Method 1:

Calculate the area of the square ABCD.

$4 x \times 4 x = 16 x^{2}$

[1]

Calculate the area of the triangle ABN, using $A = \frac{1}{2} b h$ , where $b$ is the base of the triangle and $h$ is its perpendicular height.

4-5-12

$A = \frac{1}{2} \times 4 x \times 3 x$

[1]

Simplify.

$A = 6 x^{2}$

Calculate the area of the triangle BMC.

4-5-11

$A = \frac{1}{2} \times 4 x \times 2 x$

Simplify.

$A = 4 x^{2}$

Calculate the area of the triangle MND.

4-5-13

$A = \frac{1}{2} \times x \times 2 x$

Simplify.

$A = x^{2}$

Find the area of the shaded triangle BNM by subtracting the area of triangles BMC and ABN from the area of the square ABCD.

$16 x^{2} - 6 x^{2} - 4 x^{2} - x^{2}$

[1]

Simplify.

$5 x^{2}$ [1]

Method 2:

Calculate the length NM on the right-angled triangle NDM, using Pythagoras' theorem, $a^{2} + b^{2} = c^{2}$ .

Substitute the given values into the formula.

$x^{2} + {(2 x)}^{2} = N M^{2}$

Simplify.

$\begin{array}{rcl} x^{2} + 4 x^{2} & = & N M^{2} \\ 5 x^{2} & = & N M^{2} \end{array}$

Take the square root of both sides.

$\sqrt{5} x = N M$

[1]

4-5-9

Calculate the length BM on the right-angled triangle BMC.

Substitute the given values into Pythagoras' theorem.

${(2 x)}^{2} + {(4 x)}^{2} = B M^{2}$

Simplify.

$\begin{array}{rcl} 4 x^{2} + 16 x^{2} & = & B M^{2} \\ 20 x^{2} & = & B M^{2} \end{array}$

Take the square root of both sides.

$\sqrt{20} x = B M$

[1]

Simplify.

$2 \sqrt{5} x = B M$

4-5-10

Calculate the area of the triangle BNM, using $A = \frac{1}{2} b h$ , where $b$ is the base of the triangle and $h$ is its perpendicular height.

$A = \frac{1}{2} \times 2 \sqrt{5} x \times \sqrt{5} x$

[1]

Simplify.

$5 x^{2}$ [1]

Area & Perimeter (CIE IGCSE Maths: Extended)

Exam Questions