Can you find area of the Yellow shaded triangle? | (Semicircle) | #math #maths | #geometry

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Can you find area of the Yellow shaded triangle? | (Semicircle) | #math #maths | #geometry

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[DDX]

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[aljawad]

After obtaining the value for ED = 2, consider the right triangle: OEC, where OC = radius 5 and OE = 3. Use the Pythagorean theorem to obtain the value of CE (=h).

[pwmiles]

Drop a perpendicular from C to E on AD.
AE=AB/2=(5+5+6)/2=8
OE=AE-5=3
OC^2=5^2=CE^2+OE^2=CE^2+9
CE=4
Area = AB.CE/2 = 16x4/2 = 32

[michaelkouzmin]

After we estimated ED I used Intersecting chords theorem h^2= AE*ED = (10-2)*2=16 => h=4
Ayellow = (2*5+6)*4/2=32 sq units.

[prossvay]

CA=CB
So ∆ CAB is isosceles triangle
Connect C to E (E middle AB)
AE=BE=16/2=8
Let AC=BC=a and CE=x
In ∆ACE
CE^2+AE^2=AC^2
x^2+8^2=a^2 (1)
In ∆OCE
CE^2+OE^2=OC^2
x^2+(AE-OA)^2=5^2
x^2+(8-5)^2=25
x^2+9=25
So x=4
Yellow triangle area=1/2(4)(16)=32 square units.❤❤❤

[marcgriselhubert]

H the orthogonal projection of C on (AD)
AH = AB/2 = 8 and OH = AH - AO = 8 - 5 = 3; OC = 5
In triangle OHC: CH^2 = OC^2 - OH^2 = 25 - 9 = 16
So the basis of ABC is 6 and its height is CH = 4,
so its area is (1/2).(16).(4) = 32. (VERY simple)

[DorothyMantoothIsASaint]

Did no one else do similar isoceles triangles?
∆ABC is iso with angles A and B both being alpha.
Draw OC to create iso ∆AOC since AO and OC are both radii (5)
∆ABC ~ ∆AOC since both have angles alpha
OC/AC = AC/AB
5/AC = AC/16
AC^2 = 80
Now drop the height CE to bisect AB
By the power of Pythagoras AE^2 + CE^2 = AC^2
8^2 + CE^2 = 80
CE^2 = 16
CE = 4
and finish as in the video

[devondevon]

32
Since the circle's radius = 5, and the length of AB = 10 + 6 = 16, the center of AB = 8. Label the center P

From P, draw a perpendicular line to C, but since the distance for the circle's center to P = 2,
draw a line to C, thus forming a 3-4-5 right triangle, in which 4 is the height of the triangle,
and since the base = 16, then 16* 4* 1/2 = area = 64* 1/2 = 32 Answer

[devondevon]

32
16 * 4 * 1/2 = 32 Answer
the base is 16,
a 3-4-5 right triangle can be formed left of the circl's center, hence the height of the yellow triangle = 4

very easy

[nexen]

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[quigonkenny]

As r = 5, OA = OB = r = 5 and AB = 2r = 2(5) = 10. Let ∠CAD = θ and draw CD. As A and D are ends of a diameter and C is a point on the circumference, by Thales' Theorem, ∠DCA = 90°. As cos(θ) = CA/AD = CA/10, CA = 10cos(θ).

As BC = CA, ∆BCA is an isosceles triangle, so ∠ABC = ∠CAB = θ. Draw CE, where E is the point on AD where CE is perpendicular to AD. As ∆BCA is isosceles, CE is a perpendicular bisector, and divides ∆BCA into two congruent right triangles ∆AEC and ∆CEB. AE = EB = (10+6)/2 = 8. As cos(θ) = EB/BC = 8/BC, BC = 8/cos(θ).

CA = BC
10cos(θ) = 8/cos(θ)
cos²(θ) = 8/10 = 4/5
cos(θ) = √(4/5) = 2/√5

CA = 10cos(θ) = 20/√5 = 4√5

Triangle ∆AEC:
CE² + AE² = CA²
CE² + 8² = (4√5)²
CE² = 80 - 64 = 16
CE = √16 = 4

Triangle ∆BCA:
Aᴛ = bh/2 = 16(4)/2 = 32 sq units

[marioalb]

Similarly of triangles:
h/8=2/h
h² = 16
h= 4 cm
Isosceles triangle area:
A = ½ b.h
A = ½ 16 . 4
A = 32 cm² ( Solved √ )

[unknownidentity]

Let's find the area:
.
..
...
....


First of all we calculate the length of the base AB:

AB = AO + OD + DB = r + r + DB = 5 + 5 + 6 = 16

Let M be the midpoint of AB. Since ABC is an isosceles triangle (AC=BC), the triangles ACM and BCM are congruent right triangles. Therefore the triangle OCM is also a right triangle and we can apply the Pythagorean theorem:

OC² = OM² + MC²
OC² = (AM − OA)² + MC²
r² = (AB/2 − r)² + MC²
5² = (16/2 − 5)² + MC²
5² = (8 − 5)² + MC²
5² = 3² + MC²
⇒ MC = √(5² − 3²) = √(25 − 9) = √16 = 4

Now we are able to calculate the area of the yellow triangle:

A(ABC) = (1/2)*AB*h(AB) = (1/2)*AB*MC = (1/2)*16*4 = 32

Best regards from Germany

[murdock]

φ = 30°; r = 5; ∆ ABC → AB = AO + MO + DM + BD = r + (r - k) + k + 6
AO = CO = DO = r; CM = h; sin⁡(AMC) = 1; CAM = DCM = MBC = δ →
tan⁡(δ) = k/h = h/(6 + k) → k(6 + k) = h^2
AM = BM → 6 + k = 10 - k → k = 2 → k(6 + k) = 16 → h = 4 → area ∆ ABC = 32

[michaeldoerr]

I am realy that I have learned this interesting application if Thales Theorem!!! What I have learned is that the diameter of the circle when subtended by right angle results in a difference between the given triangle side and the side adjacent to the right angle. That is why we can apply the AA similarity case. I hope that that makes sense. Why we have 32 square units as the area!!!

[jimlocke]

A variation on aljawad's solution: Start by dropping a perpendicular from C to AB, labelling the intersection as point E. Find length AB: twice the radius plus 6, or 16. In one of several ways, we determine that E is the midpoint, so AE has length 8. Length OE is AE minus the radius, 8 - 5, or 3. OC is the radius, 5. So, ΔOCE is a right triangle with a side with length 3 and hypotenuse 5, so the other side, CE, must have length 4 (from the 3-4-5 Pythagorean triple). ΔABC has base AB = 16 and height CE = 4, so area = (1/2)(16)(4) = 32, as PreMath also found.

[spafon]

Alternatively, you can use pythagorean. Since AE=8 and AO is the radius 5, OE= AE-AO=3. OC is the radius 5. By pythagorean, OE^2+EC^2=OC^2. thus 3^2+ EC^2=5^2. Hence EC^2=25-9=16, and EC=4. EC is h, the height of the triangle, and then we apply the triangle area formula as done in the solution on the video.

[ChuzzleFriends]

Because AO = DO = 5, AB = 5 + 5 + 6 = 16.
Draw an altitude CE of △ACB, where E is a point on segment AB.
Because segment CE is an altitude, E is the midpoint of segment AB and CE is the perpendicular bisector of segment AB.
So, AE = 8. Draw radius CO (C is on the semicircle).
We get a right triangle △CEO.
By the Segment Addition Postulate, EO = AE - AO = 8 - 5 = 3.
By the Pythagorean Triples, CE = 4. (EO, CE, CO) --> (3, _, 5)
Find the area of △ACB.
A = (bh)/2
= 1/2 * 16 * 4
= 8 * 4
= 32
So, the area of the yellow triangle is 32 square units.

[MegaSuperEnrique]

2:07 to 6:07 OE=3, OC=5, so CE=h=4 (3-4-5 triangle)

[himo]

OA=OC=OD=5 AB=OA+OD+DB=5+5+6=16 AE=EB=AB/2=16/2=8
OE=8-5=3 3²+CE²=5² CE=4
Yellow triangle area = 16*4/2=32