iGCSE Mathematics (0580) :E5.2 Carry out calculations involving the perimeter and area.iGCSE Style Questions Paper 2

(a) 146

(b) 51

(a) 

Since JKL is an isosceles right-angled triangle,
$
\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}
$

both the base and height are 12.8 cm
$
= \frac{1}{2} \times 12.8 \times 12.8
$
$
= \frac{1}{2} \times 163.84
$
$
= 81.92 \, \text{cm}^2
$
The diameter of the semicircle is JL = 12.8 cm
so the radius is:
$
r = \frac{12.8}{2} = 6.4 \, \text{cm}
$
The area of a semicircle is half the area of a full circle
$
\text{Area of semicircle} = \frac{1}{2} \times \pi r^2
$
$
= \frac{1}{2} \times \pi (6.4)^2
$
$
= \frac{1}{2} \times \pi \times 40.96
$
$
= \frac{1}{2} \times 128.77
$
$
= 64.39 \, \text{cm}^2
$

$
\text{Total area} = 81.92 + 64.39
$

$
= 146.31 \, \text{cm}^2
$

(b)

First Find KL
In right-angled triangle JKL, we know:
$
KL^2 = JK^2 + JL^2
$
$
KL^2 = (12.8)^2 + (12.8)^2
$
$
KL^2 = 163.84 + 163.84
$
$
KL^2 = 327.68
$
$
KL \approx 18.1 \, \text{cm}
$
$
\text{Perimeter} = JK + KL + \text{Arc length of semicircle}
$
$
r = \frac{12.8}{2} = 6.4 \, \text{cm}
$
The arc length of a semicircle is half the circumference of a full circle
$
\text{Arc length} = \frac{1}{2} \times 2\pi r = \pi r
$
$
\approx 20.11 \, \text{cm}
$
$
\text{Perimeter} = 12.8 + 18.1 + 20.11
$
$
= 51.01 \, \text{cm}
$