# 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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