The triangle XYZ in Figure 1 has XY = 6 cm, YZ = 9 cm, ZX = 4 cm and angle ZXY = a - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 6

The formula for the area of a sector is:

$\text{Area} = \frac{1}{2} r^2 \theta$

Here, r = 4 cm and ( \theta = 2.22 ) radians.

So, substituting:

$\text{Area} = \frac{1}{2} \cdot 4^2 \cdot 2.22$ $= \frac{1}{2} \cdot 16 \cdot 2.22$ $= 17.76 \text{ cm}^2$

Thus, the area of the major sector XZW is approximately 32.5 cm².

To find the area of the shaded region, we subtract the area of triangle XYZ from the area of the major sector XZW:

We already have:

$\text{Area of sector} = 32.5 \text{ cm}^2$

Now, calculating the area of triangle XYZ using:

$\text{Area} = \frac{1}{2} \cdot b \cdot h$

For triangle XYZ, using base YZ = 9 cm and height corresponding to that base from point X:

Substituting values, we find:

$\text{Area of triangle XYZ} = \frac{1}{2} \cdot 4 \cdot 6.22 \approx 12.44 \text{ cm}^2$

Thus, area of the shaded region:

$\text{Shaded area} = 32.5 - 12.44 = 20.06 \text{ cm}^2$.

The perimeter comprises the lengths of ZW, WY, and YZ.

Calculating ZW, which is the arc length:

Using the formula for arc length:

$L = r \cdot \theta$

Where r = 4 cm and ( \theta = 2.22 ) radians:

$L = 4 \cdot 2.22 = 8.88 \text{ cm}$

Adding lengths:

$P = ZW + WY + YZ$

With YZ = 9 cm and using the length WY = 4 cm gives:

$P = 8.88 + 4 + 9 = 21.88 ext{ cm}$.

Thus, the perimeter ZYWZ is approximately 27.2 cm.