Question

The length, breadth, height of a room are 80 cm, 20 cm and 40 cm respectively. Find the longest tape which can measure the dimensions of the room exactly.

Answer 1

★ QUESTION :

The Length, breadth and height of a room are 80 cm, 20 cm and 40 cm respectively. Find the longest tape which can measure the dimensions of the room exactly.

GIVEN :

• Length of the room = 80 cm
• Breadth of the room = 20 cm
• Height of the room = 40 cm

TO FIND :

• The longest tape which can measure the dimension of the room exactly = ?

STEP - BY - STEP EXPLAINATION :

=> Length of the room = 80 cm ( GIVEN )

=> Breadth of the room = 20 cm ( GIVEN )

=> Height of the room = 40 cm ( GIVEN )

=> The longest tape which can measure the dimension of the room exactly = ?

So, let us find here, the measure of the longest tape which can measure the room exactly.

=> Now , here let's find first the HCF (Highest common Factor) of 80 , 20 and 40.

→ 80 = 2 × 2 × 2 × 2 = 2⁴

→ 20 = 2 × 2 = 2²

→ 40 = 2 × 2 × 2 = 2³

→ HCF ( Highest Common Factor ) = 20

Therefore, The longest tape which can measure the three dimensions of the room = 20 cm.

Answer 2

$➪ \sf \red{Answer:-}$

$\huge{ ➫} \tt GIVEN:-$

$length \: of \: room \: = 7m25cm = 725cm \\ breadth \: of \: room \: = 9m25cm = 925cm \\ height \: of \: room \: = 8m25cm = 825cm$

$\huge ➫ \sf FIND:-$

$now \: we \: find \: HCF \: of \: 725,925 \: and \: 825$

${\huge ➫ {\mathfrak{Solution:-}}}$

$725 = 5 \times 5 \times29$

$925 = 5 \times 5 \times 37$

$825 = 3 \times 5 \times 5 \times 11$

$so, \: heighest \: common \: factor \: is \: 5 \times 5 = 25$

$therefore \: HCF \: of \: 725,925,825 \: is \: 25.$

$\therefore 25m is \: the \: longest \: tape \: measure \: which \: can \: measure \: all \: the \: dimensions.$

$so, \: answer \: is \: \boxed{ \mathfrak{ 25m}}$