# In the given figure, ABC is an isosceles right triangle, right-angled at C. Prove that $A{{B}^{2}}=2A{{C}^{2}}$_{.}

Last updated date: 21st Mar 2023

•

Total views: 304.8k

•

Views today: 3.85k

Answer

Verified

304.8k+ views

Hint: Two sides of an isosceles triangle is equal. Pythagoras theorem can be given as ${{\left( \text{hypotaneous} \right)}^{\text{2}}}\text{=}{{\left( \text{base} \right)}^{\text{2}}}\text{+}{{\left( \text{perpendicular} \right)}^{\text{2}}}$. Using these two things the desired result can be obtained.

Complete step-by-step answer:

We have an isosceles triangle ABC with$\angle C={{90}^{o}}$ and hence sides AC and BC are equal to each other, as in isosceles triangle two sides are equal. Hence, diagram of $\Delta ABC$can be represented as

Where AC = BC and $\angle C={{90}^{o}}$

Now, we need to prove the relation given as

$A{{B}^{2}}=2A{{C}^{2}}$ ………………….. (i)

So, let us calculate the value of $A{{B}^{2}}$in terms of $A{{C}^{2}}$to prove the above equation (i).

As we know any right angle triangle will follow the Pythagoras property which can be given as

${{\left( \text{hypotaneous} \right)}^{\text{2}}}\text{=}{{\left( \text{base} \right)}^{\text{2}}}\text{+}{{\left( \text{perpendicular} \right)}^{\text{2}}}$………… (iii)

Hence, we can write the equation (iii) in terms of sides of $\Delta ABC$ as

${{\left( AB \right)}^{2}}={{\left( BC \right)}^{2}}+{{\left( AC \right)}^{2}}$…………… (iv)

Now, as the given triangle is isosceles, therefore AC=BC from the figure; and hence, we can replace BC by AC in the equation (iv), so, we get

$\begin{align}

& {{\left( AB \right)}^{2}}={{\left( AC \right)}^{2}}+{{\left( AC \right)}^{2}} \\

& \Rightarrow {{\left( AB \right)}^{2}}=2{{\left( AC \right)}^{2}} \\

\end{align}$

Hence, equation (i) or the given relation is proved.

Note:

One can go wrong with the Pythagoras theorem. One may put the value of base or perpendicular in place of hypotenuse or vice-versa may also happen. Hence, be clear with the terms of the Pythagoras theorem.

One can get angles A and B as ${{45}^{o}}$. As both are equal and summation of all the angles of the triangle is ${{180}^{0}}$. Now, take $\sin {{45}^{o}}$in the triangle with respect to angle B.

Hence $\sin B=\sin {{45}^{o}}=\dfrac{AC}{AB}$

Now, put $\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}$ and square both sides.

We will get the same result as given in the question.

Complete step-by-step answer:

We have an isosceles triangle ABC with$\angle C={{90}^{o}}$ and hence sides AC and BC are equal to each other, as in isosceles triangle two sides are equal. Hence, diagram of $\Delta ABC$can be represented as

Where AC = BC and $\angle C={{90}^{o}}$

Now, we need to prove the relation given as

$A{{B}^{2}}=2A{{C}^{2}}$ ………………….. (i)

So, let us calculate the value of $A{{B}^{2}}$in terms of $A{{C}^{2}}$to prove the above equation (i).

As we know any right angle triangle will follow the Pythagoras property which can be given as

${{\left( \text{hypotaneous} \right)}^{\text{2}}}\text{=}{{\left( \text{base} \right)}^{\text{2}}}\text{+}{{\left( \text{perpendicular} \right)}^{\text{2}}}$………… (iii)

Hence, we can write the equation (iii) in terms of sides of $\Delta ABC$ as

${{\left( AB \right)}^{2}}={{\left( BC \right)}^{2}}+{{\left( AC \right)}^{2}}$…………… (iv)

Now, as the given triangle is isosceles, therefore AC=BC from the figure; and hence, we can replace BC by AC in the equation (iv), so, we get

$\begin{align}

& {{\left( AB \right)}^{2}}={{\left( AC \right)}^{2}}+{{\left( AC \right)}^{2}} \\

& \Rightarrow {{\left( AB \right)}^{2}}=2{{\left( AC \right)}^{2}} \\

\end{align}$

Hence, equation (i) or the given relation is proved.

Note:

One can go wrong with the Pythagoras theorem. One may put the value of base or perpendicular in place of hypotenuse or vice-versa may also happen. Hence, be clear with the terms of the Pythagoras theorem.

One can get angles A and B as ${{45}^{o}}$. As both are equal and summation of all the angles of the triangle is ${{180}^{0}}$. Now, take $\sin {{45}^{o}}$in the triangle with respect to angle B.

Hence $\sin B=\sin {{45}^{o}}=\dfrac{AC}{AB}$

Now, put $\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}$ and square both sides.

We will get the same result as given in the question.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE