Deploying laser beams

Geometry Level 3

Given a parallelogram $ABCD$ with area of $24\text{ cm}^2$ with points $E$ and $F$ as the midpoint of the lines $AB$ and $BC$ respectively. What is the area of the quadrilateral $EFGH$ ?

6\text{ cm}^2

4\text{ cm}^2

5\text{ cm}^2

7\text{ cm}^2

8\text{ cm}^2

3 solutions

Hadia Qadir
Jul 22, 2015

aEFGH = aDEF - aDGH aDEF = aABCD - aADE - aCDF - aBEF = 9 = 1/2 x EF x hDEF aDGH = 1/2 x GH x hDGH AI = FC = 1/2 BC = 1/2 AD => AD/DI = 2/3 AC // EF => AD/DI = HG/EF = hDGH/hDEF = 2/3 => aDGH = 1/2 x 2/3 EF x 2/3 hDEF = 4/9 x 1/2 x EF x hDEF = 4/9 x 9 = 4 aEFGH = aDEF - aDGH = 9 - 4 = 5 cm2

Tajnubal Hoque
Jun 30, 2015

At first divide the full quadrilateral into two. This gives us that ABC triangle is 12 cm^2. Now divide the shaded area into two triangles. Notice that one triangle is bigger than the other. Now compare all the triangles inside ABC. Notice that 3 are equally small and 2 are equally big. The smaller and bigger triangles add up to 12cm^2. The smaller triangles have to be 2cm^2 each and the biggers ones must be 3cm^2, as this the only ratio which helps them to add up to 12cm^2

pls elaborate the explanation.

om sai - 5 years, 11 months ago

Why should the smaller triangles be necessarily 2 or the bigger ones be necessarily 3...?? Many fractions may add up to be 12...

Istiak Reza - 5 years, 11 months ago

Sergio La Malfa
Jul 25, 2018

Deploying laser beams

3 solutions

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