Largest area Problem | AMC 8, 2003 | Problem 22

Try this beautiful problem from Geometry based Largest area.

Largest area - AMC-8, 2003 - Problem 22


The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

Problem figure
  • $A$
  • $B$
  • $C$

Key Concepts


Geometry

Circle

Square

Check the Answer


Answer:$C$

AMC-8 (2003) Problem 22

Pre College Mathematics

Try with Hints


To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares....

Can you now finish the problem ..........

area of circle =\(\pi r^2\)

can you finish the problem........

Largest  area Problem

In A:

Total area of the square =\(2^2=4\)

Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)

Area of the inscribed circle is \(\pi (1)^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In B:

Largest  area Problem - figure

Total area of the square =\(2^2=4\)

There are 4 circle and radius of one circle be \(\frac{1}{2}\)

Total area pf 4 circles be \(4 \times \pi \times (\frac{1}{2})^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In C:

finding the largest area

Total area of the square =\(2^2=4\)

Now the length of the diameter = length of the diagonal of the square=2

Therefore radius of the circle=\(\pi\) and lengthe of the side of the square=\(\sqrt 2\)

Thertefore area of the shaded region=Area of the square-Area of the circle=\(\pi (1)^2-(\sqrt 2)^2\)=\(\pi – 2\)

Therefore the answer is  C

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Covex Cyclic Quadrilateral | PRMO 2019 | Question 23

Try this beautiful problem from the Pre-RMO, 2019 based on Covex Cyclic Quadrilateral.

Covex Cyclic Quadrilateral - PRMO 2019


Let ABCD be a convex cyclic quadrilateral. Suppose P is a point in the plane of the quadrilateral such that the sum of its distance from the vertices of ABCD is the least. If {PA,PB,PC,PD}={3,4,6,8}, find the maximum possible area of ABCD.

  • is 107
  • is 55
  • is 840
  • cannot be determined from the given information

Key Concepts


Largest Area

Quadrilateral

Distance

Check the Answer


Answer: is 55.

PRMO, 2019, Question 23

Geometry Vol I to IV by Hall and Stevens

Try with Hints


Let \(\angle APB= \theta\)

area of \(\Delta PAB\)=\(\frac{1}{2}(3)(4)sin\theta\)

area of \(\Delta PAD\)=\(\frac{1}{2}(3)(6)sin(\pi-\theta)\)

area of \(\Delta PDC\)=\(\frac{1}{2}(8)(6)sin{\theta}\)

area of \(\Delta PCD\)=\(\frac{1}{2}(8)(4)sin(\pi-\theta)\)

or, area of quadrilateral ABCD is \(\frac{1}{2}(12+18+48+32)sin{\theta}\)

or, maximum area of ABCD=6+9+24+16=55.

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