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

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

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

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

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

Thus,

b= 15 cm.

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

Hope it helps you.

Answer 2

Answer:

Step-by-step explanation:

applying pythagorus theroum to find ps/a

PR^2=SR^2+PS^2

41^2=40^2+a^2

1681-1600=a^2

a^2=81

a=\sqrt(81)

9cm

b:QP^2=QS^2+PS^2

b^2=12^2+9^2

b^2=144+81

b^2=225

b=\sqrt(225)

b=15cm