Question

A ladder is leaning against the side of a 10 cm house. if the base of the ladder os 3 cm away from the house. How tall is the ladder? Draw a daigram and show all work
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Answer 1

Answer:

length of the ladder is √109 cm^2

Step-by-step explanation:

In attachement ladder is shown as a red line.

As wall is perpendicular to the ground we can use Pythagoras theorem.

 Using Pythagoras theorem,

⇒ 10^2 + 3^2 cm^2 = length of ladder^2

⇒ 100 + 9 cm^2 = length of ladder^2

⇒ 109 cm^2 = length of the ladder^2

⇒ √109 cm = length of the ladder

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{2cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(9.3,1){\large\sf{C}}\put(8,1){\line(1,0){1.2}}\put(8,1){\line(0,2){1.9}}\qbezier(9.2,1)(8.9,1.5)(8,2.9)\put(7.3,1.9){\sf{\large{10 cm}}}\put(8.4,0.7){\sf{\large{3 cm}}}\put(8.7,2){\bf{Ladder}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(8.1,1.5){\bf{House}}\end{picture}$

$\rule{120}{1}$

• Let AB be House of 10 cm
• Distance b/w House & Ladder = 3 cm
• Let AC be Ladder.

$\underline{\bigstar\:\boldsymbol{Using\: Pythagoras\:Theorem\:Here :}}$

$:\implies\sf AC^2=AB^2+BC^2\\\\\\:\implies\sf (Ladder)^2=(House)^2+(Distance\:b/w)^2\\\\\\:\implies\sf (Ladder)^2=(10\:cm)^2+(3\:cm)^2\\\\\\:\implies\sf (Ladder)^2=100\:cm^2+9\:cm^2\\\\\\:\implies\sf (Ladder)^2=109\:cm^2\\\\\\:\implies\sf Ladder=\sqrt{109\:cm^2}\\\\\\:\implies\underline{\boxed{\sf Ladder=\sqrt{109}\:cm}}$

⠀

$\therefore\:\underline{\textsf{Hence, the Ladder is \textbf{ \sqrt{\text{109}} cm} taller in height}}.$