Question

The sides of a rectangular fort are 2045m and 1135m. the height of the walls is 105m. find the length of the largest tape which can measure the three dimensions of the fort exactly

Answer 1

$\bigstar$DIAGRAM

$\setlength{\unitlength}{0.74 cm}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\put(8,5.5){2945m}\put(4,6.3){1135m}\put(11.2,7.5){105m}\end{picture}$

$\bigstar$GIVEN

• The sides of a rectangular fort are 2045m and 1135m.
• The height of the walls is 105m.

$\bigstar$TO FIND

• Length of the largest tape which can measure the three dimensions of the fort exactly.

$\sf\:Point\:to\:remember$

We need to find the H.C.F of 2945, 1135 and 105 which would be equal to the longest tape which can measure the three dimensions of the room exactly.

Let's find the prime factors of 2945, 1135 and 105.

$\longrightarrow \sf\: 2945 = 5 \times 19 \times 31 \\ \\ \longrightarrow \sf\: 1135 = 5 \times 227 \\ \\ \longrightarrow \sf\: 105= 3 \times 5 \times 7$

5 is the common prime factor.

$\therefore$$\sf\:HCF=5$

A tape of $\boxed{\sf\:5m}$ can measure the three dimensions of the fort exactly.