Area of parallelogram

Question

ABC is equilateral triangle and AM perpendicular to BC. AD and MC are equal and parallel to each other. If AM= 6v3 cm , find the area of quadrilateral ABCD.

Answer 1

Hi Sam,

 

Below is the link to a visual diagram with notations confirming Arthur's earlier solution:

 

https://dl.dropbox.com/s/yinestamoo0r4y1/Diagram_and_Solution_for_Area_of_Quadrilateral_ABCD.png?raw=1

 

God bless, Jordan.

Answer 2

draw the diagram

draw triangle ABC

draw line segment AM perpendicular to line segment BC

draw line segment AD to the left, parallel to line segment MC and equal in length to MC

draw line segment DC to form rectangle ADCM

you now also have quadrilateral ABCD

now find the area of right triangle AMB

AM and MB are the legs and AB is the hypotenuse

AM bisects BC so call MB x and AB will therefore be 2x because you have an equilateral triangle

find the area of the triangle on the right using the Pythagorean Theorem first and then A=(1/2)bh

AM^2+MB^2=AB^2

(6√3)^2+x^2=(2x)^2

(36*3)+x^2=4x^2

108=3x^2

108/3=x^2

36=x^2

x=6 cm

A=(1/2)bh

A=(1/2)(6)(6√3)

A=18√3 sq cm

now find the area of rectangle AMCD

A=bh

A=MC*AM

A=6*6√3 because MC=MB (because of AM the perpendicular bisector of BC) and we said MB was x and x=6 cm

A=36√3 sq cm

now add the two areas

A=18√3+36√3

A=54√3 sq cm is the area of quadrilateral ABCD