Question

If the figure. PQRS is a square, O is centre of the circle. If RS = 10v2, then area of shaded
region is:​

Answer 1

Answer:

Step-by-step explanation:

In ▲PQR {by pythagorus theoram}

(PR)​2=(PQ)2+(QR)2

=(10 root 2)2 + ​(10 root 2)2

=400

PR=20cm

Diameter of cemicircle=​PR=20cm

Radius=10cm

Area=1/2×π​​×r2

=1/2 ×22/7×10×10

=50×π​ cm2

Area of triangle=1/2×​base×​height

=1/2×​10 root 2×​10 root 2​

=100 cm2

Required Area=50×π-​100 cm2

hope this helps u.....

Answer 2

Concept: A circle is a closed, two-dimensional object in which all points in the plane are equally spaced apart from the centre. The symmetry line of reflection is formed by each line that traverses the circle. Additionally, it possesses rotational symmetry about the centre at all angles.

Given: PQRS is a square, O is centre of the circle. RS = $10\sqrt{2}$

To find: area of shaded region

Solution: Using the appropriate calculation, determine the needed area.

∴

$Area of sq PQRS = (side)^2 sq. units\\Area of sq PQRS = (10\sqrt{2}) ^2 sq. units\\Area of sq PQRS = 100 * 2 sq. units\\Area of sq PQRS = 200 sq. units$

Diagonal of square = $\sqrt{2}$ × side unit

                                = $\sqrt{2}$ × $10\sqrt{2}$

                                = 20 cm

diameter of circle = diagonal of square

diameter of circle = 20 cm

therefore, radius of circle = $\frac{20}{2} = 10 cm$

area of circle = $\pi r^{2}$

                      = 3.14 × 10 ×10

                      = 314 $cm^{2}$

Area of shaded region = $\frac{1}{2} * [area of circle - area of square]$

                                      = $\frac{1}{2} [314 - 200]\\\\\\\frac{1}{2} [114]\\ \\57 cm^2$

Hence, area of shaded portion = 57 $cm^{2}$.

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