#righttriangle

আজকের ভিডিওতে সহজ উদাহরণের মাধ্যমে দেখানো হয়েছে কীভাবে কোনো জটিল মুখস্থ না করেও ধাপে ধাপে অপর বাহুর দৈর্ঘ্য নির্ণয় করা যায়। বিষয়টি স্কুল, কলেজ ও চাকরির পরীক্ষার্থীদের জন্য খুবই উপকারী।

How can you find the missing side of a right triangle when the other two sides are known? In this video, you'll learn an easy step-by-step method with clear examples. Perfect for students preparing for school exams, competitive tests, and anyone who wants to strengthen their basic geometry skills.

Full video link:-

In the given figure, the hypotenuse of a right-angled triangle is trisected by two circles. If the lengths of these segments are a, b, and c respectively. Then prove:- a/b×√[c/(a+b)]=2

Let the lengths of the sides PQ, QR, and PR of △PQR be y, x, and z respectively. The length of the radius of the smaller semicircle is r.

Let ∠QPR=θ

Therefore ∠QRP=90°-θ

Draw ST and QU

Then ∠STP and ∠QUR=90°

Theorem: The angle subtended by a diameter of a circle at any point on its boundary is 90°.

So, ∠PST=90°-θ and ∠UQR=θ

PQ is the tangent and PR is the secant of the larger circle, so by theorem

PQ²=PU×PR

y²=(a+b)×z

a+b=y²/z …….(i)

△QUR~△PQR (Similar Triangles)

UR/RQ=RQ/PR

c/x=x/z

c=x²/z …….(ii)

△PST~△PQU (Similar Triangles)

PT/TU=PS/SQ

a/b=2r/(y-2r) …….(iii)

Now draw VW, here V and W are the centers of the respective circles.

In the right triangle △VQW, apply Pythagoras’ Theorem:

VW²=VQ²+QW²

Put VW=r+x/2

VQ=y-r

QW=x/2

r=y²/(x+2y)

Put this value of r in equation (iii)

a/b=2r/(y-2r)

After solving, a/b=2y/x ……(iv)

Now put all the values in the given expression:

a/b×√[c/(a+b)]= (2y/x)×(x/y)=2

Hence Proved

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I drew a #brushstroke #design based on a specific shape(This time it is a #righttriangle ). The 2nd one is an explanation of it 以特定的圖形為主題，描繪了 #毛筆字 的 #設計 (這次是 #直角三角形 )。第二張是關於這個解釋 特定の図形をモチーフに、#筆文字 設計を描いてみました(今回は直角三角形)。2枚目はその解説です (Note(JA)-> https://note.com/nox_0/n/n76e4ed8bf43a )

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