Semi Semi Circles

Geometry Level 1

In the figure above, semicircles are constructed on the hypotenuse and legs of a right angle triangle. If $S_{1 }$ , $S_{2 }$ are the area of shaded regions and $S$ is the area of right angled triangle, which of the following is true ?

S_{1 } + S_{2 } = 2S

S_{1 } + S_{2 } = S

S_{1 } + S_{2 } = 3S

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Rahul Paswan
Jan 11, 2015

Nice one both the problem n solution

Kundan Patil - 6 years, 5 months ago

why you have taken S in the RHS. i dont understand pls clarify

Sanjay Singh - 6 years, 5 months ago

Area of shaded region ( $S_{1} + S_{2}$ ) = Total area ( Area of semicircles on legs AB & BC + Area of triangle ) - Area of semicircle on hypotenuse AC

Rahul Paswan - 6 years, 5 months ago

But Area of triangle is covered in area of semicircle on hypotenuse AC

Sanjay Singh - 6 years, 5 months ago

@Sanjay Singh – Please figure this out : You have two ways to calculate total area, 1.Total area = S1 + S2 + Semicircle AC. 2.Total area = Triangle S + Semicircle AB + Semicircle BC. So, S1 + S2 + Semicircle AC = Triangle S + Semicircle AB + Semicircle BC. Then, S1 + S2 = Triangle S + Semicircle AB + Semicircle BC - Semicircle AC. Because AB^2+BC^2-AC^2=0 (Pythagorean), we get : S1 + S2 = Triangle S

Horisadi Afyama - 6 years, 4 months ago

Thank u for your solution

Kun Yi Ong - 6 years, 2 months ago

Nguyen Thanh Long
Jan 14, 2015

Assume the radius of two circles is r and R respectively. According to Pythagoras theory to get the results: $S_{1}+S_{2}=\frac{\pi*(r^{2}+R^{2})}{2} - (\frac{\pi*(r^{2}+R^{2})}{2} - \frac{1}{2}*(rR))=S$ So: $\boxed{S_{1}+S_{2}=S}$

correct relationship is S1 + S2 = S Now S = (L * B )/2 Now S1 + S2 = Area of semi circle on AB as diameter + Area of semi circle on BC as diameter - (Area of semi circle on AC as diameter - Area of right triangle) = (pi/4)* B^2 + (pi/4) L^2 - ( Pi/4) ( B^2 + L^2) + ( L B)/2 = (L*B) / 2 = S Hence Proved

Dinesh Kumar Cpwd - 6 years, 5 months ago

A nice short solution. You may simply add, $~~ \color{#D61F06}{ "second"~~r^2+R^2=AC^2}$

Niranjan Khanderia - 6 years, 5 months ago

Mike Peralta
Jan 18, 2015

ABsegment + BCsegment = AsemicircleAC - S

S1 + S2 = AsemicircleAB + AsemicircleBC - AsemicircleAC + S

            = PiAB²/8 + PiBC²/8 - PiAC²/8 + S        ,    AC² = BC² + AB²   then

            =  (PiAB² + PiBC² - Pi(BC² + AB²)) / 8 + S

            = 0 + S

S1 + S2 = S

Ayman Wahba
Jan 15, 2015

S1 + S2 = Pi(AB^2/4) + Pi(BC^2/4) - {Pi(AC^2/4) - S}

S1 + S2 = Pi/4(AB^2 + BC^2) - {Pi(AC^2/4) - S}

S1 + S2 = Pi/4(AC^2) - {Pi(AC^2/4) - S}

S1 + S2 = S

Curtis Clement
Jan 14, 2015

Let AB = 2 ${x}$ and BC = 2 ${y}$ , so by pythagoras' theorem AC = 2R = 2 $\sqrt{x^2 + y^2}$ $\Rightarrow$ R = $\sqrt{x^2 + y^2}$ so area of large semicircle = $\frac{\pi(x^2 + y^2)}{2}$ . Now we take the difference of these area to calculate the area of sectors AB and BC (lets call this A):

${A}$ = $\frac{\pi(x^2 + y^2)}{2}$ - 2 ${xy}$ (#1). Now area of semi-circle BC = $\frac{\pi\ y^2}{2}$ and area of semi-circle AB = $\frac{\pi\ x^2}{2}$ , so their sum is $\frac{\pi(x^2 + y^2)}{2}$ $\Rightarrow$ $\S_{1}$ + $\S_{2}$ = $\frac{\pi(x^2 + y^2 )}{2}$ - [ $\frac{\pi (x^2 + y^2)}{2}$ - 2 ${xy}$ ]= 2 ${xy}$ (from #1).

$\therefore$ $\S_{1}$ + $\S_{2}$ = ${S}$

Typos:-BC =2. $\color{#D61F06}{y}$ . Also add AC=R. You may use $S_1, S_2$ .

Niranjan Khanderia - 6 years, 5 months ago

Mina Yaccoub
Jan 13, 2015

Area of semicircles is directly proportional with its diameter squared

Area of semicircle with diameter AB + area of semicircle with diameter BC = area of semicircle with diameter AC (pythagorean theorem)

Subtracting the non-dashed parts around triangle ABC from the equation we are left with S1+S2 = area of triangle

Semi Semi Circles

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