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Answer:

Q5-- AQ=3

Q6---45°

Q7---90°

Q8---6

Step-by-step explanation:

∠PAQ=90°{Diagonals of rhombus bisects at 90°}

Applying Pythagoras theorem

AQ=3

Diagonals of rhombus are equal in length and they bisects each other

i.e. AP=AQ

∆APQ is an isoceles ∆

Applying Angle sum property of ∆

m∠APQ= 180°-90°/2

m∠APQ=45°

SIMILARLY m∠ANM and m∠ANP=45°

m∠MNP=m∠ANM + m∠ANP

m∠MNP=90°

PM=AP+PM

PM=6

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