Question

11. In Fig. 10.11, find the distance of D from A, unit of
length is cm, (AB = 2, BC = 4, CD = 2).
A​

Answer 1

Given:

AB = 2 units

BC = 4 units

CD = 2 units

$\angle B = \angle C = 90^\circ$

To find:

Distance of D from A = ?

Solution:

First of all, let us make one construction in the given figure.

Join the line from A to D which cuts the line BC at E.

Please refer to the attached figure now.

There are 2 right angled triangles here, $\triangle ABE,\ \triangle DCE$.

By symmetry, the line BC is divided into 2 equal parts by E i.e. BE = EC = 2.

Using pythagorean theorem in $\triangle ABE$:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AE^{2} = AB^{2} + BE^{2} \\\Rightarrow AE^{2} = 2^{2} + 2^{2}\\\Rightarrow AE = 2\sqrt2\ units$

Using pythagorean theorem in $\triangle DCE$:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow DE^{2} = CD^{2} + CE^{2} \\\Rightarrow DE^{2} = 2^{2} + 2^{2}\\\Rightarrow DE = 2\sqrt2\ units$

Now, the side AD = AE + ED = $2\sqrt2+2\sqrt2 = \bold{4\sqrt2}\ units$

So , the answer is $\bold{4\sqrt2}\ units$

Answer 2

Answer:

this may help you

Step-by-step explanation:

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