Try this beautiful problem from Geometry based on the Area of the quadrilateral.

## Area of the quadrilateral – AMC-10A, 2020- Problem 20

Quadrilateral \(ABCD\) satifies \(\angle ABC= \angle ACD=90^{\circ}\), \(AC=20\) and \(CD=30\). Diagonals \(AC\) and \(BD\) intersect at the Point \(E\) and \(AE=5\). What is the area quadrilateral \(ABCD\)?

- \(330\)
- \(340\)
- \(350\)
- \(360\)
- \(370\)

**Key Concepts**

Geometry

Triangle

Area

## Check the Answer

Answer: \(360\)

AMC-10A (2020) Problem 20

Pre College Mathematics

## Try with Hints

Given that \(AC=20\) and \(Cd=30\).Area of \(\triangle ACD=300\).Now if we have to find out the area of \(\triangle ABC\) then we can find out the area of whole \(ABCD\).Now draw altitude from $B$ to $AC$ and call the point of intersection $F$.Let \(FE=x\) .Noe \(AE=5\) then \(AF=5-X\).Now observe that the \(\triangle BEF \sim \triangle DCE\).

Can you now finish the problem ……….

Now $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$. Now, if we drawn another diagram $ABC$, we get that $(2x)^2=(5-x)(15+x)$ as we know that the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into.Now can you find out expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$

can you finish the problem……..

Now $x^2+2x-15=0$\(\Rightarrow (x-3)(x+5)=0\)\(\Rightarrow x=3,-5\) as length can not be negetive then \(x=3\).So area of \(\triangle ABC=60\)

Therefore the total Region \(ABCD\)= are of \(\triangle ABC\) + area of \(\triangle ACD\)=\(300+60\)=\(360\)

## Other useful links

- https://www.cheenta.com/problem-based-on-lcm-amc-8-2016-problem-20/
- https://www.youtube.com/watch?v=7AlfBAPWEMg