. 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
Answers
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]