Question : In a house society complex a central park has to be developed. The shape of the central park is shown in the figure. ABCD is a square & CDE is an equilateral triangle on the side CD of the square.
Find
i) ∠ADE
ii) ∠BCE
iii) ∠AED
iv) ∠AEB
v) ΔAEB is an isosceles triangle.
Doubt by Vikrant
Solution :
ABCD is square (Given)
∠ADC=∠BCD=90°
(All angles of squares are of 90°)
CBE is an equilateral triangle (Given)
∠EDC=∠ECD=60°
(All angles of an equilateral triangle are of 60°)
i) ∠ADE=∠ADC+∠EDC
∠ADE=90°+60°=150°
ii) ∠BCE=∠BCD+∠ECD
∠BCE=90°+60°=150°
iii) The base of the equilateral triangle is on one side of the square. All sides of square are equal and all sides of equilateral triangle are equal so we can say that AB=BC=CD=DA=DE=CE
In ΔADE
DA=DE (Proved above)
∠1=∠2 (Angles opposite to equal sides of a triangle are equal)
Now,
∠1+∠2+∠ADE=180° (ASP)
∠2+∠2+150°=180°
2∠2=180°-150°
2∠2=30°
∠2=30°/2
∠2=15°
∴ ∠AED=15°
Similarly ∠4=15°
OR ∠BEC=15°
iv) ∠CED=60° (Angle of equilateral triangle)
∠2+∠5+∠4=60°
15°+∠5+15°=60°
∠5+30°=60°
∠5=60°-30°
∠5=30°
∴ ∠AEB=30°
v) In ΔADE and ΔBCE
AD=BC (Opposite sides of squares are equal)
∠ADE=∠BCE (Each equals to 150°)
ED=EC (Sides of equilateral triangle are equal)
ΔADE ≌ ΔBCE (By SAS)
AE=BE (By CPCT)
Two sides of ΔAEB are equal hence we can say that ΔAEB is an isosceles triangle.