Question

The foot of a ladder rests at a distance of 36 m from a wall and the top of the ladder rests on the wall at a height of 15 m from the ground. Find the length of the ladder...


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Answer 1

Given : -

• The foot of a ladder rests at a distance of 36 m from a wall.

• the top of the ladder rests on the wall at a height of 15 m from the ground.

To Find : -

• Find the length of the ladder.

Solution : -

Diagram :

$\setlength{\unitlength}{20mm}\begin{picture}(6,2)\linethickness{0.4mm}\put(7.7,2.9){\large\sf{A}}\put(7.9,0.8){\large\sf{B}}\put(10.4,0.8){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.5,1.9){\sf{\large{15 m}}}\put(9,0.7){\sf{\large{36 m}}}\put(9.8,2.2){\sf{\large{Ladder = ?}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\multiput(8.1,2.8)(.33,-.25){8}{\line(2,1){.3}}\qbezier(11,1)(10.5,1.4)(8.35,3)\end{picture}$

Let the Wall be AB of height 15 metres & Distance b/w Ladder and Wall be BC of distance 36 metres.

AC be the Ladder.

⠀

$\underline{\bigstar\:\textsf{By Pythagoras theorem in \triangle ABC :}}$

$:\implies\sf (Hypotenuse)^2=(Height)^2+(Base)^2\\\\\\:\implies\sf (AC)^2=(AB)^2+(BC)^2\\\\\\:\implies\sf (AC)^2=(15\:m)^2+(36\:m)^2\\\\\\:\implies\sf (AC)^2=225\:m^2+1296\:m^2\\\\\\:\implies\sf (AC)^2=1521\:m^2\\\\\\:\implies\sf AC= \sqrt{1521\:m^2} \\\\\\:\implies\sf AC = \sqrt{39 \:m \times 39 \:m} \\\\\\:\implies\underline{\boxed{\sf \red{AC = 39 \:m}}}$

⠀

$\therefore\:\underline{\textsf{Hence, Length of the ladder is \textbf{\red{39 metres}}}}$

Answer 2

Answer:

Given : -

• The foot of a ladder rests at a distance of 36 m from a wall.

• the top of the ladder rests on the wall at a height of 15 m from the ground.

To Find : -

• Find the length of the ladder.

Solution : -

Diagram :

$\setlength{\unitlength}{20mm}\begin{picture}(6,2)\linethickness{0.4mm}\put(7.7,2.9){\large\sf{A}}\put(7.9,0.8){\large\sf{B}}\put(10.4,0.8){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.5,1.9){\sf{\large{15 m}}}\put(9,0.7){\sf{\large{36 m}}}\put(9.8,2.2){\sf{\large{Ladder = ?}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\multiput(8.1,2.8)(.33,-.25){8}{\line(2,1){.3}}\qbezier(11,1)(10.5,1.4)(8.35,3)\end{picture}[\tex]</p><p></p><p></p><ul><li>Let the Wall be AB of height 15 metres & Distance b/w Ladder and Wall be BC of distance 36 metres.</li><li>AC be the Ladder.</li></ul><p></p><p>[tex]\underline{\bigstar\:\textsf{By Pythagoras theorem in \triangle ABC :}}$

$\rm⟹(Hypotenuse)2=(Height)2+(Base)2$

$\rm⟹(AC)^2=(AB)^2+(BC)2$

$\rm⟹(AC)^2=(15m)^2+(36m)2$

$\rm⟹(AC)^2=225m^2+1296m2$

$\rm⟹(AC)^2=225m^2+1296m2\rm \: ac \: = \: \sqrt{1521}$

$\rm⟹(AC)= \: = \: \sqrt{1521}$

$\rm⟹(AC)= \: = \: \sqrt{39 × 39}$

$\rm⟹(AC)= \: \: {39}$

☆Hence, Length of the ladder is 39 metres