Question

question answers of chapter maps of class 7 of book gulmohar​

Answer 1

$\huge\mathcal\pink{Question:−}$

ABCD is a silver rectangle whose sides are in the ratio 1: (√ 2 - 1) . By assembling such 4 silver rectangles, we can made a regular octagon (as shown in the picture). What is the ratio of the area of silver rectangle to the area of the octagon? ABCD

$\huge\mathcal\pink{Answer:-}$

Here, ABCD is a square, each side = 1.7m

Regular Octagon PQRSTUV is inscribed in the Square ABCD . ie, all the vertices of regular octagon are lying on the Square.

As shown above every side of regular octagon = X

DC = DP + PQ + QC , Here, PQ = X & DP = QC = Y

Now, we form the equations :

2Y + X = 1.7 …………(1)

& in right triangle PDW

X ² = 2Y²

=> X = √2 Y …………….(2)

Eq (1) => 2Y + √2Y = 1.7

=> Y(2+√2 ) = 1.7

=> Y = 1.7 / (2+√2) * (2-√2) /(2-√2)

=> Y = 1.7 ( 2-√2) / 2

=> Y = [3.4 - 1.7√2 ] /2

=> Y = [ 3.4 - 1.7 * 1.414 ] /2

=> Y = [3.4 - 2.4038] / 2

=> Y = 0.9962 /2

=> Y = 0.4981 ………..(3)

So X = 1.7 - 2* 0.4981

=> X = 0.7038 ……….(4)

So, when we round up.

Y = 0.5, X = 0.7

DC = 0.5 + 0.7 + 0.5 = 1.7

ANS : side of the regular Octagon = 0.7 m (approx)

Verification:

0.7038² = 0.4981² + 0.4981²