Question

In figure 2, ABC is an isosceles triangle right angled at C with AC=4 cm . Find the length of AB.​

Answer 1

Answer:AB is 1.656

Step-by-step explanation:

In an isosceles triangle there are 2 equal sides .... So AC = BC

Angle C is 90°

Use Pythagoras theorem....

AB^2= BC^2 + AC^2

=4^2+4^2

AB=√16+16. (4^2=16)

=√32

=4√2

4*√2

4*0.414... (√2=0.414)

1.656

Answer 2

$\huge\sf\pink{Answer}$

☞ Length of AB is 2√2cm

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ ∆ABC is an isosceles triangle

✭ The measure of angle C is 90°

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ Length of AB?

$\rule{110}1$

$\huge\sf\purple{Steps}$

$\large\sf\star\: Diagram\: \star$

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{B}}\put(7.7,1){\large\sf{C}}\put(10.6,1){\large\sf{A}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(9,0.7){\sf{\large{4 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

We are here given with a right angled ∆ ABC ,right angled at C .

Then in ∆ABC ,

$\rightarrowtail\sf{\angle ABC +\angle BCA +\angle BAC =180^{\circ}}$

$\rightarrowtail\sf{x+x+90^{\circ}=180^{\circ}}$

$\rightarrowtail\sf{x+x=180^{\circ}-90^{\circ}}$

$\rightarrowtail\sf{2x =90^{\circ}}$

$\red{\rightarrowtail{\sf x =45^{\circ}}}$

So, $\sf{\angle BAC=\angle CBA=90^{\circ}}$

Now in ∆ABC,

$\sf{\dashrightarrow cos\theta=\dfrac{base}{hypotenuse}=\dfrac{AC}{AB}}$

$\sf{\dashrightarrow cos(45^{\circ})=\dfrac{4cm}{AB}}$

$\sf{\dashrightarrow \dfrac{1}{\sqrt{2}}=\dfrac{4cm}{AB}}$

$\sf{\dashrightarrow AB=\dfrac{4cm}{\sqrt{2}}}$

$\sf{\dashrightarrow AB=\dfrac{\sqrt{2}\times\sqrt{2}\times2cm}{\sqrt{2}}}$

$\orange{\sf{\dashrightarrow AB=2\sqrt{2}cm}}$

$\rule{170}3$