In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm......
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 14 | Page 44
In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.
\[s = \frac{a + b + c}{2}\]
\[ = \frac{13 + 13 + 10}{2}\]
\[ = \frac{36}{2}\]
\[ = 18 cm\]
\[\text{Area of the triangle} = \sqrt{18\left( 18 - 13 \right)\left( 18 - 13 \right)\left( 18 - 10 \right)}\]
\[ = \sqrt{2 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2}\]
\[ = 60 sq . cm\]
\[\text{Also}, \]
\[\text{Area of the triangle} = \frac{1}{2} \times base \times height\]
\[ \Rightarrow 60 = \frac{1}{2} \times 10 \times \text{height}\]
\[ \Rightarrow \text{height} = \frac{60}{5}\]
\[ \Rightarrow \text{height} = 12 cm\]
The centroid is located two third of the distance from any vertex of the triangle.
Hence, the distance between the vertex opposite the base and the centroid is 8 cm.
Explanation:-
In an isosceles triangle, the centroid lies on the median drawn from the vertex opposite to the base. Also, the centroid divides the median in the ratio 2:1.
Let AD be the median drawn from the vertex opposite to the base BC. Let G be the centroid such that BG = GC = (1/3)AD.
We can find AD using the Pythagorean theorem. Since the triangle is isosceles, we can draw an altitude from vertex A to the midpoint of BC, say E. Then, BE = BC/2 = 5 cm.
Using the Pythagorean theorem in right triangle ABE, we get:
AE^2 = AB^2 – BE^2 = 13^2 – 5^2 = 144
Therefore, AE = 12 cm.
Since AD is the median, we have:
AD = 2AE = 24 cm
Using the above information, we can find BG and GC as:
BG = GC = (1/3)AD = (1/3)×24 = 8 cm
Now, we can use the Pythagorean theorem in right triangle AGB to find the distance between vertex A and the centroid G, which is the required distance. We have:
AG^2 = AB^2 – BG^2 = 13^2 – 8^2 = 105
Therefore, AG = sqrt(105) cm.
Hence, the distance between the vertex opposite the base and the centroid is sqrt(105) cm.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 14 | Page 45