Question

6. If ABCD is an A4 sheet and BCPO is the
square, prove that AOCD is an isosceles
triangle. And find the angles marked as 1 to
8without usingaprotractor.ABCD एक A4
पेपर है, जिसमें BCPO सबसे बड़ा वर्ग है। सिद्ध करें कि
AOCD एक समद्विबाहु त्रिभुज है। और दिए गए चित्र में
चांदे का उपयोग किए बिना सभी 8 कोणों का पता लगाएं।

Answer 1

Step-by-step explanation:

Given:If ABCD is an A4 sheet and BCPO is the square.

To find: Prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 without using a protractor.

Solution:

As to draw largest square in A4 sheet,Sheet should be fold breadth wise.

BCPO is a square,its diagonal OC =CD

Thus,

In triangle OCD ; OC= CD

So,OCD is an isosceles triangle.

We know that angle opposite to equal sides are equal.

So,

$\angle CDO = \angle COD \\ \\ \angle 6 = \angle 4 + \angle5$

OC is diagonal,it bisect the angle C

$\angle 2 = \angle 3 = 45°$

By the same way

$\angle 1 = \angle4 = 45°$

$\angle CDO + \angle COD + \angle DOC = 180° \\ \\ \angle CDO + \angle COD + 45° = 180° \\ \\ 2\angle CDO = 135° \\ \\ \angle CDO = 67.5° \\ \\ \angle 6 = 67.5° \\ \\ \angle 4 + \angle 5 = 67.5°$

In Silver Rectangle AOPD,all angles are 90°

$\angle D = 90° \\ \\ \angle D = \angle 7 + \angle 6 = 90° \\ \\ \angle 7 + 67.5° = 90° \\ \\ \angle 7 = 22.5°$

Line DP||AO

So,

$\angle 7 = \angle5 = 22.5° \\ \\ (alternate \: interior \: angle)$

thus,

$\angle 6 = \angle8 = 67.5° \\ \\ (alternate \: interior \: angle)$

Thus

$\angle 7 = \angle5 = 22.5° \\ \\ \angle 6 = \angle8 = 67.5° \\ \\ \angle 1 = \angle2 =\angle 3 = \angle4 = 45°$

Hope it helps you.

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