Thank you Math Booster. Triangles ABC and QBP are similar, so QB/AB = BP/BC = QP/AC Given AB×PQ = 15×5 = 75, AC×QB is also 75. The area [ABC] is half AB×AC = AC × 15 /2 Letting BQ = QC = b 4(b^2) = 15^2 + AC^2 , by Pythagoras' theorem. Letting BP = p and AC = c c^2 = 4 (b^2) -225 p^2 - b^2 = 25 ( AC×QB is also 75 ) and b=75/c c^2 = 4 (5625/ c^2)-225 after substituting for b c^4 + 225c^2 - 22500 = 0 c^2 = - 225/2 + ( 15 /2 × (sqrt ( 225+400)= - 225/2 + 15×25/2= (6 ×25) c = 5 ×sqrt (6) Checking: b= 75/(5sqrt(6)) = 3×25×sqrt (6) / ( 30) = (2 + 1/2) sqrt 6 so BC = 5 sqrt 6 which can not be right. BC > AC. Now I will see if I got anything right. BP value is not necessary and I should have had 4 x BQ^4 in place of AC^4 , and 225 BQ^2 in place of 225 AC^2 in the quadratic equation some how. I was in a hurry to find AC before BQ. So I probably slipped up twice.
Triangles BPQ and ABC are similar. Let BQ = x and we find x/15 = √(x² + 25)/2x. From this, x = 5√3. From this, BC = 10√3. Therefore, AC=5√3. Therefore, the area of triangle ABC is equal to (15*5√3)/2 = 75√3/2.
Area of triangle ABC=1/2(15)(5√3)=1/2(75√3).❤❤❤
a=BQ,b=AC..15^2+b^2=(2a)^2...5/a=b/15...risultano a=5√3,b=15/√3...A=(15*15/√3)/2=225/2√3
Thank you Math Booster.
Triangles ABC and QBP are similar, so QB/AB = BP/BC = QP/AC Given AB×PQ = 15×5 = 75, AC×QB is also 75.
The area [ABC] is half AB×AC = AC × 15 /2
Letting BQ = QC = b 4(b^2) = 15^2 + AC^2 , by Pythagoras' theorem.
Letting BP = p and AC = c c^2 = 4 (b^2) -225
p^2 - b^2 = 25 ( AC×QB is also 75 ) and b=75/c
c^2 = 4 (5625/ c^2)-225 after substituting for b
c^4 + 225c^2 - 22500 = 0 c^2 = - 225/2 + ( 15 /2 × (sqrt ( 225+400)= - 225/2 + 15×25/2= (6 ×25)
c = 5 ×sqrt (6)
Checking: b= 75/(5sqrt(6)) = 3×25×sqrt (6) / ( 30) = (2 + 1/2) sqrt 6 so BC = 5 sqrt 6 which can not be right. BC > AC.
Now I will see if I got anything right.
BP value is not necessary and I should have had 4 x BQ^4 in place of AC^4 , and 225 BQ^2 in place of 225 AC^2 in the quadratic equation some how.
I was in a hurry to find AC before BQ. So I probably slipped up twice.
Pytagorean theorem:
c² = a² + b² = 15² + b²
Similarity of triangles:
½c / 5 = 15 / b
c = 15*5*2 / b = 150 / b
Replacing:
(150/b)² = 15² + b²
150² = b² (15² + b²)
b⁴ + 225 b² - 22500 = 0
b² = 75 --> b= 5√3 cm
Area of triangle:
A = ½b.h= ½a.b= ½*15*5√3
A = 64,95 cm² ( Solved √ )
Let BQ = QC = x.
Triangle ∆ABC:
CA² + AB² = BC²
CA² + 15² = (2x)²
CA² = 4x² - 225
CA = √(4x²-225)
As ∠BQP = ∠CAB = 90° and ∠B is common, then ∆ABC and ∆BQP are similar triangles.
BQ/QP = AB/CA
x/5 = 15/√(4x²-225)
75 = x√(4x²-225)
x²(4x²-225) = 5625
4x⁴ - 225x² - 5625 = 0
4u² - 225u - 5625 = 0
4u² - 300u + 75u - 5625 = 0
4u(u-75) + 75(u-75) = 0
(u-75)(4u+75) = 0
u = 75 | -u = -75/4- ❌ u ≥ 0
x² = 75
x = √75 = 5√3 | -x = -√75 = -5√3- ❌ x ≥ 0
CA² = 4x² - 225 = 4(75) - 225
CA² = 300 - 225 = 75
CA = √75 = 5√3
A = bh/2 = AB(CA)/2
A = 15(5√3)/2
A = (75√3)/2 ≈ 64.95 sq units
Triangles BPQ and ABC are similar. Let BQ = x and we find x/15 = √(x² + 25)/2x. From this, x = 5√3. From this, BC = 10√3. Therefore, AC=5√3. Therefore, the area of triangle ABC is equal to (15*5√3)/2 = 75√3/2.
(15)^2 (5)^2={225+25}=250/180°ABC=1.70ABC1.2^35 1.2^5^7 1.2^5^3^4 1.2^2^3^1^2^2 1^1^3^1^2 32 (ABC ➖ 3ABC+2).
این خیلی راحت بود حتی ذهنی هم میشد حساب کرد
no!!! si pone a2=x e l'equazione diventa x2-225x-75=0 con l'equazione di 2° grado si calcola x= a2 perciò a=radq di x
A questão é muito boa. Parabéns pela escolha 😅 BRASIL Dezembro de 2024🎉
Area= 64.95