Question

(C)
Radhika wants to visit her frien
it her friend who recently moved
map between Radhika's
as well as the distance known to
house. The road map between Ra
#Radhika
to a new house
home and her fri
as shown in the figure given below:
Radhika are as shown
||2.5 km
10.5 km
Radhika
house
10.5 km
4.5 km
2.5 km
reach the friend's house, the shortest distance
which Radhika has to travel, is
(a) 30.95 km
(b) 32.5 km
(c) 28.5 km
(d) 35.35 km
​

Answer 1

Answer:

option (A)

30.95

is the answerof this question

Answer 2

Answer:

The shortest distance which Radhika has to travel is 30.95 km.

Step-by-step explanation:

• In the given two triangles, ∆AMB and ∆ DMC we can say that angle AMB = angle DMC.
• so by using alternate angle theory we can say that both the triangles are similar and ratio of their sides is equal.
• so, AB/DC = AM/MC = BM/MD
• we know that AB= 21KM , BM = 14.5 KM, MC = 4.5KM , DC = 5KM . we have to find AM and MD.
• so by using the above formula we can say
• AM/MC = AB/DC

AM/4.5 = 21/5

AM = 21 × 4.5 /5

AM = 21 × 0.9 = 18.9 Km

• BM/MD = AB/DC

14.5/MD = 21/5

MD = 14.5 × 5 /21 = 3.45 Km

• so, if we observe the above attached picture we can say that there are four routes for radhika to reach her friend.
• The first route is,

R to A to M to C to F = RA + AM + MC+ CF

= 10.5 + 18.9+ 4.5 + 2.5 = 36.4

• The second route is,

R to A to M to D to F = RA + AM + MD + DF

= 10.5 + 18.9 + 3.45 + 2.5 = 35.35 Km

• The third route is,

R to B to M to D to F = RB + BM + MD + DF

= 10.5 + 14.5 + 3.45 + 2.5 = 30.95 Km

• The fourth route is,

R to B to M to C to F = RB + BM + MC + CF

10.5 + 14.5 + 4.5 + 2.5 = 32 Km

From the above calculations we can conclude that third route is the shortest route and the shortest distance traveled is 30.95Km .

# SP J3