This question was uploaded on 14/08/21 on social media accounts.
A square with side 6 units and one of its diagonal between the 2nd and 4th points are drawn. A point 4 units from 2nd point and 2 units from 3rd point is taken. Two lines are drawn from that point to 1st point and point at center of 1st and 4th point. Find the area formed by these two lines, diagonal and the line between 1st and 4th point.
Step 1:
Give names to all vertex and intersection of lines.
Step 2:
Observe triangles ADG and FBG:
Both have similar angles,
⇒
△ADG
~ △FBG⇒ AG / FB = AG / FG
⇒ AG / FG = 6 / 4
Let, AG = 6a and FG = 4a
Now,
Observe triangles EDH and FBH:
Both have similar angles,
⇒
△EDH
~ △FBH⇒ ED / FB = EH / FH
⇒ EH / FH = 3 / 4
Let, EH = 3b and FH = 4b
Step 3:
Area of triangle AEF = ½×AE×AB = ½×3×6 = 9
Also,
Area of triangle AEF = ½×FA×FE×sinθ
⇒ 9 = ½×10a×7b×sinθ
⇒ absinθ = ⁹⁄₃₅
A(AEHG) = A(1) = A(AEF) - A(2)
⇒ A(1) = 9 - ½×FG×FH×sinθ
⇒ A(1) = 9 - ½×4a×4b×sinθ
⇒ A(1) = 9 - 8×absinθ
⇒ A(1) = 9 - 8×⁹⁄₃₅
⇒ A(1) = ²⁴³⁄₃₅
Image solutions:
Solution from @geometri_hayattir
Solution from @jay_jp_35
Solution from @l.lawliet1337
Solution from @arminski42
Puzzle related to Geometry, Square, Area
