Question

In the given figure, the value of x is
(a) 4 cm
(b) 5 cm
(c) 8 cm
(d) 3 cm

Answer 1

Answer is (b)5cm

Explanation:

By Pythagorean theorem,

(AD)^2+(DC)^2=(AC)^2

I take AC=5cm and substitute in the above equation.

Then AD=√9=3cm.

But In other cases I won't get the value as a whole number measure if I substitute AC as the other given options.

Answer 2

Given: Triangle ABC, right angled at A.

           AD ⊥ BC, AC = x, DC = 4 cm and BC = 16 cm

To find: Value of x

Solution: In triangle ADC (right angled at D)

Using Pythagoras Theorem:

$AC^{2}$ = $AD^{2} +DC^{2}$

AC = x and DC = 4 cm

⇒ $x^{2}$ = $AD^{2} + (4)^{2}$

⇒ $x^{2}$ = $AD^{2} + 16$                          ...(1)

Similarly, In triangle ADB (right angled at D)

Using Pythagoras Theorem:

$AB^{2} = BD^{2} + AD^{2}$

Now, given that BC = 16 cm and DC = 4 cm. Then, BD = BC - DC = 16 - 4 = 12

⇒ $AB^{2}$ = $(12)^{2}$ + $AD^{2}$

⇒ $AB^{2} = 144 + AD^{2}$                    ...(2)

In triangle BAC (right angled at A)

$BC^{2} = AB^{2} + AC^{2}$

BC = 16 cm and AC = x

⇒ $(16)^{2} = AB^{2} + x^{2}$        

⇒ 256 = $AB^{2} + x^{2}$                       ...(3)

Putting value of $AD^{2}$ from equation (2) in equation (1)

⇒ $x^{2}$ = ($AB^{2}$ - 144) + 16                 ...(4)

Putting value of $AB^{2}$ from equation (3) in equation (4)

⇒ $x^{2}$ = [($256 - x^{2}$) - 144] + 16

⇒ $x^{2}$ = (256 - $x^{2}$ - 144) + 16

⇒ $x^{2}$ = 256 - 144 + 16 - $x^{2}$

⇒ $x^{2} + x^{2}$ = 128

⇒ 2$x^{2}$ = 128

⇒$x^{2}$ = 128/2 = 64

   $x$ = $\sqrt{64}$ = ± 8

Since, length of a side can't be negative, reject -8

Therefore, value of x(AC) = 8 cm