Question

solution :-

let the ladder BC reaches the building at c

let the height of building where the ladder reaches be ac

now according to question :-

BC = 25mAC = 24m

now we need to find the distance of the root of the ladder from the building

as we know that it is right angled triangle so we can apply Pythagoras theorem

\begin{gathered}:\implies\sf \: bc^{2} = ab^{2} + ac^{2} \\ \end{gathered}:⟹bc2=ab2+ac2​

here bc² is the hypotenuse side

and two right angle sides are ab,AC

now,

\begin{gathered}:\implies\sf \: ab^{2} = bc^{2} - ac^{2} \\ \end{gathered}:⟹ab2=bc2−ac2​

by putting values we get,

\begin{gathered}:\implies\sf \: ab^{2} = {25}^{2} - {24}^{2} \\ \end{gathered}:⟹ab2=252−242​

by squaring these values we get,

\begin{gathered}:\implies\sf \: ab^{2} = 625 - 576 \\ \end{gathered}:⟹ab2=625−576​

now by subtracting 625 - 576 we get,

\begin{gathered}:\implies\sf \: ab^{2} = 49 \\ \end{gathered}:⟹ab2=49​

take the square to opposite side it will become square root to the number

\begin{gathered}:\implies\sf \: ab \: = \sqrt{49} \\ \end{gathered}:⟹ab=49​​

we can write 49 as 7 × 7

\begin{gathered}:\implies\sf \: ab \: = \sqrt{7 \times 7} \\ \end{gathered}:⟹ab=7×7​​

now by taking the number out from the square root we can take one number so we get

\begin{gathered}:\implies\sf \: ab \: = 7m \\ \end{gathered}:⟹ab=7m​

so therefore the distance of foot of the ladder from the building is 7m

Note :-

this method can apply to equilateral triangle and isoceles right triangle also if two straight lines are given.

.
.
.
bye I'm going now
​

Answer 1

Answer:

The velocity of car increases from 50 km per hour 90 km per hour in 0.3 minute calculate the acceleration produced the velocity of car increases from 30 km per hour 90 km per hour in 0.2 minute calculate the acceleration produced?

I don't know what u have typed

Answer 2

Answer:

hi sister

is this a question or my annual exam!!

mathematics, the science of structure, order, and relation that has evolved from elemental practices of counting, measuring, and describing the shapes of objects.