How the area of largest triangle and rectangle gets equal?

0 votes

A rectangle of the largest possible area is cut out from a semi-circle of perimeter 72 cm. What is the area of the rectangle cut out? ( Take n = 22 / 7 )

- 196 cm
^{2} - 156.8 cm
^{2} - 210 cm
^{2} - 528 cm
^{2}

0 votes

It's 196cm^2.

Required area is equivalent to finding the area of largest possible triangle that can be inscribed in the semi circle.

The largest possible triangle that can be inscribed inside a semicircle is the right angled triangle whose hypotenuse is the diameter of the semicircle, and whose other two sides are equal.

The area of the largest triangle that can be inscribed in a semicircle of radius r = r^2.

Radius of the semicircle is 14cm. (perimeter of semi-circle = 2r + 22r/7 = 72)

So 14^2 = 196cm^2.

Required area is equivalent to finding the area of largest possible triangle that can be inscribed in the semi circle.

The largest possible triangle that can be inscribed inside a semicircle is the right angled triangle whose hypotenuse is the diameter of the semicircle, and whose other two sides are equal.

The area of the largest triangle that can be inscribed in a semicircle of radius r = r^2.

Radius of the semicircle is 14cm. (perimeter of semi-circle = 2r + 22r/7 = 72)

So 14^2 = 196cm^2.

0

On concatenating two semicircles of same radius that each of which will contain the largest possible rectangle in them, then the resulting circle must contain the largest possible rectangle that can be inscribed in the circle.

Suppose the largest rectangle that can be inscribed inside the circle has area ‘2a’, then the area of each of the largest rectangles that can be inscribed inside the semicircle must be ‘a’.

Also the diagonal of any rectangle divides the rectangle into two right angle triangles of exactly same area (and the dividing diagonal will become hypotenuse of each of this right angled triangle).

So if any rectangle has an area of ‘2a’ then its diagonal will divide it into two right angled triangles of area ‘a’ each.

Now two ends points of a diameter of a circle always subtends a right angle at any point in the perimeter of the circle, & if any two end points of a chord are subtending a right angle at any point in the perimeter of a circle then the chord must be a diameter of that circle.

So instead of breaking the rectangle of area ‘2a’ inscribed in the circle into two rectangles each of area ‘a’, we can also break it into two right angled triangles each of area ‘a’.

& since the rectangle is inscribed in the circle, so its diagonals must be diameters of the circle, since they are subtending right angles at the perimeter of the triangle.

& since diameters divides a circle into two semicircles, each of the semicircle will contain a right angled triangle of area ‘a’.

Also to maximize area of a triangle we must try to maximize the dimensions of the triangle within the given constraint, thus in order to create the largest possible triangle inside a semicircle, its base must lie on the diameter of the semicircle.

Now out of all the triangles that can be inscribed in a semicircle, & which have diameter of the circle as their diameter, the triangle with maximum area will be the triangle for which sin(theta) * cos(theta) is maximum where theta is the angle between the diameter of the semicircle and any one of the side remaining two sides of the triangle that is inscribed in the semicircle.

Thus we can replace largest possible rectangle with the largest possible triangle within a semicircle.

Suppose the largest rectangle that can be inscribed inside the circle has area ‘2a’, then the area of each of the largest rectangles that can be inscribed inside the semicircle must be ‘a’.

Also the diagonal of any rectangle divides the rectangle into two right angle triangles of exactly same area (and the dividing diagonal will become hypotenuse of each of this right angled triangle).

So if any rectangle has an area of ‘2a’ then its diagonal will divide it into two right angled triangles of area ‘a’ each.

Now two ends points of a diameter of a circle always subtends a right angle at any point in the perimeter of the circle, & if any two end points of a chord are subtending a right angle at any point in the perimeter of a circle then the chord must be a diameter of that circle.

So instead of breaking the rectangle of area ‘2a’ inscribed in the circle into two rectangles each of area ‘a’, we can also break it into two right angled triangles each of area ‘a’.

& since the rectangle is inscribed in the circle, so its diagonals must be diameters of the circle, since they are subtending right angles at the perimeter of the triangle.

& since diameters divides a circle into two semicircles, each of the semicircle will contain a right angled triangle of area ‘a’.

Also to maximize area of a triangle we must try to maximize the dimensions of the triangle within the given constraint, thus in order to create the largest possible triangle inside a semicircle, its base must lie on the diameter of the semicircle.

Now out of all the triangles that can be inscribed in a semicircle, & which have diameter of the circle as their diameter, the triangle with maximum area will be the triangle for which sin(theta) * cos(theta) is maximum where theta is the angle between the diameter of the semicircle and any one of the side remaining two sides of the triangle that is inscribed in the semicircle.

Thus we can replace largest possible rectangle with the largest possible triangle within a semicircle.