Question

2) In A PQR , L P Q R = 90° seg Qs I
hypotenuse PR, PS= 16, RS = 9. Find Qs​

Answer 1

Answer:

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Step-by-step explanation:

Given: ΔPQR is right angled triangle where Q is right angle

          QS ⊥ PR

          QM bisect the ∠PQR

To prove:

In ΔPQR,

QM bisector of ∠PQR

  (Property of angle bisector of a triangle)

..............(1)

Again in ΔPQR

QS ⊥ PR ⇒ ΔPQR is similar to ΔPSQ is similar to ΔRSQ

in ΔPSQ and ΔPQR

[tex\frac{PQ}{PR}=\frac{PS}{PQ}[/tex]  (both triangles are similar)

.........(2)

Similarly form  ΔPSQ and ΔPQR

..............(3)

From (1) , (2) & (3)

Hence Proved

Answer 2

Answer

          QS ⊥ PR

          QM bisect the ∠PQR

To prove:

In ΔPQR,

QM bisector of ∠PQR

  (Property of angle bisector of a triangle)

..............(1)

Again in ΔPQR

QS ⊥ PR ⇒ ΔPQR is similar to ΔPSQ is similar to ΔRSQ

in ΔPSQ and ΔPQR

[tex\frac{PQ}{PR}=\frac{PS}{PQ}[/tex]  (both triangles are similar)

.........(2)

Similarly form  ΔPSQ and ΔPQR

..............(3)

From (1) , (2) & (3)

Hence Proved