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Home » For going to a city B from city A, there is a route via city C such that AC⊥CB, AC = 2 x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
For going to a city B from city A, there is a route via city C such that AC⊥CB, AC = 2 x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
CBSE | Class 10 | Exercise 6.4 | Maths | NCERT Exemplar | Triangles
For going to a city B from city A, there is a route via city C such that AC⊥CB, AC = 2 x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

Solution:

According to the given question,

AC⊥CB,

AC\text{ }=\text{ }2x\text{ }km,

CB=2\left( x+7 \right)km and AB=26\text{ }km

As a result, we get triangle ACB right angled at C.

Now, using Pythagoras Theorem in right angled ∆ACB,

A{{B}^{2}}~=\text{ }A{{C}^{2}}~+\text{ }B{{C}^{2}}

\Rightarrow ~{{\left( 26 \right)}^{2}}~=\text{ }{{\left( 2x \right)}^{2}}~+\text{ }{{\left\{ 2\left( x\text{ }+\text{ }7 \right) \right\}}^{2}}

\Rightarrow ~676\text{ }=\text{ }4{{x}^{2}}~+\text{ }4({{x}^{2}}~+\text{ }196\text{ }+\text{ }14x)

\Rightarrow ~676\text{ }=\text{ }4{{x}^{2}}~+\text{ }4{{x}^{2}}~+\text{ }196\text{ }+\text{ }56x

\Rightarrow ~676\text{ }=\text{ }{{8}^{2}}~+\text{ }56x\text{ }+\text{ }196

\Rightarrow ~8{{x}^{2}}~+\text{ }56x-\text{ }\text{ }480\text{ }=\text{ }0

NCERT Exemplar Solutions Class 10 Maths Chapter 6 Ex. 6.4-15

Now dividing the equation by 8, we obtain,

{{x}^{2}}~+\text{ }7x-\text{ }\text{ }60\text{ }=\text{ }0

{{x}^{2}}~+\text{ }12x-\text{ }\text{ }5x-\text{ }\text{ }60\text{ }=\text{ }0

x\left( x\text{ }+\text{ }12 \right)-5\left( x\text{ }+\text{ }12 \right)\text{ }=\text{ }0

\left( x\text{ }+\text{ }12 \right)\left( x-\text{ }\text{ }5 \right)\text{ }=\text{ }0

\therefore ~x\text{ }=\text{ }-12\text{ }or\text{ }x\text{ }=\text{ }5

As the distance cannot be negative, we neglect x\text{ }=\text{ }-12

\therefore ~x\text{ }=\text{ }5

Now,

AC\text{ }=\text{ }2x\text{ }=\text{ }10km

BC\text{ }=\text{ }2\left( x\text{ }+\text{ }7 \right)\text{ }=\text{ }2\left( 5\text{ }+\text{ }7 \right)\text{ }=\text{ }24\text{ }km

As a result, the distance covered to city B from city A via city C = AC + BC

AC\text{ }+\text{ }BC\text{ }=\text{ }10\text{ }+\text{ }24

=\text{ }34\text{ }km

After the highway was constructed the distance covered to city B from city A =\text{ }BA=26\text{ }km

As a result, the distance saved =\text{ }3426=8\text{ }km.

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