Question

15. The value of X in the adjoining figure is a) 4 cm b) 5 cm c) 8 cm d) 3 cm​

Answer 1

Answer:

Option b)

               5cm

Step-by-step explanation:

○Given-

BC=16cm

DC=4cm

○To Find-

AC=x=?

○Calculations-

BC=16cm [Given]

DC=4cm [Given]

In Triangle ADC,

AC is hypotenuse

$2m,\: m^{2}+1, \: m^{2}-1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [Pythagorean \: Triplet]$

AC>AD>DC

Since,DC is smallest side and AC is largest side

Therefore,DC=2m, AD=$m^{2} -1$, AC=x=$m^{2} +1$

$DC=4=2m\\=>m=\frac{4}{2}=2$

$AD=m^{2} -1\\=>2^{2} -1\\=>4-1\\=>3$

$x=m^{2} +1\\=>2^{2} +1\\=>4+1\\=>5$

Therefore. value of x is 5 cm

Answer 2

Given:

Three right-angled triangles ADC, ADB and BAC.

DC= 4cm, BC= 16cm and AC= xcm.

To find:

The value of x.

Solution:

The Pythagoras theorem states that in a right-angled triangle ABC with right angle at B, if BC is the base, AB is the height of the triangle, then AC is the hypotenuse that is determined using the formula:

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

Also,

BD + DC = BC

BD + 4cm = 16cm

BD = 12cm

Using Pythagoras theorem in right angled triangle ADC, we have

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

${x}^{2} = {4}^{2} + {AD}^{2} \: \: \:$

${AD}^{2} = {x}^{2} - 16\: \: \: (i)$

Similarly, in right angled triangle ADB, we have

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

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

${AD}^{2} = {AB}^{2} - {12}^{2}$

${AD}^{2} = {AB}^{2} - 144 \: \: \:(ii)$

Equating (i) and (ii) we have,

${x}^{2} - 16 = {AB}^{2} - 144$

${AB}^{2} - {x}^{2} = 128 \: \: (iii)$

In right angled triangle BAC, we have

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

$256 = {AB}^{2} + {x}^{2}$ (iv)

Subtracting (iv) from (iii), we have

128 = 2 ${x}^{2}$

${x}^{2}$ = 64

x = +8, -8.

Length cannot be negative, so x ≠ -8.

Hence, the value of x is 8cm.

The correct option is c) 8cm.