Question

Plzz help me in solving 8th nd 10th question..

Answer 1

x=50
y=80
hope it helps

Answer 2

In the 8th question,
Since, QT is perpendicular to PR
QTR is a right angle triangle.
Now by angle sum property,
angle QTR + angle TQR + angle TRQ = 180°
40°+ 90° + x = 180°
x = 180° - (130°)
x = 50 ° ...... (i)

In triangle PSR,
angle PSR+ angle SPR+ angle PRS = 180°
30° + 50°+ angle PSR = 180° [from (i)]
Angle PSR = 100°

By linear pair,
angle PSQ + angle PSR =180°
y = 80°


I looked online for the correct answer of question 10, and found this, hopefully it'll help.


Answer: ABCD is a quadrilateral in which P, Q, R, & S are mid points of AB, BC, CD & AD
In Δ ACD
SR is touching mid points of CD and AD
So, SR || AC
Similarly following can be proved
PQ || AC
QR || BD
PS || BD
So, PQRS is a parallelogram.
PR and QS are diagonals of the parallelogram PQRS, so they will bisect each other.