CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that ∆ ADE ≅ ∆ BCE.
Solution:
Given, CDE is an equilateral triangle formed on the side CD of a square ABCD.
We have to show that the triangles ADE and BCE are congruent.
We know that sides of a square are equal and each angle is equal to 90 degrees.
AB = BC = CD = AD
∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°
We know that in an equilateral triangle all the sides are equal and each angle is equal to 60 degrees.
CD = CE = DE
∠CDE = ∠DCE = ∠DEC = 60°
Considering triangles ADE and BCE,
Sides of a square, AD = BC
Sides of equilateral triangle, DE = CE
∠ADE = ∠ADC + ∠CDE
∠ADE = 90° + 60°
∠ADE = 150°
∠BCE = ∠BCD + ∠DCE
∠BCE = 90° + 60°
∠BCE = 150°
∠ADE = ∠BCE
By SAS criterion, ΔADE ≅ΔBCE
Therefore, the triangles ADE and BCE are congruent.
✦ Try This: In the given figure, DE || BC. If DE = 3cm, BC = 6 and ar(ΔADE) = 15 square cm, find the area of Δ ABC.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.3 Problem 3
CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that ∆ ADE ≅ ∆ BCE
Summary:
CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). It is shown that ∆ ADE ≅ ∆ BCE by SAS criterion
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