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04 Apr 2021, 22:45
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3 rules and concepts:
(1) in an isosceles triangle, the height drawn from the Vertex between the two equal sides perpendicular to he non-equal side will Bisect the non-equal side and will act as a the Median
(2) the Median, in any triangle, divides the triangle into 2 sub-triangles that have Equal Area
(3) for an obtuse triangle such as triangle BCE, you can take the height outside of the triangle by drawing a line from Vertex B perpendicular to a straight line extended from Base Side CE
The diagonal of the square = (side) * sqrt(2)
Connect the square’s diagonal from B to D = (1) * sqrt(2)
This gives you Isosceles Triangle BED
Next, extend the Median from Vertex E to side BD.
This Median, because it is between the 2 equal sides of the Isosceles Triangle, will meet BD at a perpendicular Angle.
Call the point where the Median hits BD —- point X
If we take Side CE as the Base of Triangle BCE
Then BX will be the perpendicular height drawn from the opposite Vertex B ———> which will equal (1/2) the Diagonal of the Square ———> (1/2) * sqrt(2)
Using side CE = 1 as the Base of the Triangle, this will be the perpendicular height.
Area of triangle BCE will therefore be:
(1/2) * (1) * (1/2) * sqrt(2)
Or
Sqrt(2) * (1/4)
Answer B
Posted from my mobile device
(1) in an isosceles triangle, the height drawn from the Vertex between the two equal sides perpendicular to he non-equal side will Bisect the non-equal side and will act as a the Median
(2) the Median, in any triangle, divides the triangle into 2 sub-triangles that have Equal Area
(3) for an obtuse triangle such as triangle BCE, you can take the height outside of the triangle by drawing a line from Vertex B perpendicular to a straight line extended from Base Side CE
The diagonal of the square = (side) * sqrt(2)
Connect the square’s diagonal from B to D = (1) * sqrt(2)
This gives you Isosceles Triangle BED
Next, extend the Median from Vertex E to side BD.
This Median, because it is between the 2 equal sides of the Isosceles Triangle, will meet BD at a perpendicular Angle.
Call the point where the Median hits BD —- point X
If we take Side CE as the Base of Triangle BCE
Then BX will be the perpendicular height drawn from the opposite Vertex B ———> which will equal (1/2) the Diagonal of the Square ———> (1/2) * sqrt(2)
Using side CE = 1 as the Base of the Triangle, this will be the perpendicular height.
Area of triangle BCE will therefore be:
(1/2) * (1) * (1/2) * sqrt(2)
Or
Sqrt(2) * (1/4)
Answer B
Posted from my mobile device
06 Apr 2021, 23:34
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Solution:
In the figure,
BE=DE (given)
EC(common)
BC=CD(sides of the square) SSS congruency for triangles BCE and CDE
Let the extended CE meet diagonal BD of the square at X.
=> Area of BCE = Area of BXE - Area of BXC
Area BXE = 1/2 * XB *XE
= 1/2 * (√2/2) * (XC+CE) (XB = 1/2 of the diagonal of the square with side 1 unit here)
= 1/2 *(√2/2) * (√2/2 + 1)
= (√2 + 1 ) / 4
Area BXC = 1/4 the the area of the square with side 1 = 1/4 unit square
Area BCE = Area of BXE - Area of BXC
=(√2 + 1 ) / 4 - 1/4
= √2 / 4 (option b)
Devmitra Sen
GMAT SME
In the figure,
BE=DE (given)
EC(common)
BC=CD(sides of the square) SSS congruency for triangles BCE and CDE
Let the extended CE meet diagonal BD of the square at X.
=> Area of BCE = Area of BXE - Area of BXC
Area BXE = 1/2 * XB *XE
= 1/2 * (√2/2) * (XC+CE) (XB = 1/2 of the diagonal of the square with side 1 unit here)
= 1/2 *(√2/2) * (√2/2 + 1)
= (√2 + 1 ) / 4
Area BXC = 1/4 the the area of the square with side 1 = 1/4 unit square
Area BCE = Area of BXE - Area of BXC
=(√2 + 1 ) / 4 - 1/4
= √2 / 4 (option b)
Devmitra Sen
GMAT SME
