Question

In the given figure, \[\vartriangle ABC\] is right angled at \[A\]. \[AD\] is perpendicular to \[BC\]. If \[AB = 5cm\], \[BC = 13cm\] and \[AC = 12cm\], find the area of \[\vartriangle ABC\]. Also, find the length of \[AD\].

Answer 1

Hint: We are given a Right-Angled triangle \[ABC\]. We know the area of the Right-Angled triangle \[ = \dfrac{1}{2} \times base \times perpendicular\]. So, we will find the area of \[\vartriangle ABC\] using this formula. We are given that \[AD\] is perpendicular to \[BC\]. Also, we know that area of a triangle \[ = \dfrac{1}{2} \times base \times height\], where height is the perpendicular of the triangle. We will then equate both the areas and find the length of the perpendicular i.e. \[AD\].

Complete step by step answer:
In \[\vartriangle ABC\], we have \[base = AB\] and \[perpendicular = AC\]
We are given, \[AB = 5cm\] and \[AC = 12cm\].
So, \[base = 5cm\] and \[perpendicular = 12cm\]
We know, area of right angled triangle \[ABC\] i.e. area of \[\vartriangle ABC\] \[ = \dfrac{1}{2} \times base \times perpendicular\]
Substituting the values, we get
\[ = \dfrac{1}{2} \times 5cm \times 12cm\]
Cancelling out the terms, we get
\[ = 5cm \times 6cm\]
\[ = 30c{m^2}\]
Hence, we get, Area of \[\vartriangle ABC\]\[ = 30c{m^2} - - - - - - (1)\]
Now, we need to find the length of \[AD\].
Also, In \[\vartriangle ABC\], we can take \[base = BC\] and \[height = AD\]
We are given, \[BC = 13cm\]
Using the formula, area of a triangle \[ = \dfrac{1}{2} \times base \times height\], we have
Area of \[\vartriangle ABC\] \[ = \dfrac{1}{2} \times base \times height\]
\[ = \dfrac{1}{2} \times BC \times AD\]
\[ = \dfrac{1}{2} \times 13cm \times AD\]
\[ = \dfrac{{13}}{2}cm \times AD\]
Hence, we get, Area of \[\vartriangle ABC\] \[ = \dfrac{{13}}{2}cm \times AD - - - - - - (2)\]
Using (1) and (2), we have
\[30c{m^2} = \dfrac{{13}}{2}cm \times AD\]
Reshuffling the terms, we get
\[\left( {30c{m^2}} \right) \div \left( {\dfrac{{13}}{2}cm} \right) = AD\]
\[\dfrac{{\left( {30c{m^2}} \right)}}{{\left( {\dfrac{{13}}{2}cm} \right)}} = AD\]
Using \[\dfrac{a}{b} = a \times \dfrac{1}{b}\], we get
\[\left( {30} \right) \times \left( {\dfrac{1}{{\dfrac{{13}}{2}}}} \right)cm = AD\]
Using \[\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{b}{a}\], we get
\[\left( {30} \right) \times \left( {\dfrac{2}{{13}}} \right)cm = AD\]
Multiplying the numerator, we get
\[\left( {\dfrac{{60}}{{13}}} \right)cm = AD\]
Hence, we get
\[AD = \left( {\dfrac{{60}}{{13}}} \right)cm\]

Note:
To find the length of \[AD\], we can take \[BD = x\]cm and so \[DC = \left( {13 - x} \right)\]cm, Now, since \[\vartriangle ABD\] and \[\vartriangle ACD\] are right angled triangles, we can apply Pythagoras Theorem in both the triangles and find the value of \[{\left( {AD} \right)^2}\], in terms of \[x\] from both the equations. We will then solve for \[x\] by equating both the equations. After finding the value of \[x\], we will again use Pythagoras Theorem in \[\vartriangle ABD\] or in \[\vartriangle ACD\] to find the value of \[AD\]. Also, we need to take care of the units of area and the units of length of a side.