A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF.
Solution:
Given, regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB.
We have to find ∠BCF.
We know, measure of each interior angle of a regular pentagon = (n - 2)/n × 180°
Where n is the number of sides.
Here, n = 5
So, each interior angle = (5 - 2)/5 × 180°
= 3/5 × 180°
= 3 × 36
= 108°
So, ∠CBA = 108°
Join CF.
Each angle of square = 90°
We know, ∠FBC = 360° - (∠ABF + ∠CBA)
∠FBC = 360° - (90° + 108°)
= 360° - 198°
= 162°
Considering triangle FBC,
By angle sum property of a triangle,
∠FBC + ∠BCF + ∠BFC = 180°
162° + ∠BCF + ∠BFC = 180°
∠BCF + ∠BFC = 180° - 162°
∠BCF + ∠BFC = 18°
FBC is an isosceles triangle, so BF = BC
We know that the angles opposite to the equal sides are equal.
So, ∠BCF = ∠BFC
Now, ∠BCF + ∠BFC = 18°
∠BCF + ∠BCF = 18°
2∠BCF = 18°
Therefore, ∠BCF = 9°
✦ Try This: In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of ∠BAE.
☛ Also Check: NCERT Solutions for Class 8 Maths
NCERT Exemplar Class 8 Maths Chapter 5 Problem 182
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF.
Summary:
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. the value of ∠BCF = 9°.
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