MATHS PUZZLE # 10 FIRST LEG (Second Year)

MATHS PUZZLE# 10 (from the 6th to the 13th of December)

THE MIDDLE LOT

28 Hm²  . Can you calculate the area of the shaded lot? 
Andreu has a triangle-shaped piece of land which is divided in 7 different lots that have all the same width, as you can see in the picture below. We know the area of the whole land is  

You can send your answers to matecampanar@gmail.com

SOLUTION

Let us call:

BMT = The longer base of the trapezium

BmT = The smaller base of the trapezium

h = The height of the trapezium

B = The base of the triangle

H = The height of the triangle

We will use now Thales's theorem:

BMT/H = 4/7 → BMT = 4H/7

BmT/H = 3/7 → BmT = 3H/7 ; adding now BMT + Bmt = 4H/7 + 3H/7 = H

On the other hand , we know from the area of the triangle that = 28 = BH/2 →B = 56/H

Besides,

h = B/7 = 56/(7H) = 8/H

Hence the area of the trapezium is: A = (BMT + BmT)·h/2 = (H·8/H)/2 = 4 Hm2

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There is another more intuitive but less rigorous way to solve this problem. This is it:

The triangle is divided into six trapezia and one triangle. Some are bigger, some of them smaller. The trapezium in the middle is just between the bigger trapezia and the smaller ones.. It is logical to think thus its area to be:28/7= 4 Hm2