Trapezium Area

In trapezoid ABCD, let E be a point on DC and BE be parallel to AD s.t. ABED is a parallelogram,

then BE=AD=7, DE=EC=25 (opposite sides of a parallelogram are equal).

By Heron's formula,
Area of triangle BCE = \sqrt{s(s-a)(s-b)(s-c)}
where a,b,c are sides of triangle BCE, s is semi-perimeter of triangle BCE i.e. s=(a+b+c)/2

by no loss of generality let a=7,b=24,c=25

then s=(7+24+25)/2 =28

\Rightarrow Area of triangle BCE = \sqrt{28(28-7)(28-24)(28-25)} = 84

again, Area of triange BCE=CE \times AF /2

where BF is perpendicular on DC

\Rightarrow 84=25\times AF /2

\Rightarrow BF=6.72

Area of trapezoid ABCD= (AB+CD) \times BF/2
                                   = (25+50) \times 6.72/2
                                   = 252 sq. units  (Ans.)

If we "remove" the rectangle formed by drawing altitudes down from A and B down to CD, we form a rectangle of length 25 and height of h. Reassembling the triangles formed after the rectangle's removal gives us a triangle with edges of 7, 24, and 50-25, or 25. A 7-24-25 triangle is a right triangle, so the area is 84. Dividing by 25 and multiplying by 2 will give the height of the triangle, 168/25. Plugging this into the trapezoid area formula yields

$(\frac{168}{25})(75)(\frac{1}{2}) = \frac{168}{2} \cdot 3 = 252$

Let $E$ is midpoint of $DC$ . We have $AB=DE=25$ and $AB$ is parallel to $CD$ . So, we have $ABED$ is a parallelogram. We have $\angle ADE = \angle BEC,$ and $AD=BE=7,$ so $\Delta ADE=\Delta BEC = \Delta \Delta EBA$ . So, we have $S(ABCD)=3S(BEC).$ Use Heron's formula, we obtain $S(BEC)=84 \Rightarrow S(ABCD)=252.$

[Latex edits]

construct 3 triangles i.e right angled triangles....with bases 24 and 7 and add up their areas i.e 24x7x3x( .5 ) as the hypotenuse will be 25 in all the cases

Two parallel sides of Trapezium includes $25$ and $50=(x+25+(25-x))$ and let the perpendicular distance between the parallel sides be y. So, by Pythagoras Theorem, $y^2=7^2-x^2$ and $y^2=24^2-(25-x)^2$ and thus equating $y^2$ from both equations we get $x=1.96$ thus y comes out to be $6.72(=y)$ . So, the area of the trapezium will be $Area=\frac 1 2 \times Sum of Parallel Sides \times Perpendicular Distance b/w Them$ $Area=\frac 1 2 \times 75 \times 6.72=252$

construct heights from vertices A & B to CD at M and N, also AM=BN (parallel lines are at equal distances). now:- AM^2+MD^2=49 and BN^2+NC^2=24^2, also CD-AB=MD-NC; 25-MD=NC. now simply putting the value in the above two equations and then subtracting them; =>576-49=(25-MD)^2-MD^2+AM^2-BN^2 =>527=625-50MD =>MD=1.96 =>49-MD^2=AM^2 =>AM=6.72 =>AM/2(AB+CD)=3.36*75=252

Since the area of a trapezoid is (a+b)h/2 ,where a and b are the length of the parallel sides,and h is the height.All are given except the height. in order to find height we use,

   square root((-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)/2|b-a|,the denominator is not included in the square root.

where c and d are the legs. which gives the height to be 336/50,simplified to 168/25.

Giving the area of the trapezoid to be 252

AB is parallel to DC. Draw two perpendiculars AQ & BP to CD from A & B. Now we have two right angle triangles having height H. BC & AD will be the hypotenuse for the triangles.

Let DQ = a. So, CP = CD - 25 - a = 25 - a. Now in triangle AQD, H^2 = 24^2 - (25 - a)^2 => H^2 = 50a - a^2 - 49 --------(1)

In triangle BCP, H^2 = 7^2 - a^2 = 49 - a^2 --------(2) from (1) & (2), a = 1.96 So, H = 6.72

Area ABCD = (25+50) X 6.72 / 2 = 252

CD=y+z+25 therefore x+y=25

By Heron theorem the area of two side triangules of trapezium is: 84 where semiperimeter is 28 Now we can get the heigth that will be (84*2)/25=6.72

So the area of a trapezium is (B+b)h/2=(75)(6.72)/2=252

assume that CD=50=x+y+25

x is adjacent of CD which is hypotenuse

y is adjacent of AD which is also hypotenuse

the square trapesium altitude or Z^2 = CD^2-x^2 = AD^2-y^2 while x+y=25 (look at the first line)

therefore CD^2-AD^2 = X^2-Y^2

=24^2-7^2= (x+y)(x-y)

x-y=21.08 solve this equation with the first equation and you'll get x=23.04

Z^2=24^2-23.04^2

z=6.72

area = 6.72*(25+50)/2= 252