Question

I will mark the brainlest - helppp please
In the figure below, the square JKLM is inscribed within a circle and △JMN is a right-angled isosceles triangle. The point marked O is the centre of the circle.

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Answer 1

Step-by-step explanation:

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.

CPQR is a square

∴CP=PQ=PR=RC

ΔABC is an isosceles triangle

∴AC=BC

⇒AR+RC=CP+BP

⇒AR=BP ……..(1) [∵RC=CP]

In ΔARQ and ΔQPB

AR=BP

∠ARQ=∠QPB=90

o

QR=PQ

∴ΔARQ≅ΔQPB

⇒AQ=QB

Answer 2

Given:

ΔJMN is a right-angled triangle. Square JKLM is inscribed within a circle.

Point O is the center.

To Find:

Shaded area

Solution:

The square JOMN is also a circle so,

 ∠O = ∠J = ∠M = ∠N =90°

JN = NM   (right angled isosceles triangle)

JO = OM ( has same radius)

Now,

   JN = JO = 1cm  (radius of the quadrant)

Area of the shaded region = area of square JOMN - area of the quadrant

                                            = (1)² - 3.14×1×90/360

                                            = 1 - π/4cm²

Therefore, the area of the shaded region is 1 - π/4cm².

and in the given options c). is the correct answer.