Question

5 marks long answer type question

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Answer 1

Given, AC perpendicular to CB km,CB=2(x+7)km and AB=26km.

Now, in ABC by pythogoras theorem,

AB^2=AC^2+BC^2

(26)^2=(2x)^2+{2(x+7)}^2

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

On dividing by 8, we get

x=-12,x=5

Since distance cant be negative

so, x=5

Now, AC=2x=10km

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

The distance covered to reach city B from city A via city C

=AC+BC

=10+24

=34km

Distance covered to reach city B from city A after the construction of the highway

= BA=26km

Hence,the required saved distance is 34-26 I.e., 8km.

Answer 2

◦•●◉✿[ welcome to the concept of Mathematics ]✿◉●•◦

Given AC  =  2x km  , CB =  2 ( x + 7 ) km

And

AC perpendicular to CB , So 

< ACB  = 90

And as direct highway between cities A , and B  .

AB  =  26 km....

now ,apply Pythagoras theorem in triangle ACB

$ab {}^{2} = ac {}^{2} + cb {}^{2}$

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

after solving you get

$x {}^{2} + 7x - 60 = 0$

$x {}^{2} + 12x - 5x - 6 0= 0$

$(x - 5)(x + 12)$

$x = 5 \: \: or \: - 12$

but distance cannot be negative

then X = 5

thus ,

AB = 10km

CB = 24 km .

total distance of old route = 34 km.

total distance by new highway = 26 km

then,

total distance save in reaching city B from city A after the construction of highway =( 34-26)km= 8km

⏩ yr answer is 8 km

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