If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of

How can we determine the triangle heights? I've read through the above and haven't been able to grasp it. I'm sure its very simple, so thanks in advance, and excuse my oversight.

No problem!!


Notice that Triangle BAE is right angled at A and Angle A=90..
So BA is perpendicular to AE...so Area of ABE will be 1/2*3*4=6
Area of Bigger Triangle CAD (right angled at A) =1/2*6*8=24

So area of Trapezium is Area of CAD -Area of BAE= 24-6=18

Consider the triangle ABE,
The two sides are 3 and 4. Hence the third side would be 5 (3 - 4 - 5 right triangle)
Area = 1/2 * 3 * 4 = 6

In triangle ABC,
The sides of this triangle are twice the sides of triangle ABE, hence area would be 2^2 times.
Area of ABC = 4*6 = 24

Area od the trapezoid = 24 - 6 = 18

Correct Option: B

Hi Bunuel, Could you help to explain the area? Why we could use the side to calculate the area of triangle?

While formula for area of triangle = (1/2)*(Base)*(Height)

ABE and ACD are right triangles: area = 1/2*(leg1)(leg2), which is the same as (1/2)*(Base)*(Height), because legs in a right triangle are perpendicular to each other (so we can consider one of them to be base and another to be the height).

Hi Bunuel,

Why are we considering the height as 8 and 4 can u pls tell me ?

How can 8 and 4 be the height?

A height (altitude) of a triangle is the perpendicular from a vertex to the opposite side. Because legs (non-hypotenuse sides) in a right triangle are perpendicular to each other we can consider one of them to be the base and another to be the height.

Can we not directly use formula of trapezium rather than finding area of smaller triangle and bigger triangle and then subtracting?

The area of a trapezoid is (a + b)/2*h, where a and b is the lengths of the parallel sides and h is the length of the height. To apply it directly we are missing the lengths of BE and the height.

Yes.

Any side of a triangle can be deemed the base.
Each base has a corresponding height.
No matter which base-height combination is used, the area of the triangle must be the same value.
Implication:
Regardless of which side of a triangle is deemed the base, bh must always yield the same product.

Triangle ABE:
If AB=3 is considered the base, then AE=4 is the corresponding height.
If BE=5 is considered the base, let AF = the corresponding height.
Since each base-height combination must yield the same product, we get:
\(5AF = 3*4\)
\(AF = \frac{12}{5}\)

Triangle ACD:
If AC=6 is considered the base, then AD=8 is the corresponding height.
If CD=10 is considered the base, let AG = the corresponding height.
Since each base-height combination must yield the same product, we get:
\(10AG = 6*8\)
\(AG = \frac{48}{10} = \frac{24}{5}\)

Trapezoid BEDC:
height \(= AG - AF = \frac{24}{5} - \frac{12}{5} = \frac{12}{5}\)
area = (average of the two bases)(height) \(= \frac{BE+CD}{2} * \frac{12}{5} = \frac{5+10}{2} * \frac{12}{5} = 18\)