For going to a city $ \mathrm{B} $ from city $ \mathrm{A} $, there is a route via city $ \mathrm{C} $ such that $ \mathrm{AC} \perp \mathrm{CB} $, $ \mathrm{AC}=2 x \mathrm{~km} $ and $ \mathrm{CB}=2(x+7) \mathrm{km} $. It is proposed to construct a $ 26 \mathrm{~km} $ highway which directly connects the two cities $ \mathrm{A} $ and $ \mathrm{B} $. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

AcademicMathematicsNCERTClass 10

Given:

For going to a city \( \mathrm{B} \) from city \( \mathrm{A} \), there is a route via city \( \mathrm{C} \) such that \( \mathrm{AC} \perp \mathrm{CB} \), \( \mathrm{AC}=2 x \mathrm{~km} \) and \( \mathrm{CB}=2(x+7) \mathrm{km} \). It is proposed to construct a \( 26 \mathrm{~km} \) highway which directly connects the two cities \( \mathrm{A} \) and \( \mathrm{B} \).

To do:

We have to find how much distance will be saved in reaching city B from city A after the construction of the highway.

Solution:

In \( \triangle \mathrm{ACB} \), by Pythagoras theorem,

$A B^{2} =A C^{2}+B C^{2}$

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

$676 =4 x^{2}+4(x^{2}+49+14 x)$

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

$676 =8 x^{2}+56 x+196$

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

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

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

$x^{2}+12 x-5 x-60=0$

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

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

$x=5$ or $x=-12$ which is not possible since the distance cannot be negative.

$x=5$

Therefore,

$A C=2 x$

$=2(5)$

$=10 \mathrm{~km}$

$B C=2(x+7)$

$=2(5+7)$

$=24 \mathrm{~km}

Hence, the distance covered to city $B$ from city $A$ via city $C$

$=A C+B C$

$=10+24$

$=34 \mathrm{~km}$

Distance covered to city $B$ from city $A$ after the highway is constructed $=B A=26 \mathrm{km}$
Therefore, the distance saved $=34-26$

$=8 \mathrm{~km}$

Hence, the required saved distance is $8\ km$.

raja
Updated on 10-Oct-2022 13:28:12

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