The question asks for the area of a triangle with a given base and two equal side lengths. This describes an isosceles triangle.
The area of a triangle is calculated using the formula:
$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $
We need to find the height ($h$) of the triangle.
In an isosceles triangle, the altitude drawn to the base bisects the base.
Using the Pythagorean theorem ($a^2 + b^2 = c^2$):
$ 5^2 + h^2 = 6^2 $
$ 25 + h^2 = 36 $
$ h^2 = 36 - 25 $
$ h^2 = 11 $
$ h = \sqrt{11} \text{ cm} $
Now, substitute the base and the calculated height into the area formula:
$ \text{Area} = \frac{1}{2} \times b \times h $
$ \text{Area} = \frac{1}{2} \times 10 \text{ cm} \times \sqrt{11} \text{ cm} $
$ \text{Area} = 5 \times \sqrt{11} \text{ sq. cm} $
The area of the triangle is $5\sqrt{11}$ sq. cm.
In the given figure, if \(RS = 3\sqrt{3} \text{ cm}\) and \(RPS\) is an equilateral triangle, then find the value of \(QR\)
For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?

In a right angled triangle, with angle at A being , the side $AB$ is of length 4cm and $BC$ is 15 cm. What is the length of side $AC$?

What is the angle at $B$ in the triangle $ABC$, if the angle made at $K$ is $30^\circ$ and the triangles shown in the figure are similar?

What is \(AB + BC\) equal to?
In the given figure, if \(RS = 3\sqrt{3} \text{ cm}\) and \(RPS\) is an equilateral triangle, then find the value of \(QR\)
For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?
