Question

A piece of cardboard is in the shape of a quadrant of a circle of radius 7 cm bounded by perpendicular radii OX and OY points A and B lie on OX and OY respectively such that OA equals to 3 cm and OB equals to 4 centimetre if the triangular part of the cardboard is removed calculate the area and the perimeter of the remaining part


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Answer 1

I think it is clear and you have understood this solution

Answer 2

Answer:

Area of Remaining Part is 32.5 cm² and Perimeter of remaining part is 23 cm

Step-by-step explanation:

Given: Radius of circle, r = 7 cm

           OA = 3 cm & OB = 4 cm

To find: Area & perimeter of segment left by removing ΔAOB from

             sector/quadrant XOY

Figure is attached.

$Area\;of\:Quadrant=\frac{90^{\circ}}{360^{\circ}}\times\pi r^2$

                       $=\frac{1}{4}\times\frac{22}{7}\times7^2$

                       $=\frac{1}{2}\times11\times7$

                       $=38.5\:cm^2$

ΔAOB is right angled triangle.

⇒ Area of Triangle = $\frac{1}{2}\times OA\times OB$

                               =  $\frac{1}{2}\times3\times4$

                               =  $3\times2$

                               =  6 cm²

⇒ Area of Segment = 38.5 - 6 = 32.5 cm²

Perimeter of Segment = Length of Arc AB + Length of line segment AB

                                          + AX + BY

From ΔABC,

using Pythagoras theorem,

AB² = OA² + OB²

AB² = 3² + 4²

AB² = 9 + 16

AB² = 25

AB = √25

AB = 5 cm

Length of arc AB = $\frac{90^{\circ}}{360^{\circ}}\times2\pi r$

                            = $\frac{1}{4}\times2\times\frac{22}{7}\times7$

                            = 11 cm

AX = OX - OA = 7 - 3 = 4 cm

BY = OY - OB = 7 - 4 = 3 cm

⇒ Perimeter of Segment = 11 + 5 + 4 + 3 = 23 cm

Therefore, Area of Remaining Part is 32.5 cm² and Perimeter of remaining part is 23 cm