abcd is a parallelogram with sides ab=12 cm,bc=10 cm and diagonal ac=16 cm .find the area of the parallelogram also find the distance between its shortest sides
119.8cmsquare;11.98cm
119.8cm²and 11.98cm
To find the area of a parallelogram, you can use the formula: Area = base × height.
In this case, we need to find the height of the parallelogram. The height is the perpendicular distance between the base and the opposite side. Since we are given the length of sides AB and BC, we can use the formula for the height of a parallelogram:
height = (2 × Area) / base
First, let's find the area. We know that the diagonal AC divides the parallelogram into two congruent triangles. We can use the formula for the area of a triangle, which is half the product of the base and height.
We have two congruent triangles, so the area of each triangle is:
area of triangle = (1/2) × base × height
Given that AC is the diagonal and its length is 16 cm, we can use the Pythagorean theorem to find the length of AB:
(AB)^2 + (BC)^2 = (AC)^2
(AB)^2 + 10^2 = 16^2
(AB)^2 + 100 = 256
(AB)^2 = 256 - 100
(AB)^2 = 156
AB = √156
Now we can substitute the values into the formula for the area:
area of triangle = (1/2) × base × height
area of triangle = (1/2) × √156 × height
Since the two triangles are congruent, the area of the parallelogram is twice the area of one of the triangles:
Area = 2 × area of triangle
Now, let's solve for height:
height = (2 × Area) / base
height = (2 × (2 × area of triangle)) / √156
Substituting the values we have:
height = (2 × (2 × (1/2) × √156 × height)) / √156
Cross-multiplying, we get:
height × √156 = 4 × √156 × height
Dividing both sides by √156:
height = 4
Finally, we can substitute the values for the base and height into the formula for the area:
Area = base × height
Area = √156 × 4
Area = 4√156
Therefore, the area of the parallelogram is 4√156 square units.
Now, let's find the distance between the shortest sides of the parallelogram. Since ABCD is a parallelogram, AB is parallel to CD, and BC is parallel to AD. The distance between two parallel lines is the perpendicular distance between them.
In this case, the perpendicular distance between AB and BC is the height of the parallelogram, which we have already found to be 4 cm.
Therefore, the distance between the shortest sides of the parallelogram is 4 cm.