Question

in cyclic quadrilateral PQRS, 7LP=2LR Then LP=​

Answer 1

Answer:

Answer

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Area of PQRS=4

3

...(1)

⇒

2

1

(ad+bc)sinP=4

3

⇒(2d+bc)sinP=8

3

....(2)

In △PQR, by sine formula,

sinP

2

3

=2R;

R= radius of circumcircle △PQR which is same as that for △PQR=2 (given)

⇒sinP=

2

3

⇒cosP=±

2

1

...(3)

⇒±

2

1

=

2(2)(d)

4+d

2

−12

⇒±2d=d

2

−8

⇒d

2

∓2d−8=0⇒d=2,4

∴ From (2) and (3) and (4);

bc=16−2d=12,8

⇒(Qr).(SR)=12,8

solution

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Answer 2

Answer:

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