Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
544 views
in Perimeter and Area of Plane Figures by (23.5k points)
closed by

Find the area of regular octagon with each side ‘a’ cm.

(a) 2a2 (1 + √2) 

(b) √2a (1 + π) 

(c) a2 (√2 + 2) 

(d) None of these

1 Answer

+1 vote
by (24.0k points)
selected by
 
Best answer

(a) 2a2 (1 + √2) 

Let ABCDEFGH be the regular octagon of side a cm. 

Now if we produce the sides of the octagon on both the sides, we get a square PQRS. 

Given, BC = DE = FG = HA = a cm. 

Also, 

BQ = QC = DR = RE = FS = SG = HP = PA = \(\frac{a}{\sqrt2}\)

( BQC, DRE, FSG, HPA are rt. ∠d Δs)

∴ Side of square = \(\frac{a}{\sqrt2}\) + a + \(\frac{a}{\sqrt2}\) = a (1 + √2) cm.

∴ Area of square = a2(1 + √2)2 cm2 = a2(3 + 2√2) cm2 

Each of the shaded Δs, APH, BCQ, DER, FGS is an isosceles right angled Δ, whose area

\(\frac{1}{2}\times\frac{a}{\sqrt2}\) x \(\frac{a}{\sqrt2}\) cm2\(\frac{a^2}{4}\) cm2

∴ Area of octagon = Area of square PQRS – Total area of shaded isosceles Δs = a2 (3 + 2√2) - a2

2a2 (1 + √2) cm2.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...