I drew DQ and since triangle DQA is a right triangle because DQ = sqrt(18), AQ = sqrt(8) and AD = sqrt(26) its area is 6. Now to find the area of triangle PDQ... I drew the perpendicular from P to DQ and showed its length to be (3/4)sqrt(2). I found this length by extending QD to the point F such that EF and QF are perpendicular. From here I used similar triangles to find the distance from P to DQ. Result: Area = 6 - 9/4 = 15/4.
Thank you for your patient teaching. There is a humility, that I have seen this before and did not remember how it would be solved. BC = 5 Completing a rectangle DCBS , AS = 1 and [DCBS] = 3 by 5 = 15 * triangle APQ = 15 - [APD] - [PQDC] - [ABQ] [ABQ] =2.2/2 = 2 [APD] = 1.5/2 = 2 +1/2 The quadrilateral PQDC's area seems a greater problem than the first one ----------------------------------------------------- 4 +1/2 < [PQDC] < 9 and nohing has been gained yet from knowing that DE = 4 so finding theta = CQE tan (theta) = 7/3 Angle AQP = 180 - 45- theta . Letting FCD ba a triangle congruent to ECQ: Point F is where DA produced meets CB produced and BAF is a reflection of BAQ (PD =DQ the drawing is very much out of scale EP=PF and I do not usually use rough paper and pen when working out here so perhaps I might start..) Joining DM to the midpoint of EP then DP=AP = DQ DM = AQ and they are parallel [APQ] is a quarter of [DMAQ] and these parallels arise by the fact that ED=4 [ADQ] is twice [APQ] **Triangle ADQ = 15- 2-(2+1/2) - [CDQ] [CDQ] = 4 +1/2 [APQ] = (1/2) (15- 2 - (2+1/2) - (4 +1/2) = (1/2) (15-9) = 3 [APQ] = 3 _______________________________________________________This answer is wrong , because the assumption that EAF be a straight line was wrong.
Drop perpendicular from A to F on EC. ABCF is rectangle with CF = AB = 2 → DF = CD−CF = 3−2 = 1 Drop perpendicular from P to R on EC. Let DR = x, PR = y. △PRD ~ △AFD by AA [∠PDR = ∠ADF (same) & ∠PRD = ∠AFD = 90° (construction)] PR/DR = AF/DF y/x = 5/1 y = 5x △PRE ~ △QCE by AA [∠PER = ∠QEC (same) & ∠PRE = ∠QCE = 90° (construction/given)] PR/ER = QC/EC y/(4+x) = 3/7 5x/(4+x) = 3/7 35x = 12 + 3x x = 12/32 = 3/8 y = 5x = 15/8 △EDP has base ED = 4 and height = y = 15/8 △ECQ has base EC = 7 and height CQ = 3 [CDPQ] = [ECQ] − [EDP] = (1/2 × 7 × 3) − (1/2 × 4 × 15/8) = 21/2 − 15/4 = 27/4 Trapezoid ABCD has bases AB = 2 and CD = 3 and height BC = 5 [ABCD] = 1/2 × (2+3) × 5 = 25/2 △ABQ has base AB = 2 and height BQ = 2 [ABQ] = 1/2 × 2 × 2 = 2 *[APQ] = [ABCD] − [ABQ] − [CDPQ] = 25/2 − 2 − 27/4 = 15/4*
Once known the length of AQ, MQ and PQ, as in the video, I found the area with trigonometry doing: Area(AQP)=1/2*AQ*MQ*sin 45 + 1/2*MQ*PQ*sin alpha Sin alpha =3/sqrt58 AQ=2sqrt2 PQ can be found with Pythagorean theorem on EQC with EQ=8a
Congratulations. Another resolution would be ◣︎AEQ is right-angled at A, so |AEQ| = ½•2 √︎2•5 √︎2=10. EP/PQ=5/3 and EP+PQ=√︎58 ⇒︎ PQ=3√︎58/8, height of the ΔAEQ in relation to EQ is 20/√︎58 so |APQ| = ½•3√︎58/8•20/√︎58=15/4😉
Suppose A(-2,5),B(0,5),C(0,0)D(-3,0),E(-7,0)Q(0,3) The equation of line AD is y=5x+15 and the equation of line EQ is y=(3/7)x+3 The point of intersection of line AD and EQ is P(-21/8,15/8) and from this the area of triangle APQ is equal to |1/2[-2(3-15/8)-5(0+21/8)+1(0+63/8)]|=15/4
F es la proyección ortogonal de P sobre DC. Pendiente de EQ =3/7 y de DA =5/1=5. DF=b---> 5b={3/7)(4+b)---> b=3/8---> PF=5*3/8 =15/8. ; distancia de P a QC =FC =3-b=21/8. Área APQ =ABCD-ABQ-QCP-DPC =[(2+3)(2+3)/2]-(2*2/2)-(3*21/2*8)-(3*15/2*8)= 3,75 u². Gracias y saludos.
Following I show how I tried to solve the problem with exploitation of line equations in (x,y). Let's start... [APQ] = [ABCD] - [ABQ] - ([EQC] - [EDP]) = (1/2)*(3+2)*5 - (1/2)*2*2 - (1/2)*7*3 + [EDP] = 25/2 - 4/2 -21/2 + [EDP] = [EDP] To find [EDP] see the following points and lines... E(0,0), A(5,5), D(4,0), Q(7,3), P(px,py), line thru E and Q, line thru A and Q. Line EQ... (y-0)/(x-0) = (3-0)/(7-0) y = (3/7)*x Line AD... (y-0)/(x-4) = (5-0)/(5-4) y = 5x - 20 (3/7)*px = 5*px -20 3*px = 35*px - 140 32*px = 140 8*px = 35 px = 35/8 py = (3/7)*(35/8) py = 15/8 [EDP] = (1/2)*4*py = 2*(15/8) = 15/4 . So, ... [APQ] = [EDP] = 15/4 .
I drew DQ and since triangle DQA is a right triangle because DQ = sqrt(18), AQ = sqrt(8) and AD = sqrt(26) its area is 6.
Now to find the area of triangle PDQ...
I drew the perpendicular from P to DQ and showed its length to be (3/4)sqrt(2).
I found this length by extending QD to the point F such that EF and QF are perpendicular.
From here I used similar triangles to find the distance from P to DQ.
Result: Area = 6 - 9/4 = 15/4.
Thank you for your patient teaching. There is a humility, that I have seen this before and did not remember how it would be solved.
BC = 5
Completing a rectangle DCBS , AS = 1 and [DCBS] = 3 by 5 = 15
* triangle APQ = 15 - [APD] - [PQDC] - [ABQ] [ABQ] =2.2/2 = 2 [APD] = 1.5/2 = 2 +1/2
The quadrilateral PQDC's area seems a greater problem than the first one -----------------------------------------------------
4 +1/2 < [PQDC] < 9 and nohing has been gained yet from knowing that DE = 4 so finding theta = CQE tan (theta) = 7/3
Angle AQP = 180 - 45- theta .
Letting FCD ba a triangle congruent to ECQ: Point F is where DA produced meets CB produced and BAF is a reflection of BAQ
(PD =DQ the drawing is very much out of scale EP=PF and I do not usually use rough paper and pen when working out here so perhaps I might start..)
Joining DM to the midpoint of EP then DP=AP = DQ DM = AQ and they are parallel [APQ] is a quarter of [DMAQ] and these parallels arise by the fact that ED=4
[ADQ] is twice [APQ]
**Triangle ADQ = 15- 2-(2+1/2) - [CDQ] [CDQ] = 4 +1/2
[APQ] = (1/2) (15- 2 - (2+1/2) - (4 +1/2) = (1/2) (15-9) = 3
[APQ] = 3 _______________________________________________________This answer is wrong , because the assumption that EAF be a straight line was wrong.
Drop perpendicular from A to F on EC. ABCF is rectangle with CF = AB = 2 → DF = CD−CF = 3−2 = 1
Drop perpendicular from P to R on EC. Let DR = x, PR = y.
△PRD ~ △AFD by AA [∠PDR = ∠ADF (same) & ∠PRD = ∠AFD = 90° (construction)]
PR/DR = AF/DF
y/x = 5/1
y = 5x
△PRE ~ △QCE by AA [∠PER = ∠QEC (same) & ∠PRE = ∠QCE = 90° (construction/given)]
PR/ER = QC/EC
y/(4+x) = 3/7
5x/(4+x) = 3/7
35x = 12 + 3x
x = 12/32 = 3/8
y = 5x = 15/8
△EDP has base ED = 4 and height = y = 15/8
△ECQ has base EC = 7 and height CQ = 3
[CDPQ] = [ECQ] − [EDP] = (1/2 × 7 × 3) − (1/2 × 4 × 15/8) = 21/2 − 15/4 = 27/4
Trapezoid ABCD has bases AB = 2 and CD = 3 and height BC = 5
[ABCD] = 1/2 × (2+3) × 5 = 25/2
△ABQ has base AB = 2 and height BQ = 2
[ABQ] = 1/2 × 2 × 2 = 2
*[APQ] = [ABCD] − [ABQ] − [CDPQ] = 25/2 − 2 − 27/4 = 15/4*
Once known the length of AQ, MQ and PQ, as in the video, I found the area with trigonometry doing:
Area(AQP)=1/2*AQ*MQ*sin 45 + 1/2*MQ*PQ*sin alpha
Sin alpha =3/sqrt58
AQ=2sqrt2
PQ can be found with Pythagorean theorem on EQC with EQ=8a
I got the same result though I first got it wrong. [AQP] = [ABCD] - [ABQ] - [CDPQ]. [ABCD] = 5 * (3+2)/2 = 25/2. [ABQ] = 2*2/2 = 2. [CDPQ] = [ECQ] - [EDP]. [ECQ] = 7*3/2 = 21/2. For [EDP] we need the height. By algebra, 3x/7 = (x-4) * 5 --> 3x/7 = 5x - 20 --> x*(5-3/7) = 20 --> 20 = ((5*7-3)/7) = (35-3)/7 = 32/7 --> x = 20 * 7/32 = 20/32 * 7 = 5/8 * 7 = 35/8 --> Height - 3x/7 = 35/8 * 3/7 = 5*3/8 = 15/8. Then [PED] = ED * Height/2 = 4*15/8/2 = 2*15/8 = 15/4. Then [ABQ] = 25/2 - 2 - (21/2 - 15/4) = 25/2 - 4/2 - (21/2 - 15/4) = 21/2 - 21/2 + 15/4 = 15/4.
Congratulations. Another resolution would be ◣︎AEQ is right-angled at A, so |AEQ| = ½•2 √︎2•5 √︎2=10. EP/PQ=5/3 and EP+PQ=√︎58 ⇒︎ PQ=3√︎58/8, height of the ΔAEQ in relation to EQ is 20/√︎58 so |APQ| = ½•3√︎58/8•20/√︎58=15/4😉
By analytic geometry it would be the area of the triangle with vertices A(5,5), Q(7,3), P(35/8,15/8).
別解法 Ｐから底辺と垂線に垂線を引き、（７、３）、(1、5)二つの直角三角形の相似から、それぞれの長さが求まる。全体の台形から三角形と四角形の面積を引けば求まる。
Suppose A(-2,5),B(0,5),C(0,0)D(-3,0),E(-7,0)Q(0,3) The equation of line AD is y=5x+15 and the equation of line EQ is y=(3/7)x+3 The point of intersection of line AD and EQ is P(-21/8,15/8) and from this the area of triangle APQ is equal to |1/2[-2(3-15/8)-5(0+21/8)+1(0+63/8)]|=15/4
A(-2,5)
Thank you I corrected it but it doesn't affect me because I didn't use it anymore.
I find this to be a Brilliant solution : )
PD=l..l/sinarctg(3/7)=4/sin(arctg5-arctg(3/7)),l=3√26/8(t.seni)..h=lsinarctg5=15/8...Ablue=5*5/2-2*2/2-(3*7/2-4h/2)=25/2-2-27/4=(50-8-27)/4=15/4
F es la proyección ortogonal de P sobre DC.
Pendiente de EQ =3/7 y de DA =5/1=5.
DF=b---> 5b={3/7)(4+b)---> b=3/8---> PF=5*3/8 =15/8. ; distancia de P a QC =FC =3-b=21/8.
Área APQ =ABCD-ABQ-QCP-DPC =[(2+3)(2+3)/2]-(2*2/2)-(3*21/2*8)-(3*15/2*8)= 3,75 u².
Gracias y saludos.
Following I show how I tried to solve the problem with exploitation of line equations in (x,y). Let's start...
[APQ] = [ABCD] - [ABQ] - ([EQC] - [EDP]) = (1/2)*(3+2)*5 - (1/2)*2*2 - (1/2)*7*3 + [EDP] = 25/2 - 4/2 -21/2 + [EDP] = [EDP]
To find [EDP] see the following points and lines... E(0,0), A(5,5), D(4,0), Q(7,3), P(px,py), line thru E and Q, line thru A and Q.
Line EQ... (y-0)/(x-0) = (3-0)/(7-0) y = (3/7)*x
Line AD... (y-0)/(x-4) = (5-0)/(5-4) y = 5x - 20
(3/7)*px = 5*px -20 3*px = 35*px - 140 32*px = 140 8*px = 35 px = 35/8
py = (3/7)*(35/8) py = 15/8
[EDP] = (1/2)*4*py = 2*(15/8) = 15/4 . So, ... [APQ] = [EDP] = 15/4 .
Drop a perpendicular from A onto EC and then it is all about similar triangles.
EQとABを延長させて交わる点をFを置く。
△CQE∽△BQFであり、BQ:QC=2:3より、BF=7*2/3=14/3。
AF=AB+BF=2+14/3=20/3。AF:ED=20/3:4=5:3より、AP:PD=5:3。
△ADQ=台形ABCD-△ABQ-△DCQ=25/2-2-9/2=6
△APQ=△ADQ*5/8=6*5/8=15/4
(4)^(3)^2 (2)^2={16+9+4}=29{90°A+90°B+90°C+90°D}=360°ABCD/29=10.70ABCD 1^7 13^4 1^3^2^2 1^3^1^2 32 (ABCD ➖ 3ABCD+2).
QR=2+2/5=12/5 ED : QR = 12/5 : 4 = 3 : 5 3*3/8=9/8
area of triangle APQ = AQR + QPR = 12/5*2*1/2 + 12/5*9/8*1/2 = 12/5 + 27/20 = 75/20 = 15/4
Your point R is his point M it seems. How do you get 9/8 ( which side) while getting area QPR
@@vcvartak7111
QC=3 QRP∞EDP RP : PD = 3 : 5
3h+5h=3 8h=3 h=3/8
so height of QRP is 3h=9/8 .