Question

for going to a city B from city A, there is route via city C such that AC is perpendicular to CB,AC = 2x km and CB = 2(x+7) km.It is proposed to construct a 26km highway which directly connect the two cities A and B . Find how much distance will be saved in reach city B from city A after the Constitution of highway?

Answer 1

Answer:

The distance of 8 km will be saved.

Step-by-step explanation:

Given information: AC⊥CB, AC = 2x km and CB = 2(x+7) km.

Pythagoras theorem: In a right angled triangle

$hypotenuse^2=leg_1^2+leg_2^2$

Using Pythagoras in triangle ABC,

$(AB)^2=(AC)^2+(CB)^2$

$(26)^2=(2x)^2+(2(x+7))^2$

$676=4x^2+4x^2+56x+196$

$0=8x^2+56x-480$

Taking out GCF.

$0=8(x^2+7x-60)$

Splitting the middle term we get

$0=8(x^2+12x-5x-60)$

$0=8(x(x+12)-5(x+12))$

$0=8(x+12)(x-5)$

Using zero product property we get

$x+12=0\Rightarrow x=-12$

$x-5=0\Rightarrow x=5$

The value of x is either -12 or 5.

If x=-12, then

$AC=2x=2(-12)=-24$

The distance can not be negative. So the only possible value of x is 5.

If x=5 then

$AC=2x=2(5)=10$

$CB=2(x+7)=2(5+7)=2(12)=24$

The distance between A and B before the Constitution of highway.

$AB=AC+CB=10+24=34$

The distance between A and B after the Constitution of highway is 26 km. So, the distance saved after the Constitution of highway is

$34-26=8$

Therefore, the distance of 8 km will be saved.

Answer 2

Answer:According to the question,

AC⊥CB,

AC = 2x km, CB = 2 (x + 7) km and AB = 26 km

Thus, we get ∆ ACB right angled at C.

Now, from ∆ACB, Using Pythagoras Theorem,

AB2 = AC2 + BC2

⇒ (26)2 = (2x)2 + {2(x + 7)}2

⇒ 676 = 4x2 + 4(x2 + 196 + 14x)

⇒ 676 = 4x2 + 4x2 + 196 + 56x

⇒ 676 = 82 + 56x + 196

⇒ 8x2 + 56x – 480 = 0

Dividing the equation by 8, we get,

x2 + 7x – 60 = 0

x2 + 12x – 5x – 60 = 0

x(x + 12)-5(x + 12) = 0

(x + 12)(x – 5) = 0

∴ x = -12 or x = 5

Since the distance can’t be negative, we neglect x = -12

∴ x = 5

Now,

AC = 2x = 10km

BC = 2(x + 7) = 2(5 + 7) = 24 km

Thus, the distance covered to city B from city A via city C = AC + BC

AC + BC = 10 + 24 = 34 km

Distance covered to city B from city A after the highway was constructed = BA = 26 km

Therefore, the distance saved = 34 – 26 = 8 km.