Question

5 IN APQR, PM is perpendicular to QR. If PQ=6 cm, QR =12 cm and PR =10 cm, find the area of APQR. Also, find the length of PM.​

Answer 1

Answer:

Given: PQ=10cm, PR=24cm

Let QR be x cm.

In right angled triangle QPR,

(Hypotenuse)

2

=(Base)

2

+(Perpendicular)

2

[By Pythagoras theorem]

⇒(QR)

2

=(PQ)

2

+(PR)

2

⇒x

2

=(10)

2

+(24)

2

⇒x

2

=100+576=676

⇒x=

676

=26cm

Thus, the length of QR is 26cm.

Step-by-step explanation:

Given: PQ=10cm, PR=24cm

Let QR be x cm.

In right angled triangle QPR,

(Hypotenuse)

2

=(Base)

2

+(Perpendicular)

2

[By Pythagoras theorem]

⇒(QR)

2

=(PQ)

2

+(PR)

2

⇒x

2

=(10)

2

+(24)

2

⇒x

2

=100+576=676

⇒x=

676

=26cm

Thus, the length of QR is 26cm.