Can You Solve Grade 9 Homework Question From Japan

GET MY EBOOKS
•••••••••••••••••••••••
150 Challenging Puzzles with Solutions : payhip.com/b/y...
Function : payhip.com/b/y...
Sequence and Series : payhip.com/b/B...
Differentiation : payhip.com/b/W...
Indefinite Integration : payhip.com/b/8...
Definite Integration + Area under the Curve : payhip.com/b/1...
Trigonometry : payhip.com/b/8...
OTHER CHAPTERS : COMING SOON.....
--------------------------------------------------------------------------------
Join the channel to become a member
/ @mathbooster

[@marioalb9726]

½ b.h = 75 = ½ (y+6).x
(y+6).x = 150
(y+6) /x = y/6
x = 6 (y+6)/y
6(y+6)² = 150.y
6 (y² + 12y +36) = 150y
6y² + 72y + 216 = 150y
6y² - 78y + 216 = 0
y² - 13y + 36 = 0
y = 9 cm ; x = 10 cm
x+y = 19 cm ( Solved √ )

[@harikatragadda]

Area(ABC) = Area(BPC)+ Area(APC)
75=½(Y+6)*6 + ½X*6
Y+6+X=25
X+Y=19

[@WahranRai]

*Thales' theorem* is a fundamental theorem of Euclidean geometry as important as *Pytahgore's theorem*
Why don't you use it instead of always going through the similarities of triangles.
I want to remind you that the cases of similarities use Thales' theorem
Directly using Thales' theorem : BQ / BC = BP/ BA = PQ / X ....

[@tnguyen5725]

Area (ABC) = Area (BPQ) + Area(PQCA)
75 = 6*y/2 +(6+x)*6/2 =3*(x + y) + 18 => x+y = 19

[@ناصريناصر-س4ب]

The area of the triangle is 75, so x(y+6)=150. According to Thales’s theorem, (y)/(y+6)=6/x, which results in 6(x+y)=114, so x+y=19.

[@michaeldoerr5810]

The answer is x + y = 19. At the 3.40 mark, I now realize that I have been thinking HL similarity a little imprecisely. I should have thought of HL similarity as simply the pair of common sides being set equal to the pair of sides subtended by the 90° angle. I hope that this is an improvement. Please let me know when you can.

from Morocco thank you for your interesting videos and for your clear and complete explanations

[@suaebshuvo9011]

Area of triangle PBQ is half of 6×y=3y. Let, PM a perpendicular on AC. So, PM=6 and area of PQCM=6×6=36. Since PM is perpendicular on AC, AM=x-6. Hence, area of PAM= half of 6×(x-6)=3x-18.
So, area of PBQ+PQCM+PAM=75
3y+36+3x-18=75
3x+3y=57
3(x+y)=57
x+y=57/3
x+y=19
What do you guys think?

[@gaiatetuya92]

サムネを見て解けない問題を出してはいけない。三角形ABCの面積が７５であることをサムネにも必要だ。

[@RealQinnMalloryu4]

(6)^2 (6^2= {36+36}=72 180°ABCPQY/72x=2.36 2.6^6 2.3^2^3^2 1.1^13^2 3^2(ABCPQYX ➖ 3ABCPQYX+2).

[@CharlesChen-el4ot]

1/2*6*y +36+1/2*6*(x-6)
= 3y+36+3x-18
=3x+3y+18
= 1/2*(y+6)*x
xy +6x =6x+6y+36
xy = 6y +36
xy = 6 *(y+6)

[@himo3485]

(6+Y)X/2=75 6X+XY=150
6/Y=X/(6+Y) 36+6Y=XY 6X+36+6Y=150 6(X+Y)=114 X+Y=19

[@jairoeveliogordillomarin5780]

(6Y)/2+[(6+X)6]/2=75
6(X+Y)=114
X+Y=19 u.

[@marcelowanderleycorreia8876]

Very cool!!👍

[@shaozheang5528]

y is 19

[@yakupbuyankara5903]

X+Y=19

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC→ AB = AP + BP; AC = AD + CD = AD + 6 = x; BC = AQ + CQ = y + 6
sin⁡(BQP) = sin⁡(PDA) = sin⁡(BCA) = 1; APD = PBQ = ABC = δ; x(y + 6) = 150 → x + y = ?
tan⁡(δ) = 6/y = (x - 6)/6 → x = 6(y + 6)/y; x(y + 6) = 150 → x = 150/(y + 6) = 6(y + 6)/y →
y1 = 9; y2 = 4 → y > 6 → y = 9 → 4 = x - 6 → x = 10 → x + y = 19

[@خبرةوبراعة]

Easy