Let AB=AD=a and BC=b Perimeter=14 So a+1+b+a=14 2a+b=13 (1) Connect D to E (E on AB) In ∆ ADE AE^2+DE^2=AD^2 AE=a-1 ; DE=b ; AD=a (a-1)^2+b^2=a^2 a^2-2a+1+b^2=a^2 b^2=2a-1 (2) (1)&(2) (13-2a)^2=2a-1 So a=5 ; a=17/2 b=13-10=3; b=13-17=-4
@olivier34723 ай бұрын
I remove the rectangle to the right, I am left with a triangle perimeter equals 12. This is a standard Pythagorean triplet, 3 4 5. Cqfd. Area equals 9, about 10s in your head 🎉
That’s very nice Thanks Sir Thanks PreMath Good luck with glades ❤❤❤❤
@unknownidentity28463 ай бұрын
Let's find the area: . .. ... .... ..... From the known perimeter P of the trapezoid we obtain: P = AB + BC + CD + DA 14 = AB + BC + 1 + AB ⇒ BC = 13 − 2*AB Now we add point E on AB such that BCDE is a rectangle. In this case ADE is a right triangle and we can apply the Pythagorean theorem: DA² = AE² + ED² AB² = AE² + ED² AB² = (AB − BE)² + ED² AB² = (AB − CD)² + ED² AB² = (AB − CD)² + BC² AB² = (AB − 1)² + (13 − 2*AB)² AB² = AB² − 2*AB + 1 + 169 − 52*AB + 4*AB² 0 = 4*AB² − 54*AB + 170 0 = 2*AB² − 27*AB + 85 AB = [27 ± √(27² − 2*4*85)]/(2*2) = [27 ± √(729 − 680)]/4 = (27 ± √49)/4 = (27 ± 7)/4 Let us check the two possible solutions: AB = (27 + 7)/4 = 34/4 = 17/2 ⇒ BC = 13 − 2*AB = 13 − 2*(17/2) = 13 − 17 = −4 AB = (27 − 7)/4 = 20/4 = 5 ⇒ BC = 13 − 2*AB = 13 − 2*5 = 13 − 10 = 3 Therefore the only useful solution is: AB = 5 BC = 3 CD = 1 DA = 5 Now we are able to calculate the area of the trapezoid: A = (1/2)*(AB + CD)*BC = (1/2)*(5 + 1)*3 = (1/2)*6*3 = 9 Best regards from Germany
@jimlocke93203 ай бұрын
For ΔADE, x² = (x - 1)² + y², x² = x² - 2x + 1 + y², -2x + 1 + y² = 0. Perimeter = x + 1 + y + x = 2x + y + 1 = 14, 2x + 1 + y = 14. Add the 2 equations: 2x + 1 + y -2x + 1 + y² = 14, y² + y - 12 = 0, (y + 4)(y - 3) = 0, with roots y = -4 and y = 3. Discard the negative root. Substitute y = 3 into 2x + 1 + y = 14 and x = 5. The 2 bases of the trapezoid are 1 and x = 5 and the height is 3. As found in the video, the area is 9 square units.
@SkinnerRobot3 ай бұрын
Brilliant! Doing it your way, the numbers are so much easier to work with, and it is easier to tell which of the quadratic roots to discard.
@jimlocke93203 ай бұрын
@@SkinnerRobot Thanks for the compliment!
@MrPaulc2223 ай бұрын
ABD is isosceles. I suppose AB and AD can be called x (formerly Twitter). Call BC, y. These variables give 2x + y = 13. Drop a perpendicular from D to a new point F on line AB. DCFB is a rectangle with sides 1 and y, so an area of y un^2 AFD is a triangle of sides (x-1), y, and x. (x-1)^2 + y^2 = x^2 x^2 - 2x + 1 + y^2 = x^2. y^2 = x^2 - x^2 + 2x - 1 Therefore, y^2 = 2x - 1.. This should supply all the side lengths of the trapezoid in terms of the same variable. With x it is 2x + 1 + sqrt(2x - 1) = 14 Probably no need to calculate with y. Rearrange to 2x - 13 = -sqrt(2x - 1) Square both sides for 4x^2 - 52x + 169 = 2x - 1 Rearrange: 4x^2 - 54x + 170 = 0 All even numbers so 2x^2 - 27x + 85 = 0 Quadratic formula: (27+or-sqrt(729 - 4*2*85))/4 = x (27+or-sqrt(49))/4 x = 20/4 or 34/4, so 5 or 8.5. On the surface, both look viable, so test with the perimeter of 14. x = 5 gives 5 + 5 + 1 + 3 = 14, so no need to test the other value. Now for the trapezoid area: The rectangle DCFB is 3*1 so an area of 3. The triangle on the left is a standard 3,4,5, so an area of 6. 3+6 = 9 un^2. Yes, it looks like we mainly went down the same path.
@sorourhashemi32493 ай бұрын
thanks. Easy. DA=AB=a, CB=b, 2a+b+1=14==> b=13-2a. Connect a line perpendicular to AB which is "b" and mark it as "M", so AM = a-1. We have (13-2a)^2+(a-1)^2= (13-2a)^2==>a=5, b=3, area =9
@allanflippin24533 ай бұрын
How do you predict that factoring a quadratic equation could be successful? In real world problems, it seems that approach is rarely successful because real world numbers aren't "nice" like the numbers in sample equation. But basically, I feel there must be some analysis done first before deciding "factoring looks good for this equation".
@allanflippin24533 ай бұрын
@@RAG981 Thanks. That was my guess. Factoring is really cool, if you actually know it to be possible.
@tedn68553 ай бұрын
I did differently. I defined length y as sqrt(2x-1) from pythagoras. Defined 2x-1 as u. Set up perimeter as 2x+1+y=14. Substituted for u and got u+2+sqrtu =14. Solved for u got two answers 16 and 9. Solved for x got 2 8.5 and 5. Solved for y and got either 4 or 3. Checked original equaition only x 5 and corresponding y of 3 were correct. Area is 9.
@MrPaulc2222 ай бұрын
2x + y + 1 = 14, so 2x + y = 13 where x is the equal sides and y is the vertical. Drop a vertical from D to a point E on the base which gives right triangle ADE with sides x-1, y, and x. x^2 - (x-1)^2 = y^2 x^2 - (x^2 - 2x + 1) = y^2 x^2 - x^2 + 2x - 1 = y^2 2x - 1 = y^2 x + x + 1 + sqrt(2x - 1) = 14 2x + 1 + sqrt(2x - 1) = 14 sqrt(2x - 1) = 13 - 2x 2x - 1 = 169 - 52x + 4x^2 170 - 54x + 4x^2 = 0 Re-order and reduce: 2x^2 - 27x + 85 = 0 Quadratic formula: (27+or-sqrt(729 - 4*2*85))/4 = x (27+or-sqrt(49))/4 = x x = 34/4 or 20/4 Try x = 20/4 = 5 5 + 5 + 1 + sqrt(9) = 14, so x = 5 CB = 3 Rectangular part on the right = 3*1=3. Triangle on left is a 3,4,5, so area 6, 6+3=9 Answer: 9 un^2
@marcgriselhubert39153 ай бұрын
Be c = AB = AD, then if H is the orthogonal projection of D on (AB) we have h^2 = DH^2 = AD^2 - DH^2 = (c^2 - (c-1)^2 = 2.c -1 The perimeter of the trpezoïd is then 2.c+1+sqrt(2.c-1) = 14, and sqrt(2.c-1) = 13-2.c, then 2.c-1 = 169-52.c+4.c^2 with 13-2.c>=0 We then have 4.c^2-54.c+170 = 0 with c
@nandisaand52873 ай бұрын
I solved x as a function of y using Pythagoras, then plugged it into perimeter to solve y, then went back to solve x. I got lucky, my factoring was easier. If Id gone Professor's route, I'd have had go quadratic: (X-1)² + Y² = X² ... X = ½(Y² + 1) --------- Perimeter: 1 + Y + 2X = 14 1 + Y + 2•½•(Y² + 1) = 14 ... Y² + Y - 12 = 0 (Y + 4)(Y - 3)=0 Y = 3 (-4 rejected) X = ½(3² + 1) X = 5
@quigonkenny3 ай бұрын
Let AD = AB = x and CB = h. By the formula for the area of a teapezoid, that area will be A = h(x+1)/2. As the perimeter of the trapszoid is 14: 1 + h + x + x = 14 h + 2x = 13 h = 13 - 2x Drop a perpendicular from D to E on AB. As DEBC is a rectangle, EB = CD = 1 and ED = CB = h = 13-2x. Triangle ∆AED: AE² + ED² = AD² (x-1)² + (13-2x)² = x² x² - 2x + 1 + 4x² - 52x + 169 = x² 4x² - 54x + 170 = 0 2x² - 27x + 85 = 0 2x² - 10x - 17x + 85 = 0 2x(x-5) - 17(x-5) = 0 (x-5)(2x-17) = 0 x = 5 | x = 17/2 ❌ 2x < 13 h = 13 - 2x = 13 - 2(5) = 3 A = h(x+1)/2 A = 3(5+1)/2 A = 3(6)/2 = 3(3) = 9 sq units
Solution Since the perimeter is the sum of the sides, we will have a + a + b + 1 = 14 2a + b = 13 b = 13 - 2a ... ¹ Pythagorean Theorem (a - 1)² + b² = a² a² - 2a + 1 + b² = a² b² = 2a - 1 b = √(2a - 1) ... ² Equaling ¹ to ² 13 - 2a = √(2a - 1) (13 - 2a)² = [√(2a - 1)]² 169 - 52a + 4a² = 2a - 1 4a² - 54a + 170 = 0 (÷2) 2a² - 27a + 85 = 0 a = [27 ± √(729 - 680)]/4 a = (27 ± 7)/4 a' = 17/2 ... Rejected a" = 5 ... Accepted Verification in our equation ¹: b = 13 - 2a b = 13 - 2 (17/2) b = 13 - 17 b = - 4 🔴 So, we rejected a = 17/2 b = 13 - 2a b = 13 - 2 (5) b = 13 - 10 b = 3 🟢 So, we accepted a = 5 Final Step A = ½ h (m + n) m = a = 5 n = 1 h = b = 3 A = ½ 3 (5 + 1) Purple Trapezoid Area = 9 Square Units
RUSSIA! Решение состоит в том, чтобы найти всестороны трапеции. Обозничим AD=AB, через а, из этого условия находим СВ=14-2а-1=13-2а; из D опустим перпендикуляр на АВ, обозначив точкой К; DK=CB=h; АК=а-1; h^2=a^2-(a-1)^2=(13-2a)^2^, a^2-a^2+2a-1=169-52a+4a^2, 4a^2-54a+170=0; 2a^2-27a+85=0; а1=(27+\/49)/4=8,5, нам не подходит, так как высота не может быть отпицательным числом, поэтому берём а2=(27-7)/4=5; находим h=13-10=3; теперь находим площадь = (5+1)*3/2=9. Area=9.
@riteshkumar56443 ай бұрын
Hii sir
@michaeldoerr58103 ай бұрын
The area is 9 units squared. Another easier than it looks problem, and similar to the today's @MathBooster problem!!!
@alster7242 ай бұрын
Easy
@wasimahmad-t6c3 ай бұрын
5+ 5 +3+1==14)(3×4=12÷2=6+1×3=9area
@PrithwirajSen-nj6qq3 ай бұрын
The given figure adduces that BC is greater than CD, and less than AB and AD Now Let we take AB=AD=x And BC =y Now 2x + y =14-1=13 > y=13 -2x Now x may not be 1/2/3/4 as then y will be greater than x Hence x will be 5 and y will be 3 (x may not be 6 as then y will be 1 that is less than AB and AD but not greater than CD) Now area = 1/2* BC( AB + CD) =1/2*3*(5+1) =9 sq units
@wasimahmad-t6c2 ай бұрын
4×3÷2=6+3=9
@yakovspivak9623 ай бұрын
S = 9
@wackojacko39623 ай бұрын
@ 7:57 I don't reject but reserve x=17/2 as an unlikely outcome to occur to avoid unpleasant consequences in the afterlife. 🙂
@suchitroybr3 ай бұрын
Pm
@Birol7313 ай бұрын
My way of solution ▶ Let's define E: E ∈ [AB] and [EB] ⊥ [DE] [DC]=[EB] [EB]= 1 [AE]=x [AD]= [AB] ⇒ [AD]= x+1 b) Let's calculate the height of the trapezoid: [BC] or [ED] for the triaangle ΔAED, by applying the Pythagorean theorem we get: [AE]²+[ED]²= [DA]² [AE]= x [ED]=h [DA]= x+1 ⇒ x²+h²= (x+1)² x²+h²= x²+2x+1 h²= 2x+1 h= √2x+1 c) P(ABCD)= 14 length units [AB]+[BC]+[CD]+[DA]= 14 [AB]= x+1 [BC]= √2x+1 [CD]= 1 [DA]= x+1 ⇒ (x+1)+(√2x+1)+1+(x+1)= 14 2x+3+√2x+1= 14 √2x+1= 11-2x (√2x+1)²= (11-2x)² 2x+1= 121-44x+4x² 4x²-46x+120=0 = 2x²-23x+60=0 Δ= 529-4*2*60 Δ= 49 √Δ= 7 ⇒ x₁= (23+7)/4 x₁= 15/2 x₂= (23-7)/4 x₂= 4 for x₁= 15/2 2x+3+√2x+1 = 2*(15/2)+3+√2(15/2)+1 = 15+3+4 = 22 ❌ P > 14 length units ! ⇒ x= 4 d) A(ABCD)= [[AB]+[DC]]*[BC]/2 [AB]=x+1 = 5 [DC]= 1 [BC]= √2x+1 = 3 ⇒ A(ABCD)= (5+1)*3/2 A(ABCD)= 9 square units