Question

ABC is an isosceles right triangle, right angled at C. prove that AB²=2AC²​

Answer 1

Answer:

AB² = 2AC²

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure,

• △ABC is an isosceles right triangle.

• m∠C = 90°

• Seg AC = Seg BC

We have to prove that, AB² = 2AC².

Now, in △ABC, m∠C = 90°.

We know that,

Pythagoras theorem states that in a right-angled triangle, the square of length of the hypotenuse is equal to sum of squares of lengths of two remaining sides.

∴ ( AB )² = ( AC )² + ( BC )²

⇒ AB² = AC² + BC²

⇒ AB² = AC² + ( AC )² - - - [ Given ]

⇒ AB² = AC² + AC²

⇒ AB² = 2AC²

Hence proved!

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Additional Information:

1. Types of triangle:

A. Based on angles:

1. Acute angled triangle ( 3 angles with measures < 90° )

2. Right angled triangle ( 1 angle measures 90° )

3. Obtuse angled triangle ( 1 angle measures > 90° )

B. Based on sides:

1. Equilateral triangle ( 3 sides are equal in length )

2. Isosceles triangle ( 2 sides are equal in length )

3. Scalene triangle ( No side is equal to other sides )

2. Isosceles Right triangle:

A triangle with two equal sides and the included angle between the sides measures 90° ( right angle ) is an isosceles right triangle.

3. Properties of Isosceles Right triangle:

1. The longest side is called hypotenuse.

2. Two sides other than hypotenuse are equal in length.

3. Two angles other than right angle are each of 45°.

4. The square of hypotenuse is equal to twice of square of one of the remaining sides.

Answer 2

$\huge \mathfrak { Given \: data : }\\ \mathfrak \blue{∆ \: ABC \: is \: a \: right \: ∆ \: and \: also \: isosceles \: triangle.,} \\ \huge \mathfrak{ To \: prove:} \\ \mathfrak \blue{ AB²=2AC² }\\ \huge \mathfrak{ Step 1: }\\ \huge \mathfrak{Proof:} \\ \mathfrak \blue{Here,} \\ \mathfrak \blue{Hypotenuse = AB } \\ \mathfrak \blue{Also, as \: it \: is \: given \: that,∆ ABC is isosceles, }\\\huge \mathfrak{ Step 2: }\\ \mathfrak \blue{AC = BC [equal \: sides \: of \: isosceles \: A] }\\ \mathfrak \blue{ Using \: Pythagoras \: theorem, }\\\huge \mathfrak{ Step 3:} \\ \mathfrak \blue{ In \: ∆ ABC, }\\ \mathfrak \blue{ we \: have ;} \\ \mathfrak \blue{AB²= AC² + BC² AB²= AC² + AC² [AC = BC] }\\ \mathfrak \blue{ \mathrm{AB}^{2}=2 \mathrm{AC}^{2}}$