Let the legs of the isosceles right-angled triangle each have length \( x \). The hypotenuse is:
\[ x\sqrt{2} \]
Therefore, the perimeter is:
\[ x + x + x\sqrt{2} = 2x + x\sqrt{2} \]
We are given that the perimeter is:
\[ 12 + 6\sqrt{2} \]
Setting up the equation:
\[ 2x + x\sqrt{2} = 12 + 6\sqrt{2} \]
Factor out \( x \):
\[ x(2 + \sqrt{2}) = 12 + 6\sqrt{2} \]
Observing that \( x = 6 \) satisfies the equation:
\[ 6(2 + \sqrt{2}) = 12 + 6\sqrt{2} \]
Thus, the length of each leg is \( 6 \) meters.
The area of a right-angled triangle is given by:
\[ \text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 \]
Substituting the values:
\[ \text{Area} = \frac{1}{2} \times 6 \times 6 \]
\[ = \frac{36}{2} = 18 \text{ m}^2 \]
Thus, the area of the triangle is 18 m².
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$