Question

prove that the are of a semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on other two sides


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Answer 1

Step-by-step explanation:

MATHS

Prove that the area of the semi-circle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semi circles drawn on the other two sides of the triangle.

In ΔPQR

∠PQR=90

∘

[Given]

QS⊥PR [From vertex Q to hypotenuse PR]

∴QS

2

=PS×SR (i) [By theorem]

Now , in ΔPSQ we have

QS

2

=PQ

2

−PS

2

[By Pythagoras theorem]

=6

2

−4

2

=36−16

=QS

2

=20

⇒QS=2

5

cm

QS

2

=PS×SR (i)

⇒(2

5

)

2

=4×SR

⇒

4

20

=SR

⇒SR=5cm

Now , QS⊥PR

∴∠QSR=90

∘

⇒QR

2

=QS

2

+SR

2

[By Pythagoras theorem]

=(2

5

)

2

+5

2

=20+25

⇒QR

2

=45

⇒QR=3

5

cm

Hence , QS=2

5

cm,RS=5cm and QR=3

5

cm.

Answer 2

➖➖➖➖➖➖➖➖➖

• Let RST a right triangle at S & RS = y
• ST = x

Three semi circles are draw on the sides RS, ST and RT

respectively,

A1, A2 and A3.

Hence proved

➖➖➖➖➖➖➖➖➖

hope it helps