Question

A 9.85 m ladder is placed against a wall. The height to the top of the ladder is 5 m more than the distance between the wall and the foot of the ladder. Find the height to the top and the distance between the wall and the foot of the ladder.​ find base and height
base...... m
height.....m

Answer 1

Given:

Length of the ladder = 9.85 m

The height to the top of the ladder is 5 m more than the distance between the wall and the foot of the ladder.

To find:

The height to the top and the distance between the wall and the foot of the ladder.​

Solution:

let x be the distance between the wall and the foot of the ladder. Then the height to the top of the ladder is (x+5).

Pythagoras theorem: In a right angle triangle,

$Hypotenuse^2=Base^2+Perpendicular^2$

In the given situation, hypotenuse is the length of ladder, i.e., 9.85 m. The base is x m and the height is (x+5) m.

Using the Pythagoras theorem, we get

$(9.85)^2=x^2+(x+5)^2$

$97.0225=x^2+x^2+10x+25$

$0=2x^2+10x+25-97.0225$

$0=2x^2+10x-72.0225$

Here, $a=2, b=10,c=-72.0225$. Using the quadratic formula, we get

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-10\pm \sqrt{(10)^2-4(2)(-72.0225)}}{2(2)}$

$x=\dfrac{-10\pm \sqrt{676.18}}{4}$

Approximating the value, we get

$x=\dfrac{-10\pm 26}{4}$

$x=\dfrac{-10+26}{4},\dfrac{-10-26}{4}$

$x=\dfrac{16}{4},\dfrac{-36}{4}$

$x=4,-9$

Distance cannot be negative so $x\neq -9.$

Now we have $x=4$

$x+5=4+5$

$x+5=9$

Therefore, the base is 4 m and the height is 9 m.