Question

PQR is an isosceles triangle, right angled at P.Bisector of angle Q and R meet PR and PQ at Sand T respectively. If QR (in cm) is a two digitgreatest prime number, then (PQ + PR + PS + PT)is equal to(1) 388 cm(2) 194 cm(3) 97 cm(4) 291 cm​

Answer 1

If QR (in cm) is a two-digit greatest prime number, then (PQ + PR + PS + PT)is equal to

option (2): 194 cm .

Step-by-step explanation:

Given data:

As shown in the figure attached below,

PQR is an isosceles right-angled triangle where angle P = 90°

QS is the bisector of angle Q

RT is the bisector of angle R

QR is the 2-digit greatest prime number.

To find:

The value of (PQ + PR + PS + PT) in cm

Solution:

Step 1:

Since PQR is an isosceles ∆,

So, PQ = PR …. [∵ two sides of an isosceles triangle are equal in length] …. (i)

Also, QR = 97 ← 2-digit greatest prime number …. (ii)

Now,

By applying Pythagoras theorem, in isosceles right-angled ∆PQR, we have

QR² = PQ² + PR²

⇒ 97² =  2 * PQ² …. [substituting values from (i) & (ii)]

⇒ PQ = 97/√2  

⇒ PQ = 68.58 ≈ 68.6 cm

∴ PQ = PR = 68.6 cm …. (iii)

Step 2:

Since PQ = PR  

∴ ∠Q = ∠R ….. [∵ angles opposite to equal sides are also equal]

Using the angle sum property in ∆PQR, we have

∠P  + ∠Q + ∠R = 180°

⇒ 2∠Q = 180° – 90° = 90°

⇒ ∠Q = 90°/2 = 45°

∴ ∠Q = ∠R = 45°  

Now,

∠PRT = ½ * ∠R = ½ * 45° = 22.5° ….. [∵ RT is the bisector of angle R]  

And,

∠PQS = ½ * ∠Q = ½ * 45° = 22.5° ... [∵ QS is the bisector of angle Q]  

Step 3:

Applying trigonometric properties of triangle, in ∆ PRT and ∆ PQS, we have

tan θ = perpendicular/base = PT / PR = PS/PQ

 

here, θ = Angle PRT(from ∆ PRT) = Angle PQS (from ∆ PQS) = 22.5°

substituting the value of θ and the value of PQ & PR from (iii), we get

⇒ tan 22.5°  =  PS/68.6  = PT / 68.6

⇒ 0.414 = PS/68.6 = PT / 68.6

⇒ PS = PT = 0.414* 68.6

⇒ PS = PT = 28.41 cm ….. (iv)

Step 4:

Thus, substituting the values from eq. (iii) & (iv), we get

PQ + PR + PS + PT  

= 68.6 + 68.6 + 28.41 + 28.41

= 194.02 cm

≈ 194 cm