Question

. In the figure, angle PQR = 60° and ray QT bisects angle PQR
B is any point on ray QT.
seg BA perpendicular ray QP and seg BC perpendicular ray QR.
If BC = 8, find the perimeter of
ABCQ​

Answer 1

Answer:

Step 1:

∠ PQR = 60°

QT bisects angle PQR

∴  ∠AQB = ∠CQB = ½ * 60° = 30°

Step 2:

Since BA ⊥ QA & BC ⊥ QR

∴ ∠QCB = ∠QAB = 90°

From the figure attached below, consider ∆ QCB, applying angle sum property

∠QBC + ∠QCB + ∠CQB = 180°

⇒ ∠QBC = 180° - [90° + 30°] = 60°

Similarly, for ∆ AQB, by applying angle sum property, we get

∠QBA = 60°

Step 3:

Side BC = 8 ….. [given] …. (i)

Now,  

According to the 30°-60°-90° triangle rule, we have the ratio of the sides of the triangle as 1:√3:2.[assuming short length is always opposite to shortest angle(30°) and longest length is always opposite to greater angle(90°)]

Therefore,  

For ∆ QCB:

BC = ½ * QB  

⇒ QB = 2 * 8 = 16

∴AQ = √3/2 * QB = √3/2 * 16 = 8√3 ….. (ii)

And,

For ∆ AQB:

AB = ½ * QB = ½ * 16 = 8 …. (iii)

∴QC = √3/2 * QB = √3/2 * 16 = 8√3 …. (iv)

Step 4:

Thus,  

The perimeter of quadrilateral ABCQ is,

= AB + BC + QC + AQ  

Substituting the values from (i), (ii), (iii) & (iv)

= 8 + 8 + 8√3 + 8√3

= 16 + 16√3

= 16[1 + √3]