\(Area= \frac{1}{2}\times Base \times Height\)
\(=\frac{1}{2}\times 5\times 12=30\; cm^2\)
Also, area of triangle = \(\frac{1}{2}\times AD\times BC\)
\(30=\frac{1}{2}\times AD\times 13\)
\(\frac{30\times 2}{13}=AD\)
\(AD=4.6\;cm\)
Given:
AB = 5cm and AC = 12cm
Step 1: Calculate the Area of \(\Delta ABC\)
Since \(\Delta ABC\) is a right-angled triangle at A, we can use the lengths of AB and AC to find the area.
The area of a right-angled triangle is given by:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
Thus, the area is:
\(\text{Area} = \frac{1}{2} \times 5 \, \text{cm} \times 12 \, \text{cm}\)
\(\text{Area} = \frac{1}{2} \times 60 \, \text{cm}^2\)
\(\text{Area} = 30 \, \text{cm}^2\)
Step 2: Calculate the Length of AD
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(30 \, \text{cm}^2 = \frac{1}{2} \times 13 \, \text{cm} \times AD\)
\(30 \, \text{cm}^2 = \frac{13}{2} \times AD\)
\(30 \times 2 = 13 \times AD\)
\(60 = 13 \times AD\)
\(AD = \frac{60}{13}\)
\(AD = 4.62 \, \text{cm}\)
So, the answer is 4.62 cm
In triangle \( PQR \), the lengths of \( PT \) and \( TR \) are in the ratio \( 3:2 \). ST is parallel to QR. Two semicircles are drawn with \( PS \) and \( PQ \) as diameters, as shown in the figure. Which one of the following statements is true about the shaded area \( PQS \)? (Note: The figure shown is representative.)