ABCD is a square with each side lenght 1 .Suppose BPC is an equilateral triangle ,find the area if the triangle BPD ...
Thank you Mr.Henry ... :)
To find the area of the triangle BPD, we can break it down into two smaller triangles BPC and CPD.
Since BCP is an equilateral triangle, we know that each angle in BCP is 60 degrees.
Now, let's calculate the area of triangle BPC:
The formula to find the area of an equilateral triangle is A = (s^2 * sqrt(3)) / 4, where s is the length of the side of the triangle.
In this case, the side length of triangle BPC is 1 (since it is half the length of the square ABCD). Plugging this value into the formula, we have:
A_BPC = (1^2 * sqrt(3)) / 4 = sqrt(3) / 4.
Similarly, we can calculate the area of triangle CPD. Since CPD is a right triangle and we know the length of one of the legs (CP), we can use the formula for the area of a right triangle: A = (base * height) / 2.
The base of triangle CPD is 1 (since it is the length of one side of the square) and the height is CP.
To find CP, we can use the Pythagorean theorem: CP^2 = BP^2 + BC^2. Since BP = 1 (length of the square side) and BC = CP/2 (since triangle BPC is an equilateral triangle), we have:
CP^2 = 1^2 + (CP/2)^2.
Solving this equation gives us: CP^2 = 1 + (CP^2) / 4.
Multiplying both sides by 4, we get: 4 * CP^2 = 4 + CP^2.
Simplifying further, we have: 3 * CP^2 = 4.
Dividing by 3, we find: CP^2 = 4/3.
Taking the square root of both sides, we get: CP = 2 / sqrt(3).
Now, we can calculate the area of triangle CPD:
A_CPD = (1 * CP) / 2 = (1 * (2 / sqrt(3))) / 2 = (2 / sqrt(3)) / 2.
Combining the areas of triangles BPC and CPD, we get the area of triangle BPD:
A_BPD = A_BPC + A_CPD = sqrt(3) / 4 + (2 / sqrt(3)) / 2 = (sqrt(3) + 2sqrt(3)) / 4 = 3sqrt(3) / 4.
So, the area of triangle BPD is (3sqrt(3) / 4).
As = 1 * 1 = 1 sq. unit. = Area of the
sq.
The 2 diagonals divide the sq. into 4
congruent triangles. The area of each triangle = 1/4 of total area:
A = As/4 = 1/4 sq. units = BPC and BPD.