Math, asked by sunilarya267, 12 hours ago

32. In the given figure, the side PQ of a right triangle APQR where OP = 6 cm, OR = 8 cm and QR - 26 cm, is : (a) 21 cm (b) 23 cm (c) 24 cm (d) 36 cm

Answers

Answered by sonu12346450

Answer:

answer is 26 cm

Step-by-step explanation:

I hope its help u

Answered by 231001ruchi

ΔPQR is right Angled Δ with ∠POR=90∘

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cm

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2⇒PQ=(26)2−(10)2

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2⇒PQ=(26)2−(10)2=676−100

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2⇒PQ=(26)2−(10)2=676−100=576

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2⇒PQ=(26)2−(10)2=676−100=57624cm

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2⇒PQ=(26)2−(10)2=676−100=57624cm∴

ΔPQR is right Angled Δ with ∠POR=90∘∴(PO)2+(OR)2=PR⇒PR=62+82=100=10cmand Δ PQR is right Angled Δ with ∠QPR=90∘∴(PQ)2+(PR)2=(QR)2⇒PQ=(26)2−(10)2=676−100=57624cm∴ Ar of ΔPQR=21base×height