Question

Solve please!!
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from wall.

A tree is at a height of 5m from the ground and its top touches the ground at a distance of 12m from the base of the tree. Find the original height of the tree

Answer 1

★ How to do :-

Here, we are given with two problems with a case study. In this we are given with a problem of finding the measurements of the objects to other objects and the actual height of the tree. This sum can be solved by a concept called Pythagoras theorem. It is applied only for the problems of right-angled triangles. In first problem, it forms a right angle as the ladder is hypotunose. In the second problem, the broken part of the tree forms the hypotunose. So, let's solve!!

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➤ Solution (1) :-

${\tt \leadsto \underline{\boxed{\tt {AB}^{2} = {AC}^{2} + {BC}^{2}}}}$

Apply the values of those sides.

${\tt \leadsto {15}^{2} = {12}^{2} + {BC}^{2}}$

Find the square numbers of given values.

${\tt \leadsto 225 = 144 + {BC}^{2}}$

Shift the value from LHS to RHS

${\tt \leadsto {BC}^{2} = 225 - 144}$

${\tt \leadsto {bc}^{2} = 81}$

Now, find the VALUE of BC.

${\tt \leadsto BC = \sqrt{81}}$

${\tt \leadsto \orange{\boxed{\tt BC = 9 \: \: metres}}}$

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➤ Solution (2) :-

${\tt \leadsto \underline{\boxed{\tt {AB}^{2} = {AC}^{2} + {BC}^{2}}}}$

Apply the value of those sides.

${\tt \leadsto {AB}^{2} = {5}^{2} + {12}^{2}}$

Find the square numbers of given values.

${\tt \leadsto {AB}^{2} = 25 + 144}$

${\tt \leadsto {AB}^{2} = 169}$

Now, find the value of AB

${\tt \leadsto AB = \sqrt{169}}$

${\tt \leadsto AB = 13 \: \: metres}$

Actual height of tree :-

${\tt \leadsto 13 + 5}$

${\tt \leadsto \orange{\boxed{\tt 18 \: \: metres}}}$

Hence solved !!