Question

coplanar squares abgh and bcdf are adjacent, with cd=10 units and ah=5 units. point e is on segments ad and gb what is the area of triangle abe, in square units? express your answer in a common fraction.

Answer 1

Answer-

$\boxed{\boxed{\text{Area}_{ABE}=\dfrac{25}{3}\ unit^2}}$

Solution-

ΔABE ~ ΔACD by AA (Angle-Angle) similarity, as

1. m∠BAE = m∠CAD (as ∠A is common to both)
2. m∠ABE = m∠ACD = 90° (As each angles of a square is 90°)

According to the similarity of triangles,

$\Rightarrow \dfrac{AB^2}{AC^2}=\dfrac{\text{Area}_{ABE}}{\text{Area}_{ACD}}$

$\Rightarrow \dfrac{AB^2}{(AC)^2}=\dfrac{\text{Area}_{ABE}}{\frac{1}{2}\times AC\times CD}$

$\Rightarrow \dfrac{AB^2}{(AB+BC)^2}=\dfrac{\text{Area}_{ABE}}{\frac{1}{2}\times (AB+BC)\times CD}$

$\Rightarrow \dfrac{5^2}{(5+10)^2}=\dfrac{\text{Area}_{ABE}}{\frac{1}{2}\times (5+10)\times 10}$

$\Rightarrow \dfrac{5^2}{15^2}=\dfrac{\text{Area}_{ABE}}{\frac{1}{2}\times 15\times 10}$

$\Rightarrow \dfrac{25}{225}=\dfrac{\text{Area}_{ABE}}{75}$

$\Rightarrow \text{Area}_{ABE}=\dfrac{25\times 75}{225}$

$\Rightarrow \text{Area}_{ABE}=\dfrac{25}{3}\ unit^2$