ABCDE is a regular pentagon. On the outside of AB, we construct a square ABFG. What is the measure (in degrees) of ∠EAG?
Angle EAB + Angle BAG
108 + 90 = 198 degrees
The above answer is incorrect. Its is:
360-(108+90)= 360-198=162
in the diagram ,abc and chk are congruent .given that ab =14 cm ac=18 cm ,hk= 27 cm,bac=chk=114 and abc = 38 ,find chk the length of kb
To find the measure of ∠EAG, we can start by considering the properties of a regular pentagon and the square that is constructed on one of its sides.
1. A regular pentagon has five equal sides and five equal angles. Each angle of a regular pentagon measures 108 degrees (360 degrees divided by 5).
2. In a square, all four angles are right angles, which means each angle measures 90 degrees.
Now let's apply this information to our problem.
1. Since AB is a side of the regular pentagon ABCDE, angle B measures 108 degrees.
2. In the square ABFG, angles B and F are right angles and therefore measure 90 degrees.
3. Angle EAG is an exterior angle of the triangle ABE.
Here's how we can find the measure of ∠EAG:
1. The sum of the measures of the exterior angles of any polygon is always 360 degrees.
2. In triangle ABE, the sum of angles A, B, and E is 180 degrees (Triangle Angle Sum Theorem).
3. Since angle B is already known to be 108 degrees, angle A + angle E + angle G = 180 degrees - 108 degrees = 72 degrees (Subtracting known angles from 180 degrees).
4. The remaining two angles in triangle ABE, A and E, are equal since triangle ABE is an isosceles triangle with sides AE and BE being congruent.
5. Therefore, angle A = angle E = 72 degrees ÷ 2 = 36 degrees.
6. Finally, angle EAG is equal to angle A because they are corresponding angles formed by a transversal (EA) intersecting two parallel lines (AB and FG).
Hence, the measure of ∠EAG is 36 degrees.