The sides of certain triangles are given below. Determine which of them are right angled triangles,
(i)a = 7cm,b = 24cm and c = 25
(ii)a=9cm,b=16cm, and c=18cm
(iii)a=1.6cm,b=3.8cm, and c=4cm
(iv) a=8cm, b=10cm, and c=6cm
Answer
665.4k+ views
Hint: We will use the Pythagoras theorem to check if the triangle is a right angled triangle or not. The Pythagoras theorem states that \[{H^2} = {P^2} + {B^2}\] where H is the hypotenuse, P is the perpendicular and B is the base of the right angled triangle.
Complete step-by-step answer:
We know that the longest side of the triangle in a right angled triangle is the hypotenuse so we will consider the longest side in each part to be the hypotenuse and take either of the sides to be the perpendicular or base.
a = 7cm,b = 24cm and c = 25cm
The longest side in this triangle is \[c\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{c^2} = {a^2} + {b^2}\]
L.H.S:
\[
{c^2} \\
= {25^2} \\
= 625 \\
\]
RHS \[
{a^2} + {b^2} \\
= {7^2} + {24^2} \\
= 49 + 576 \\
= 625 \\
\]
L.H.S=R.H.S
Hence, Pythagoras theorem is proved.
Thus, it is a right angled triangle.
\[\left( {ii} \right)a = 9cm{\text{ }},b = 16cm{\text{ }}and{\text{ }}c = 18cm\]
The longest side in this triangle is \[c\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{c^2} = {a^2} + {b^2}\]
L.H.S:
\[
{c^2} \\
= {18^2} \\
= 324 \\
\]
RHS \[
{a^2} + {b^2} \\
= {9^2} + {16^2} \\
= 81 + 256 \\
= 337 \\
\]
L.H.S\[ \ne \]R.H.S
Hence, Pythagoras theorem is not proved.
Thus, it is not a right angled triangle.
\[\left( {iii} \right)a{\text{ }} = 1.6{\text{ }}cm,{\text{ }}b = 3.8{\text{ }}cm{\text{ }}and{\text{ }}c = 4cm\]
The longest side in this triangle is \[c\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{c^2} = {a^2} + {b^2}\]
L.H.S:
\[
{c^2} \\
= {4^2} \\
= 16 \\
\]
RHS \[
{a^2} + {b^2} \\
= {(1.6)^2} + {(3.8)^2} \\
= 2.56 + 14.44 \\
= 17 \\
\]
L.H.S\[ \ne \]R.H.S
Hence, Pythagoras theorem is not proved.
Thus, it is not a right angled triangle.
\[\left( {iv} \right)a = 8cm,{\text{ }}b = 10cm{\text{ }}and{\text{ c = }}6cm\]
The longest side in this triangle is \[b\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{b^2} = {a^2} + {c^2}\]
L.H.S:
\[
{b^2} \\
= {10^2} \\
= 100 \\
\]
RHS \[
{a^2} + {c^2} \\
= {8^2} + {6^2} \\
= 64 + 36 \\
= 100 \\
\]
L.H.S=R.H.S
Hence, Pythagoras theorem is proved.
Thus, it is a right angled triangle.
Therefore, (i) and (iv) are right angled triangles.
Note: In the last part, we observe that the hypotenuse has changed to b since it is the longest side out of all the sides in the given triangle. We need to remember that the hypotenuse is the longest side in any right angled triangle and if we are told to check for a right angled triangle, we will always use the Pythagoras theorem to do so.
Complete step-by-step answer:
We know that the longest side of the triangle in a right angled triangle is the hypotenuse so we will consider the longest side in each part to be the hypotenuse and take either of the sides to be the perpendicular or base.
a = 7cm,b = 24cm and c = 25cm
The longest side in this triangle is \[c\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{c^2} = {a^2} + {b^2}\]
L.H.S:
\[
{c^2} \\
= {25^2} \\
= 625 \\
\]
RHS \[
{a^2} + {b^2} \\
= {7^2} + {24^2} \\
= 49 + 576 \\
= 625 \\
\]
L.H.S=R.H.S
Hence, Pythagoras theorem is proved.
Thus, it is a right angled triangle.
\[\left( {ii} \right)a = 9cm{\text{ }},b = 16cm{\text{ }}and{\text{ }}c = 18cm\]
The longest side in this triangle is \[c\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{c^2} = {a^2} + {b^2}\]
L.H.S:
\[
{c^2} \\
= {18^2} \\
= 324 \\
\]
RHS \[
{a^2} + {b^2} \\
= {9^2} + {16^2} \\
= 81 + 256 \\
= 337 \\
\]
L.H.S\[ \ne \]R.H.S
Hence, Pythagoras theorem is not proved.
Thus, it is not a right angled triangle.
\[\left( {iii} \right)a{\text{ }} = 1.6{\text{ }}cm,{\text{ }}b = 3.8{\text{ }}cm{\text{ }}and{\text{ }}c = 4cm\]
The longest side in this triangle is \[c\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{c^2} = {a^2} + {b^2}\]
L.H.S:
\[
{c^2} \\
= {4^2} \\
= 16 \\
\]
RHS \[
{a^2} + {b^2} \\
= {(1.6)^2} + {(3.8)^2} \\
= 2.56 + 14.44 \\
= 17 \\
\]
L.H.S\[ \ne \]R.H.S
Hence, Pythagoras theorem is not proved.
Thus, it is not a right angled triangle.
\[\left( {iv} \right)a = 8cm,{\text{ }}b = 10cm{\text{ }}and{\text{ c = }}6cm\]
The longest side in this triangle is \[b\], so we consider it to be the hypotenuse.
By the Pythagoras Theorem, we need to prove
\[{b^2} = {a^2} + {c^2}\]
L.H.S:
\[
{b^2} \\
= {10^2} \\
= 100 \\
\]
RHS \[
{a^2} + {c^2} \\
= {8^2} + {6^2} \\
= 64 + 36 \\
= 100 \\
\]
L.H.S=R.H.S
Hence, Pythagoras theorem is proved.
Thus, it is a right angled triangle.
Therefore, (i) and (iv) are right angled triangles.
Note: In the last part, we observe that the hypotenuse has changed to b since it is the longest side out of all the sides in the given triangle. We need to remember that the hypotenuse is the longest side in any right angled triangle and if we are told to check for a right angled triangle, we will always use the Pythagoras theorem to do so.
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