Answer
Verified
423.6k+ views
Hint: Take the points (1, 0), (0, 1), (1, 1) as A, B, C respectively, obtains the values of AB, BC and CA using formula \[\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}\], if \[\left( {{x}_{1}},{{x}_{2}} \right)\] and \[\left( {{y}_{1}},{{y}_{2}} \right)\] are points. Then try to find similarities and relations to get desired results.
Complete step-by-step answer:
Let us take A as (1,0), B as (0,1) and C as (1,1).
Then we will find length of AB, BC and CA using following formula:
If points are $\left( {{x}_{1}},{{y}_{2}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then its distance is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .
So, if A is (1,0) and B is (0,1), then
\[AB=\sqrt{{{\left( 0-1 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}}=\sqrt{1+1}\]
Hence, $AB=\sqrt{2}$
Now if B is (0,1) and C is (1,1), then
\[BC=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 1-1 \right)}^{2}}}=\sqrt{{{1}^{2}}+{{0}^{2}}}\]
Hence, BC = 1.
Now, if C is (1,1) and A is (1,0), then
\[CA=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}}=\sqrt{{{0}^{2}}+{{1}^{2}}}\]
Hence, CA = 1.
So, we can say that $BC=CA\ne AB$.
Therefore, we can say that it is an isosceles triangle and not an equilateral triangle because in isosceles triangle two sides are equal.
Now we will apply Pythagoras theorem,
$\begin{align}
& B{{C}^{2}}+C{{A}^{2}}=A{{B}^{2}} \\
& {{1}^{2}}+{{1}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
\end{align}$
2 = 2.
Here the left hand side is equal to the right hand side. So, the triangle ABC is a right angled triangle.
Therefore, triangle ABC is an acute angled isosceles triangle and not an obtuse angled triangle.
Hence, the correct answer is option (a) and (c).
Note: Another approach is by plotting the points (1, 0), (0, 1) and (1,1) on the graph and joining the points one will just compare the lengths of triangles to understand what will be the answer. Students often stop their solution once they get to know that the triangle is an isosceles triangle. We should check for acute and obtuse angled conditions too.
Complete step-by-step answer:
Let us take A as (1,0), B as (0,1) and C as (1,1).
Then we will find length of AB, BC and CA using following formula:
If points are $\left( {{x}_{1}},{{y}_{2}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then its distance is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .
So, if A is (1,0) and B is (0,1), then
\[AB=\sqrt{{{\left( 0-1 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}}=\sqrt{1+1}\]
Hence, $AB=\sqrt{2}$
Now if B is (0,1) and C is (1,1), then
\[BC=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 1-1 \right)}^{2}}}=\sqrt{{{1}^{2}}+{{0}^{2}}}\]
Hence, BC = 1.
Now, if C is (1,1) and A is (1,0), then
\[CA=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}}=\sqrt{{{0}^{2}}+{{1}^{2}}}\]
Hence, CA = 1.
So, we can say that $BC=CA\ne AB$.
Therefore, we can say that it is an isosceles triangle and not an equilateral triangle because in isosceles triangle two sides are equal.
Now we will apply Pythagoras theorem,
$\begin{align}
& B{{C}^{2}}+C{{A}^{2}}=A{{B}^{2}} \\
& {{1}^{2}}+{{1}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
\end{align}$
2 = 2.
Here the left hand side is equal to the right hand side. So, the triangle ABC is a right angled triangle.
Therefore, triangle ABC is an acute angled isosceles triangle and not an obtuse angled triangle.
Hence, the correct answer is option (a) and (c).
Note: Another approach is by plotting the points (1, 0), (0, 1) and (1,1) on the graph and joining the points one will just compare the lengths of triangles to understand what will be the answer. Students often stop their solution once they get to know that the triangle is an isosceles triangle. We should check for acute and obtuse angled conditions too.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE