Quadrilaterals and Polygons
Let the height be h. Base be 20 top be 12.
Draw perpendiculars from the edges of the top to the base.
We have left overs of the base on both the sides. call them x and y.
Now X+Y =8
Now sum of the diagonals is nothing but h^2 + (20-y)^2 + h^2 + (20-x)^2
it comes out to be 800-40(x+y)+2 h^2 + X^2 + Y^2
We have h^2+X^2 = 9^2 and h^2+ Y^2 = 15^2 and X + Y = 8
Ans = 786.
Please solve this question:
ABCD is a trapezium, such that AB, DC are parallel and BC is perpendicular to them. If angle DAB = 45o, BC = 2 cm and CD = 3 cm then AB = ?
(A) 5 cm (B) 4 cm
(C) 3 cm (D) 2 cm
A dog is tied to the corner of a house with a regular hexagonal base that measures 6 ft on each side. If the rope is 12 ft in length, what is the area in square feet of the region outside the house that the dog can reach?
Hey sir, posting a question below. Though I've solved the question with the approach mentioned in the image (solution with blue pen), but is there any way to solve it faster? If somehow we could prove triangle PBC as a right angled triangle, it's just a 10sec question. But my question is, how could we do that?
In regular octagon ABCDEFGH
Find the area of ACEG, if side of octagon is 2 cm