Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

Area of trapezoid | AMC 8, 2011|Problem 20

Try this beautiful problem from Geometry: The area of trapezoid.

The area of the trapezoid - AMC-8, 2011 - Problem 20


Quadrilateral  ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?

area of trapezoid
  • $700$
  • $750$
  • $800$

Key Concepts


Geometry

Trapezoid

Area of Triangle

Check the Answer


Answer:$750$

AMC-8(2011) Problem 20

Pre College Mathematics

Try with Hints


Draw altitudes from the top points A and B to CD at X and Y points

Can you now finish the problem ..........

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times\) (height between AB and CD)

can you finish the problem........

a trapezoid

Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle .

Using The Pythagorean theorem on \(\triangle ADX and \triangle BYC\) ,

\((DX)^2+(AX)^2=(AD)^2\)

\(\Rightarrow (a)^2+(12)^2=(15)^2\)

\(\Rightarrow a=\sqrt{(15)^2-(12)^2}=\sqrt {81} =9\)

and

\((BY)^2+(YC)^2=(BC)^2\)

\(\Rightarrow (12)^2+(b)^2=(20)^2\)

\(\Rightarrow b=\sqrt{(20)^2-(12)^2}=\sqrt {256} =16\)

Now ABYX is a Rectangle so \(XY=AB=50\)

\(CD=DX+XY+YC=a+XY+b=9+50+16=75\)

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750\)

.

Subscribe to Cheenta at Youtube


Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com