Question

In the adjoining figure, if <ACB = <BAD and AD perp BC, AC = 15 cm , AB = 20 cm and BC = 25 cm, then find the length of AD ?​

Answer 1

Given:

• ∠ACB = ∠BAD
• AD ⊥ BC
• AC = 15 cm, AB = 20 cm, BC = 25 cm

To Find:

• AD = ?

Solution:

Consider Δ ACD and Δ BAD

⇒ ∠ ACD = ∠ BAD ( A ) (Given in the question)

⇒ ∠ ADC = ∠ BDA ( A ) (90°)

Hence by AA similarity criterion,

Δ ADC ~ Δ BDA

Hence the corresponding sides are proportional. Hence we get:

$\boxed{ \bf{\dfrac{AC}{AB} = \dfrac{AD}{BD} = \dfrac{CD}{AD} }}$

Let's first consider (AC/AB = AD/BD)

$\implies \dfrac{15}{20} = \dfrac{AD}{BD}\\\\\\\implies \dfrac{3}{4} = \dfrac{AD}{BD}\\\\\\\implies \dfrac{3\:BD}{4} = AD \:\:...(1)$

Now consider the second case (AC/AB = CD/AD)

$\implies \dfrac{15}{20} = \dfrac{CD}{AD}\\\\\\\implies \dfrac{3}{4} = \dfrac{CD}{AD}\\\\\\\implies AD = \dfrac{4\:CD}{3} \:\: ...(2)$

Now we know that:

⇒ BC = CD + BD

⇒ 25 = CD + BD

⇒ BD = 25 - CD ...(3)

Substituting (3) in (1) we get:

$\implies \dfrac{ 3(25 - CD)}{4} = AD\\\\\\\implies AD = \dfrac{75 - 3\:CD}{4} \:\: ...(4)$

Now since the LHS of (4) and (2) are equal, we equate the RHS. Hence we get:

$\implies \dfrac{ 75 - 3\:CD}{4} = \dfrac{4\:CD}{3}\\\\\\\implies \dfrac{ 75 - 3\:CD}{4} - \dfrac{4\:CD}{3} = 0\\\\\\\text{Taking LCM we get:}\\\\\\\implies \dfrac{3(75 - 3\:CD) - 4(4\:CD)}{3 \times 4} = 0\\\\\\\implies \dfrac{225 - 9\:CD - 16\:CD}{12} = 0\\\\\\\implies 225 - 25\:CD = 0\\\\\implies 225 = 25\:CD\\\\\implies CD = \dfrac{225}{25} \\\\\\\implies \boxed{CD = 9}$

Substituting Value of CD in (2) we get:

$\implies AD = \dfrac{ 4 \:CD}{3}\\\\\\\implies AD = \dfrac{ 4 \times 9}{3}\\\\\\\implies AD = \dfrac{36}{3}\\\\\\\implies \boxed{ \bf{ AD = 12\:cm}}$

Hence the length of AD is equal to 12 cm.

Answer 2

$\Large{\underbrace{\underline{\bf{QUESTION\:}}}}:$

In the attachment figure, if $\rm{\angle{ACB}\:=\:\angle{BAD}}$ and $\rm{AD\:\perp\:BC}$, AC = 15 cm, AB = 20 cm and BC = 25 cm, then find the length of AD?

${\bf{Given\::}}$

• $\bf{\angle{ACB}\:=\:\angle{BAD}}$

• $\rm{AD\:\perp\:BC}$

• AC = 15 cm

• AB = 20 cm

• BC = 25 cm

$\\ {\bf{To\: Find\::}}$

• The length of AD.

$\\ {\bf{Solution\::}}$

In ∆ABD & ∆CAD,

⠀⠀$\bf{1)\:\angle{ACD}\:=\:\angle{BAD}}\:{\left[\:\because\:\angle{ACB}\:=\:\angle{ACD}\right]}$

⠀⠀❷ $\rm{AD\:\perp\:BC}$ $\bf{\angle{ADB}\:=\:\angle{CDA}}$

Hence,

⠀⠀∆ABD $\cong$ ∆CAD⠀[By A-A criterion]

$\therefore\:\bf{\dfrac{AB}{CA}\:=\:\dfrac{BD}{AD}\:=\:\dfrac{AD}{CD}\:}---(1)$

Now,

We take,

$:\implies\:\tt{\dfrac{BD}{AD}\:=\:\dfrac{AD}{CD}\:}$

$:\implies\:\tt{(AD)^2\:=\:BD\times{CD}\:}$

As we know that,

• BD = BC - CD

$:\implies\:\tt{(AD)^2\:=\:(BC\:-\:CD)\times{CD}\:}$

$:\implies\:\tt{(AD)^2\:=\:BC\times{CD}\:-\:(CD)^2\:}$

$:\implies\:\tt{(AD)^2\:+\:(CD)^2\:=\:BC\times{CD}\:}$

In ∆CAD,

• AD² + CD² = AC²

$:\implies\:\tt{(AC)^2\:=\:BC\times{CD}\:}$

$:\implies\:\tt{CD\:=\:\dfrac{(AC)^2}{BC}\:}$

$:\implies\:\tt{CD\:=\:\dfrac{(15)^2}{25}\:}$

$:\implies\:\tt{CD\:=\:\dfrac{225}{25}\:}$

$:\implies\:\bf{CD\:=\:9\:cm}$

From equation (1), we take

$:\implies\:\tt{\dfrac{AB}{CA}\:=\:\dfrac{AD}{CD}\:}$

$:\implies\:\tt{AD\:=\:\dfrac{AB}{CA}\times{CD}\:}$

$:\implies\:\tt{AD\:=\:\dfrac{20}{15}\times{9}\:}$

$:\implies\:\tt{AD\:=\:\dfrac{4}{3}\times{9}\:}$

$:\implies\:\tt{AD\:=\:4\times{3}\:}$

$:\implies\:\bf\pink{AD\:=\:12\:cm}$

∴ The length of AD is 12 cm.