Question

Q4 please help thanks

Answer 1

Answer:

$\large\boxed{A==\dfrac{3-\sqrt6}{12}\ cm^2}$

Step-by-step explanation:

The shaded region is the triangle with base b and height h.

$b=BD-CD\to b=\dfrac{\sqrt2}{2}-\dfrac{\sqrt3}{3}=\dfrac{3\sqrt2}{(2)(3)}-\dfrac{2\sqrt3}{(2)(3)}=\dfrac{3\sqrt2-2\sqrt3}{6}\\\\h=AD\to h=\dfrac{\sqrt2}{2}$

The formula of an area of a triangle:

$A=\dfrac{bh}{2}$

Substitute:

$A=\dfrac{\frac{3\sqrt2-2\sqrt3}{6}\cdot\frac{\sqrt2}{2}}{2}=\left(\dfrac{3\sqrt2-2\sqrt3}{6}\right)\left(\dfrac{\sqrt2}{2}\right)\left(\dfrac{1}{2}\right)\\\\\text{use the distributive property}\ a(b+c)=ab+ac\\\\=\dfrac{(3\sqrt2-2\sqrt3)(\sqrt2)}{(6)(2)(2)}=\dfrac{(3\sqrt2)(\sqrt2)-(2\sqrt3)(\sqrt2)}{24}\\\\\text{use}\ \sqrt{a}\cdot\sqrt{a}=a\ \text{and}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{(3)(2)-2\sqrt6}{24}=\dfrac{2(3-\sqrt6)}{24}=\dfrac{3-\sqrt6}{12}$

Answer 2

Answer:

see explanation

Step-by-step explanation:

The area of the shaded triangle = area of ΔABD - area of ΔADC

A of ΔABD = $\frac{1}{2}$ × AD × BD

A = $\frac{1}{2}$ × $\frac{\sqrt{2} }{2}$ × $\frac{\sqrt{2} }{2}$

   = $\frac{2}{8}$ = $\frac{1}{4}$

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A of ΔACD = $\frac{1}{2}$ × AD × DC

A = $\frac{1}{2}$ × $\frac{\sqrt{2} }{2}$ × $\frac{\sqrt{3} }{3}$

A = $\frac{\sqrt{6} }{12}$

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shaded area = $\frac{1}{4}$ - $\frac{\sqrt{6} }{12}$

               

= $\frac{3}{12}$ - $\frac{\sqrt{6} }{12}$ = $\frac{3-\sqrt{6} }{12}$