Find area of Yellow shaded region inscribed in a right triangle | Step-by-step tutorial

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Find area of Yellow shaded region inscribed in a right triangle | Step-by-step tutorial

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[bobjordan]

Sweet, I figured it out thanks to all of the previous posts done by PreMath. Getting very confident with this, thanks! I like doing a couple or three daily to keep mentally fit...

[ybodoN]

There is a general formula for the radius of a circle inscribed in a triangle:
_r = √((s − a)(s − b)(s − c) / s)_ where _s_ is the semiperimeter and _a, b_ and _c_ are the sides of the triangle.
In the special case of a right triangle, this formula reduces to _½ (a + b − c)_ where _c_ is the hypotenuse.
Another formula for the special case of a right triangle is _r = ab/(a + b + c)_ where _c_ is the hypotenuse.

[peterbenoit]

Nice to discover I can still do this question without paper, pen, or calculator.

[theoyanto]

Nice mixture of calculations, great example

[vaggelissmyrniotis]

Wow!!I didn't expect to find the same result with all these calculations but in the end i was correct!

[spiderjump]

using pythg theorem, we can find CB = 51
Then we can find that CD = CE = 34
AD = 51 and EB = 17
drop a line from D perpendicular to AB, let the point of intersection be G
DG and AG can be calculated using similar triangles.
Area of triangle ADG can then be found
Trapezium DEBG 's area can be found.
add the areas to find answer.

[MrPaulc]

I got this one by (mostly) using your method, but with some variation near the end.

[palomoreytactico]

i learn more theorems in this videos thanks

[wackojacko]

The harmony of congruence is music too my ears! 🙂

[privateinformation]

I used BE as x as well, but I found the ratio between AD and CD, and CE and BE to find out the area instead of using trigonometry.

[marioalb]

Height of triangle
h² = 85² - 68²
h = 51 cm

Radius of inscribed circle:
(68-r)+(51-r)=85
2r = 68+51-85
r= 17 cm

Side of isosceles triangle:
s = 51 - 17 = 34 cm

Area of isosceles triangle:
A = ½ s² sin α
A = ½ 34² . 68 / 85
A = 462, 40 cm²

Area of bigger right triangle:
A = ½ b. h
A = ½ 68 . 51
A = 1734 cm²

Yellow Shaded Area :
A = A₁ - A₂
A = 1734 - 462, 4
A = 1271, 6 cm² ( Solved √ )

[RobG]

Why not look for LCD of hypotenuse and side, 17, to recognize this as a 3•4•5 right triangle?
Ooh, I commented too soon.
Thanks for all your thorough videos.

[jimlocke]

Save a step by letting x be length CD or equal (by two tangent theorem) length CE. To avoid the use of the trigonometric sin() function, drop a perpendicular from point D to CB and call the intersection point G. Using CE as the base of ΔCDE, length DG is the height, call it h. ΔCDG and ΔCAB are similar. Therefore, CD/CA=DG/AB, so 34/85 =h/68 and h=27.2. So area ΔCDE = (1/2)(27.2)(34) = 462.4 As PreMath did, wrap up by subtracting 462.4 from 1734 to get the yellow area, 1271.6.

[misterenter-izrz]

By pythagorean theorem, the height of the triangle is 51, then the area of the 0:03 triangle is 34x51, it is also equal to r(85+68+51)/2=102r, where r is the radius, so r=34x51/102=17 hence the area of the rhombus is 17x34, yhe length of two diagonals of the rhombus are 17xroot 5 and 17x34/17xroot 5=34/root 5, then the height of the white triangle is sqrt(4x17^2-17^2/5)=17 sqrt(19/5), therefore the answer is 68x51/2-34/root approximately 😅.

[vidyadharjoshi]

BC = 51. Let BF = BE = X. 68 - X + 51 - X = 85. X = 17. EC = 34, Angel C = 53.13
Area CDE = 34*34*sin 53.13*0.5 = 462.4
Yellow Area = Area ABC - 450 = 1734 - 462.4 = 1271.6

[alancs]

My solution:

- Found BC using Pythagoras, which is 51

- ABC's area, which is 1, 734

- Circle's radius, using Area = Radius x Semiperimeter. Radius equals 17

- CE by subtracting 17 from 51, which equals CD and 34

- With two similar triangles using the circle's center and a line from center to C, found DE and CDE's height, in order to find CDE's area, which is 2312/5

- Finally, to find yellow area, subtracted 2312/5 (CDE's area) from 1, 734 (ABC's area), which equals 6358/5 or 1, 271, 6

[giuseppemalaguti]

Risulta Ho rifatto i calcoli A=289*22/5

[michaelkouzmin]

A fork within the solution radius of a circle inscribed in a triangle (r=x in terms of the Teacher) r = 2*Aabc/(AB+BC+CB) = 2*1734/(86+68+51) =17

[JSSTyger]

DC = DE = x and FB = EB = r
68-r = 85-x and so x = r+17
85² = 68²+(2r+17)²
85² = 68²+4r²+17²+68r
7225 = 4624+4r²+289+68r
2312 = 4r²+68r
4r²+68r-2312 = 0
r = [-68+sqrt(68²-4(4)(-2312))]/8
r = [-68+204]/8 = 17 and x = 34
So the triangle has side lengths of 68, 51, and 85
Angle C = sin-1(68/85) = sin-1(4/5) = 53.13°
DE = 2xsin(53.13/2) = 30.41
Blue area = (30.41)(xcos(53.13/2)/2 = 462.4
Yellow area = (1/2)(68)(51)-462.4 = 1271.6

[devondevon]

Below 6.2k, Learn how to find the area of the blue. However, since the area of the triangle is 34 * 51= 1734 if the area of the blue is 462.4 then the area of the yellow is 1735 -462.4 = 1271.6