Math, asked by nehajadhav44, 9 months ago

8. Ashok wants to go by car from village A to village B.
He can (i) go straight from A to B (ii) first go to C due
east and then from C
to B due north. The
distance between A
and B exceeds the
distance between A
and C by 9 km. The
distance between A
and B is 9 km less
than twice the distance between B and C. Ashok
decides to go straight from A to B. Find the distance
between (i) A and B (ii) A and C (iii) C and B.​

Answers

Answered by Natsukαshii
7

AB = 45 units.

AC = 36 units

And, CB = 27 units.

Step-by-step explanation:

See the attached diagram.

The distance between A and C = a and between C and B = b (say)

Now, the direct distance between A and B = \sqrt{a^{2} + b^{2}}.

From the given conditions we can write the equations

\sqrt{a^{2} + b^{2}} - a = 9 .......... (1)

And, 2b - 9 = \sqrt{a^{2} + b^{2}} ........ (2)

Now, from equations (1) and (2) we can write,

2b - 9 = 9 + a

⇒ 2b - a = 18 .......... (3)

Again, we can write from equations (1) and (2),

\sqrt{a^{2} + b^{2}} - a = 2b - \sqrt{a^{2} + b^{2}}

Squaring both sides we get,

a^{2} + b^{2} + a^{2} - 2a\sqrt{a^{2} + b^{2}} = 4b^{2} + a^{2} + b^{2} - 4y \sqrt{a^{2} + b^{2}}

⇒ a^{2} - 4b^{2} = 2\sqrt{a^{2} + b^{2}} (a - 2b)

⇒ a + 2b = 2\sqrt{a^{2} + b^{2}} {Since (a - 2b) ≠ 0, from equation (3)}

Again, squaring both sides we get,

a^{2} + 4b^{2} + 4ab = 4a^{2} + 4b^{2}

⇒ 4ab = 3a²

⇒ 4b = 3a {Since a ≠ 0}

⇒ a = \frac{4}{3}b ............... (4)

Now, solving equations (3) and (4) we get

2b - \frac{4}{3}b = 18

⇒ b = 27

And, a = \frac{4}{3}\times 27 = 36

Therefore, AB = \sqrt{36^{2} + 27^{2}} = 45 units.

AC = a = 36 units

And, CB = b = 27 units. (Answer)

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Attachments:
Answered by amitnrw
3

Given : Ashok wants to go by a car from village A to B.he can

(i) go straight from A to B

(ii) first go to C due east and then from C to B due north.

the distance between A and B exceeds the distance between A and C by 9 km.

the distance between A and B is 9km less than twice the distance between B and C.

To Find :   distance between (i) A and B (ii) A and C (iii) C and B

Solution:

Distance between AC  = x   km

=> Distance between  AB   = x + 9

2 ( BC)  - 9  = x + 9

=> 2 (BC) = x + 18

=> BC = x/2 + 9

AC² + BC²  = AB²

=> x² + (x/2 + 9)² = (x + 9)²

=> x²  + x²/4 + 9x  + 81  = x² + 18x + 81

=>  x²/4 = 9x

=> x/4 = 9

=> x = 36

Distance between AC  = 36   km

Distance between  AB   = 36 + 9  = 45 km

Distance between  BC  = x/2 + 9 = (36/2) + 9 = 27 km

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