Point "Q" is located at 45° from "O", then the coordinates are x=4, y=4 R² = x² + y² = 2*4² R = √32 = 4√2 cm Labelling segment OP = m Tangent secant theorem: 4.m = 2² ---> m=1 Cosine rule: (2r)²=1²+R²-2Rcosα 4r² = 1 + (4√2)² - 8√2cos45° 4r² = 1 + 32 - 8 = 25 r = 2,5 cm A = ½πr² = 9,82 cm² ( Solved √ )
Point "Q" is located at 45° from "O", then the coordinates are x=4, y=4 Labelling segment OP = m Tangent secant theorem: 4.m = 2² ---> m=1 Pytagorean theorem: (2r)² = 4² + (4-m)² = 4²+3² r = 2,5 cm A = ½πr² = 9,82 cm² ( Solved √ )
The area is (25/8)pi units squared. At the 4:25 mark, I think that when HL congruence is established, it is established due to a common angle and common side being justified by right angle. I think that ALL congruence postulates require a common side and angle to be established. And then you combined that with the circle theorem and that justified and that exact sides. I hope that that is a sufficient explanation that shows you that I SHOULD be able to do this. And I hope that my comments on the other past videos indicate that!!!
Let M be the center of the semicircle. Let the radius of semicircle M be r and the radius of quarter circle O be R. Let OP = x. Draw MS. As MS is a radius of semicircle M, MS = r. As OB is tangent to semicircle M at S, ∠OSM = 90°. Let N be the point on MS where PN is parallel to OB and perpendicular to MS. As ∠POS and ∠OSN equal 90° and ∠NPO and ∠SNP equal 90° by construction, POSN is a rectangle, NS = OP = x, and NP = OS = 2. Triangle ∆PNM: NP² + MN² = MP² 2² + (r-x)² = r² 4 + r² - 2xr + x² = r² x² - 2xr + 4 = 0 --- [¹] Let T be the point on OA where QT is perpendicular to OA and parallel to OB. As ∠TOQ = 90°/2 = 45°, as arcs AQ and QB are congruent, and OQ = R, QT = Rsin(45°) = R/√2, and OT = Rcos(45°) = R/√2. As QT and NP are parallel, and QP intersects both, ∠PQT and ∠QPN are alternate interior angles, and thus congruent. As ∠QTP = ∠PNM = 90°, ∆QTP and ∆PNM are similar triangles. QT/QP = NP/MP (R/√2)/2r = 2/r R/√2 = (2r)2/r = 4 R = 4√2 MN/MP = TP/QP MN/r = (4-x)/2r MN = r(4-x)/2r = (4-x)/2 = 2 - x/2 MS = MN + NS r = (2-x/2) + x = 2 + x/2 --- [²] x² - 2xr + 4 = 0 x² - 2x(2+x/2) + 4 = 0 x² - 4x - x² + 4 = 0 4x = 4 x = 1 r = 2 + x/2 r = 2 + (1)/2 = 5/2 Semicircle M: [M] = πr²/2 = π(5/2)²/2 = (25/4)π/2 = 25π/8 sq units
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Point "Q" is located at 45° from "O", then the coordinates are x=4, y=4
R² = x² + y² = 2*4²
R = √32 = 4√2 cm
Labelling segment OP = m
Tangent secant theorem:
4.m = 2² ---> m=1
Cosine rule:
(2r)²=1²+R²-2Rcosα
4r² = 1 + (4√2)² - 8√2cos45°
4r² = 1 + 32 - 8 = 25
r = 2,5 cm
A = ½πr² = 9,82 cm² ( Solved √ )
شكرآ لكم على المجهودات
يمكن استعمال OS=ST=2
QT=OT=4
OP /OS =QT/TS
OP=1
PQ^2=SP^2 + SQ^2 =4a^2
a^2 = 25/4
a=5/2
Point "Q" is located at 45° from "O", then the coordinates are x=4, y=4
Labelling segment OP = m
Tangent secant theorem:
4.m = 2² ---> m=1
Pytagorean theorem:
(2r)² = 4² + (4-m)² = 4²+3²
r = 2,5 cm
A = ½πr² = 9,82 cm² ( Solved √ )
The area is (25/8)pi units squared. At the 4:25 mark, I think that when HL congruence is established, it is established due to a common angle and common side being justified by right angle. I think that ALL congruence postulates require a common side and angle to be established. And then you combined that with the circle theorem and that justified and that exact sides. I hope that that is a sufficient explanation that shows you that I SHOULD be able to do this. And I hope that my comments on the other past videos indicate that!!!
φ = 30° → sin(3φ) = 1; TO = TQ = 4 = SO + ST = 2 + 2; SC = QC = PC = a = ?
sin(STQ) = 1 → QS = 2√5 = QW + SW = √5 + √5 → sin(SWC) = 1; TQS = CSW = δ →
sin(δ) = √5/5 → cos(δ) = 2√5/5 = √5/a → a = 5/2 → area semicircle = 25π/8
btw: PS = √5; PO = 1; OQP = γ → cos(γ) = 7√2/10 → sin(γ) = √2/10 → tan(γ) = 1/7
Join PQ to intersect CS at K. Join PS & QS. < POK= SOK= 45. KSO is an isosc. triangle, OS= MS= 2. Triangles POS & STQ are similar. PO/ ST=OS/TQ= 1/2. 2PQ= ST=2, PQ=1. In triangle QPO, CK= 1/2 PO= 1/2. CS= a= CK+ KS=1/2 + 2= 5/2..
Let M be the center of the semicircle. Let the radius of semicircle M be r and the radius of quarter circle O be R. Let OP = x.
Draw MS. As MS is a radius of semicircle M, MS = r. As OB is tangent to semicircle M at S, ∠OSM = 90°.
Let N be the point on MS where PN is parallel to OB and perpendicular to MS. As ∠POS and ∠OSN equal 90° and ∠NPO and ∠SNP equal 90° by construction, POSN is a rectangle, NS = OP = x, and NP = OS = 2.
Triangle ∆PNM:
NP² + MN² = MP²
2² + (r-x)² = r²
4 + r² - 2xr + x² = r²
x² - 2xr + 4 = 0 --- [¹]
Let T be the point on OA where QT is perpendicular to OA and parallel to OB. As ∠TOQ = 90°/2 = 45°, as arcs AQ and QB are congruent, and OQ = R, QT = Rsin(45°) = R/√2, and OT = Rcos(45°) = R/√2.
As QT and NP are parallel, and QP intersects both, ∠PQT and ∠QPN are alternate interior angles, and thus congruent. As ∠QTP = ∠PNM = 90°, ∆QTP and ∆PNM are similar triangles.
QT/QP = NP/MP
(R/√2)/2r = 2/r
R/√2 = (2r)2/r = 4
R = 4√2
MN/MP = TP/QP
MN/r = (4-x)/2r
MN = r(4-x)/2r = (4-x)/2 = 2 - x/2
MS = MN + NS
r = (2-x/2) + x = 2 + x/2 --- [²]
x² - 2xr + 4 = 0
x² - 2x(2+x/2) + 4 = 0
x² - 4x - x² + 4 = 0
4x = 4
x = 1
r = 2 + x/2
r = 2 + (1)/2 = 5/2
Semicircle M:
[M] = πr²/2 = π(5/2)²/2 = (25/4)π/2 = 25π/8 sq units
Замечательное решение с объяснением. Спасибо большое. Thanks.
(2)^2=4 90°A0PSQD/4=22.2AOPSQD 2^11.2 1^11^1.2 1^1.2 1.2 (AOPSQD ➖ 2AOPSQ+1).
This question, if I'm not mistaken, has already been asked on this channel.
The answer should be 2pi sq units
PO should be 2 cm
Draw parallel of OB from Q, intersecting AO at H. OH=HQ=OS*2=4. OS^2=OP*OH, so OP=1. r^2=(r-OP)^2+OS^2, so r=5/2