Question

Square board side 10cm standing vertically is tilted to the left so that the bottom right corner is raised 6 cm from the ground. by what distance is the top left Corner lowered from its original position.Pls don't post irrelevant answers..If u know the answer u can answer this questions with explanations...​

Answer 1

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In the figure we see that,

$\sf{\longrightarrow \angle ACB+\angle ACE+\angle DCE=180^o}$

Since $\sf{\angle ACE}$ is angle of the square ABCD,

$\sf{\longrightarrow \angle ACB+90^o+\angle DCE=180^o}$

$\sf{\longrightarrow \angle ACB+\angle DCE=90^o\quad\quad\dots(1)}$

From triangle ABC,

$\sf{\longrightarrow \angle CAB=90^o-\angle ACB}$

By (1),

$\sf{\longrightarrow \angle CAB=\angle DCE}$

From triangle CDE,

$\sf{\longrightarrow \angle CED=90^o-\angle DCE}$

By (1),

$\sf{\longrightarrow \angle CED=\angle ACB}$

Now, consider triangles ABC and CDE.

• $\sf{\angle CAB=\angle DCE}$

• $\sf{\angle ACB=\angle CED}$

• $\sf{AC=CE=10\ cm}$

This implies by ASA congruency that,

$\sf{\longrightarrow \triangle ABC\cong\triangle CDE}$

Therefore, since $\sf{\angle CAB=\angle DCE,}$

$\sf{\longrightarrow DE=AB}$

$\sf{\longrightarrow DE=\sqrt{(AC)^2-(BC)^2}}$

$\sf{\longrightarrow DE=\sqrt{10^2-6^2}}$

$\sf{\longrightarrow DE=\sqrt{100-36}}$

$\sf{\longrightarrow DE=\sqrt{64}}$

$\sf{\longrightarrow DE=8\ cm}$

Hence, the distance by which the top left corner gets lowered is,

$\sf{\longrightarrow x=10-DE}$

$\sf{\longrightarrow x=10-8}$

$\sf{\longrightarrow\underline{\underline{x=2\ cm}}}$

Hence 2 cm is the answer.

Answer 2

Step-by-step explanation:

Given :-

• Square board side 10 cm standing vertically is tilted to the left.

• the bottom right corner is raised 6 cm from the ground.

To Find :-

• what distance is the top left Corner lowered from its original position.

Solution :-

Suppose the coordinate of the square (before rotation ) be A(0,0) , B(10,0) , C(10,10) , D(0,10)

• D' is lowered by a distance = 10 - 8 = 2 cm.

A square board side 10 centimeters standing vertically.

It's one side corner length was x cm from ground

Then,

x² = 10² - 6²

x² = 100 - 36

x² = 64

x = 8cm

Then distance is the top left corner lowered from its original position

= 10 - 8

= 2 cm

Hence The 2cm is the top left Corner lowered from its original position.