Question

A semiconductor arch-shaped tunnel measures 36 cm from one end to 156 cm in height.
Raised in such a way that his head is attached to the arch. So find the width of the arch​

Answer 1

The width of the semiconductor arc is 730 cm

Step-by-step explanation:

Given as :

A semiconductor arc-shaped tunnel

The height of person = h = 156 cm

The distance of person from one end  = 36 cm

Let The distance of person from one end  = x cm

Let The width of arc = w  = (x + 36) cm

According to question

From figure

Height of person = h = AO = 156 cm

The distance of person from one end  = BA = 36 cm

In ΔAOB

OB² = OA² + AB²

y² = (156 cm)² + (36 cm)²

Or, y² = 24336 cm² + 1296 cm²

Or, y² = 25632            .........1

Again

In ΔAOC

OC² = OA² + AC²

z² = (156 cm)² + (x cm)²

Or, z² = 24336 cm² + x² cm²         ,,,,,,,,,2

And

In ΔCOB

BC² = OC² + OB²

(x + 36 cm)² = (z cm)² + (y cm)²

Or, x² + 36² + 72 x = z² + y²        ..........3

From eq 1 , 2 , 3

x² + 1296 + 72 x = 25632  + x²  + 25632.

Solving the equation

( x² - x² ) + 72 x = 51264 - 1296

Or, 0 + 72 x = 49968

∴     x = $\dfrac{49968}{72}$

i.e  x = 694 cm

Now,

The width of bow = CB

i.e   CB = CA + AB

Or,   w = x cm + 36 cm

Or,   w = 694 cm + 36 cm

∴     w = 730 cm

So, The width of semiconductor arc-shaped = w = 730 cm

Hence, The width of semiconductor arc-shaped is 730 cm Answer