Area of a triangle

Geometry Level pending

In the above figure, A B C D ABCD and D E F G DEFG are squares. Given that D G = 6 DG=6 , find the area of △ B G E \triangle BGE .


The answer is 18.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

We know that E G ∥ B D EG \parallel BD . Construct rectangle B K H G BKHG as shown in my figure.

A B G E = 1 2 ( E G ) ( B K ) = 1 2 ( E G ) ( D H ) A_{BGE}=\dfrac{1}{2}(EG)(BK)=\dfrac{1}{2}(EG)(DH)

The area of △ E D G = 1 2 ( E G ) ( D H ) \triangle EDG=\dfrac{1}{2}(EG)(DH) .

Therefore, area △ B G E = \triangle BGE= area △ E D G \triangle EDG .

Finally,

A B G E = A E D G = 1 2 ( 6 ) ( 6 ) = A_{BGE}=A_{EDG}=\dfrac{1}{2}(6)(6)= 18 \boxed{18}

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Nice. You were right, Marvin. Sorry it took me so long to understand this.

Marta Reece - 3 years, 11 months ago
Marta Reece
Jun 17, 2017

Area of yellow triangle is two squares minus three triangles

a 2 + 6 2 − 1 2 ( a + 6 ) a − 1 2 × 6 2 − 1 2 ( a − 6 ) a = a 2 + 36 − a 2 2 − 3 a − 18 − a 2 2 + 3 a = 18 a^2+6^2-\dfrac12(a+6)a-\dfrac12\times6^2-\dfrac12(a-6)a=a^2+36-\dfrac{a^2}2-3a-18-\dfrac{a^2}2+3a=\boxed{18}

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