Question

in triangle pqr, PS is perpendicular to QR. find the sides marked a and b if PR=41, QS=12 and SR=40​

Answer 1

Answer:. B=15

A=9

(41)2-(40)2=a2

So a=9

(12)2+(9)2=b2

So b=15

Step-by-step explanation: please mark me as brainliest

Answer 2

Step-by-step explanation:

Given that:

In triangle PQR, PS is perpendicular to QR. find the sides marked a and b. if PR=41, QS=12 and SR=40

To find: a and b

Solution:

In ∆PQR,

As PS is perpendicular.

∆PSR is right angle triangle,right angle at S

here Base(SR)=40 cm

Hypotenuse(PR)=41 cm

Perpendicular(PS)=a=?

To find PS,apply Pythagoras theorem in ∆PSR

$( {PR)}^{2} = ( {SR)}^{2} + ( {PS)}^{2} \\ \\ ( {41)}^{2} = ( {40)}^{2} + {a}^{2} \\ \\ {a}^{2} = 1681 - 1600 \\ \\ {a}^{2} = 81 \\ \\ \bold{\red{a = 9} }$

Thus,

PS= a= 9cm

Now,

∆PSQ is right angle triangle,right angle at S

here Base(SQ)=12 cm

Hypotenuse(QP)=b=?

Perpendicular(PS)=9 cm

Apply Pythagoras theorem in ∆PSQ

$( {QP)}^{2} = ( {SQ)}^{2} + ( {PS)}^{2} \\ \\ ( {QP)}^{2} = {b}^{2} = ( {12)}^{2} + ( {9)}^{2} \\ \\ {b}^{2} = 144 + 81 \\ \\ {b}^{2} = 225 \\ \\ b = \sqrt{225} \\ \\ \bold{\green{b = 15}}$

Thus,

b= 15 cm.

Final Answer: a= 9 cm,b= 15 cm

Hope it helps you.