The base of a triangle is 18 cm and height is 20 cm. Find the height of another triangle of double the area having base 24 cm.
30 cm
Area of the first triangle = \(\frac{1}{2} \times 18 \times 20 = 180\) cm2.
The second triangle has double this area, so its area = \(2 \times 180 = 360\) cm2.
Using area \(= \frac{1}{2} \times \text{base} \times \text{height}\) with base 24 cm: \(360 = \frac{1}{2} \times 24 \times h\).
So \(360 = 12h\), giving \(h = 30\) cm.
Hence, the height of the second triangle is 30 cm.
The shorter side of a rectangle is 15 cm less than the longer side. The numerical value of its area is equal to 5 times the numerical value of its perimeter. What is the length (in cm) of its longer side?