = 20\]
\[\therefore A{D^2} = 20\]
Using Pythagoras theorem in \[\vartriangle BDA\], we get
\[ \Rightarrow A{B^2} = A{D^2} + B{D^2}\]
On putting the values, we get
\[ \Rightarrow A{B^2} = 20 + {4^2}\]
On simplifying we get
\[ \Rightarrow A{B^2} = 36\]
Taking square root of both the sides, we get
\[ \Rightarrow AB = 6{\text{ }}cm\]
Therefore, \[AB = 6{\text{ }}cm\].

Note:
We can also solve this problem by another method.
In \[\vartriangle ABC\] and \[\vartriangle DBA\], we have
\[\angle A = \angle D\] (Both \[{90^ \circ }\])
\[\angle B = \angle B\] (Common)
Since, in the triangle \[\vartriangle ABC\] and \[\vartriangle DBA\], we have two angles equal. Therefore, the third angle also must be equal.
\[\therefore \angle C = \angle BAD\]
Therefore, from angle-angle-angle similarity, \[\vartriangle ABC\] and \[\vartriangle DBA\] are similar.
As we know, if two triangles are similar, then their corresponding sides are in proportion. This means that the ratio of their corresponding sides is equal to each other.
Therefore, we can write in \[\vartriangle ABC\] and \[\vartriangle DBA\],
\[ \Rightarrow \dfrac{{BC}}{{AB}} = \dfrac{{AB}}{{BD}} = \dfrac{{AC}}{{AD}}\]
Therefore, we have \[\dfrac{{BC}}{{AB}} = \dfrac{{AB}}{{BD}}\].
On cross multiplication, we get
\[ \Rightarrow A{B^2} = BC \times BD\]
Putting the given values of \[BC = 9{\text{ }}cm\] and \[BD = 4{\text{ }}cm\], we get
\[ \Rightarrow A{B^2} = 9 \times 4\]
On multiplication we get
\[ \Rightarrow A{B^2} = 36\]
On taking the square roots of both the sides, we get
\[ \Rightarrow AB = 6\]
Therefore, \[AB = 6{\text{ }}cm\].