KCC's Quizzes AQQ266 about a trapezoid area

As lines AB and CD are parallel, angle ABD = angle BDC = tan-1(5/10) = 26.6 degrees.

All three angles in a triangle sum to 180 degrees, so angle BCD is 63.4 degrees.

If we bisect triangle BDC with a line from B to a point perpendicular to line DC, then the smaller right-angled triangle has a side 5cm long.

Angle BCD is 63.4 degrees, so the other angle must be 26.6 degrees.

The other side will therefore be tan(26.6) * 5 = 0.5 * 5 = 2.5 cm.

The area of the small triangle is 0.5 * 5 * 2.5 = 6.25 cm^2

The area of the rectangle is 5 * 10 = 50 cm^2

The total area of the trapezoid id therefore 65.25 cm^2

The answer is 225/4.

The two triangles ABD and BDC are similar.

BD= 5*sqrt(5) (Pitagora).

Then BC=5*sqrt(5)/2 (because AB/AD=2)

You will then get the trapezoid area A by summing the areas of the two triangles.

A= 10*5/2 + 25*5/4= 225/4

First we can find out the angle DBA and DCB so sum of these two will be 90 degree then we can find tan(DBA) = 5/10 and tan(DBC) = 5/a. 

So a will be 5/2 and basically what a is so when I will draw a parallel line of AD going from B is named as O so a is OC means DC - 10.

DB will be sqrt(100+25) which is sqrt(125) 

Now we can find out BC which is CD(square) - DB(square) after calculations so BC is 5*sqrt(5) /2

• Area total is ABOD + BOC which is 10*5 + 1/2*5*5/2 which is around 65.25

* Area of the Trapezoid = Sum of the area of 2 right triangles

* For right triangles whose legs has the ratio of 1:2, Area = (shorter leg)² or Area = (longer leg/2)². Therefore, the area of the smaller triangle is equal to 25cm².

Area (small triangle) = bh/2 = (5*10)/2 = 25cm² or,

A = (shorter leg)² = 5² = 25cm².

* Since the 2 triangles are similar, the ratio of the legs of the larger triangle is also 1:2. We will first get the hypothenuse of the smaller circle by Pythagorean theorem.

c = √(5² + 10²) = 5√5cm

* Area (large triangle) = (c/2)² = (5√5/2)² = 31.25cm²

* Therefore, the area of the trapezoid = 25cm² + 31.25cm² = 56.25cm².

It is 56.25 cm^2.

If one draws a vertical line from B to DC (parallel to AD) and calls the intersection point E on DC, BE would be 5cm.
Like AD.

Acc to geometricmean theorem (known as Oeklid rules too)  BE^2 = DE * CE and DE=10cm (same length and parallel to AB )

--> CE=2.5cm

Then the trapezoidal area (AB+CD)*AD/2 = (10+12.5)*5/2 = 56.25 cm^2

Two angles having their sides perpendicular are equal if both are less than 90 degree or both are greater than 90 degrees.
So the two triangles are proportiannal, so we can write that the ration of their diagonal = their ratio of their longer side: 
D.C / B.D     =  B.D / A.B
from which we deduce, given the know values :  10* D.C  = 125,   or  D.C  =12.5
The average length of the trapezoid  is thus  11.25
The area of trapezoid is thus  5 * 11.25  = 56.25

Area of triangle ADB equals:  0.5x5x10 = 25 sq.cm

Length of common side DB equals:  sqrt(25+100) = 11.18 cm

Angle BDC equals:  90 – arctan(10/5) = 26.565 deg

Length of BC equals:  tan(26.565)x11.18 = 5.59 cm

Area of triangle BDC equals:  0.5x5.59x11.18 = 31.248 sq.cm

Area of trapezoid ABCD equals:  25+31.248 = 56.248 sq.cm

I got ~55,59 cm^2

Edit: I used the same strategy as retiredEE but rounded differently

Thanks Martin for your prompt reply. But you just made a little typo when you sum 50cm² to 6.25cm². It should be 56.25 cm² and not 65.25cm²....

Anyhow, congratulations!

Bingo Gaetano, you get it, congratulations!