Question

In a triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. If a chord joins A with the point of intersection D of the hypotenuse and the semicircle, then the length of AC equals to:

A. $\dfrac{{{\text{AB}}{\text{.AD}}}}{{\sqrt {{\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{D}}^2}} }}$
​B. $\dfrac{{{\text{AB}}{\text{.AD}}}}{{{\text{AB + AD}}}}$​
C. $\sqrt {{\text{AB}}{\text{.AD}}}$
​D. $\dfrac{{{\text{AB}}{\text{.AD}}}}{{\sqrt {{\text{A}}{{\text{B}}^2}{\text{ - A}}{{\text{D}}^2}} }}$

Answer 1

Hint- Proceed the solution of this question first by visualising the two triangles where we can show two angles can either common or equal i.e. in both the triangle ABC and DAC one angle is equal to and another angle is common hence using AA criterion, triangles ABC and DAC can be proved similar therefore with the help of corresponding side of similar triangle (CSST) ratio, we can find the desired length of side.

Complete step-by-step answer:

In the question it is given that a triangle ABC, right angled at A, on the leg AC as diameter, a semicircle as shown in below figure
Hence ∠BAC= ${90^0}$

Hence ∠BAC= ${90^0}$

⇒∠ADC = ${90^0}$ … ....Angle inscribed in a semicircle
In ΔABC and ΔDAC
⇒∠C is the common angle.
⇒∠BAC=∠ADC= ${90^0}$

Hence we know that
The AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other. Therefore,
⇒∠ADC = ${90^0}$ … ....Angle inscribed in a semicircle
In ΔABC and ΔDAC
⇒∠C is the common angle.
⇒∠BAC=∠ADC= ${90^0}$

Hence we know that
The AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other. Therefore,
⇒ΔABC∼ΔDAC .... A. A test of similarity

Hence we know that if two triangles are similar then those triangles have proportional corresponding sides and all the corresponding angles are the same i.e. corresponding side of similar triangle (C.S.S.T).

∴ $\dfrac{{{\text{AB}}}}{{{\text{AD}}}} = \dfrac{{{\text{BC}}}}{{{\text{AC}}}} = \dfrac{{{\text{AC}}}}{{{\text{DC}}}}$ ...C.S.S.T ………….(1)
As we represent these lengths in the above figure
⇒AB= x, AD = y
⇒AC = l
And using Pythagoras theorem
BC=$\sqrt {{{\text{x}}^2} + {l^2}}$ and DC= $\sqrt {{l^2}{\text{ - }}{{\text{y}}^2}}$
So putting these values in above expression (1)

∴$\dfrac{{\text{x}}}{{\text{y}}} = \dfrac{{\sqrt {{{\text{x}}^2} + {l^2}} }}{l} = \dfrac{l}{{\sqrt {{l^2}{\text{ - }}{{\text{y}}^2}} }}$
So on equalising 1st and 2nd term
$\dfrac{{\text{x}}}{{\text{y}}} = \dfrac{{\sqrt {{{\text{x}}^2} + {l^2}} }}{l}$
On cross multiplication
$\Rightarrow l.{\text{x = y}}\sqrt {{l^2} + {{\text{x}}^2}}$
On squaring both side

$\Rightarrow {\left( {l.{\text{x}}} \right)^2}{\text{ = }}{{\text{y}}^2}{\left( {\sqrt {{l^2} + {{\text{x}}^2}} } \right)^2}$
$\Rightarrow {l^2}.{{\text{x}}^2}{\text{ = }}{{\text{y}}^2}.\left( {{l^2} + {{\text{x}}^2}} \right)$
Opening the small bracket and further solving
$\Rightarrow {l^2}.{{\text{x}}^2}{\text{ = }}{{\text{y}}^2}.{l^2} + {{\text{y}}^2}.{{\text{x}}^2}$
Take the common term on left hand side
$\Rightarrow {l^2}.\left( {{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}} \right){\text{ = }}{{\text{x}}^2}{\text{.}}{{\text{y}}^2}$
$\Rightarrow {l^2}.{\text{ = }}\dfrac{{{{\text{x}}^2}{\text{.}}{{\text{y}}^2}}}{{\left( {{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}} \right)}}$
On taking square root on both side
$\Rightarrow \sqrt {{l^2}} {\text{ = }}\sqrt {\dfrac{{{{\text{x}}^2}{\text{.}}{{\text{y}}^2}}}{{\left( {{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}} \right)}}}$
$\Rightarrow l{\text{ = }}\dfrac{{{\text{x}}{\text{.y}}}}{{\sqrt {\left( {{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}} \right)} }}$

Now replace all the length by their usual name
$\Rightarrow l = \dfrac{{{\text{AB}}{\text{.AD}}}}{{\sqrt {{\text{A}}{{\text{B}}^2}{\text{ - A}}{{\text{D}}^2}} }}$

Note- Whenever we came up of such types of question, we should mainly focus on the diagram and its construction and if in a diagram there is more than one triangles and number of unknown sides length are more, then we should remember that definitely there will be use of concept of either congruent triangle or similar triangle because once these theorems proved then we can find lots of variables by CPCT and CSST rule.
And, with the help of the given condition in question we can find what is possible to prove whether it is a similar triangle or congruent triangle.