10.2.2 Perimeter and Area, PT3 Practice - Mathematics

Question 4:
In diagram below, PQUV is a square, QRTU is a rectangle and RST is an equilateral triangle.

The perimeter of the whole diagram is 310 cm.
Calculate the length, in cm, of PV.

Solution:

\begin{array}{l} P V = V U = T S = S R = Q P \\ Given perimeter of the whole diagram = 310 cm \\ P V + V U + U T + T S + S R + R Q + Q P = 310 \\ P V + P V + 50 + P V + P V + 50 + P V = 310 \\ 5 P V + 100 = 310 \\ 5 P V = 210 \\ P V = 42 cm \end{array}

Question 5:
In diagram below, ABCD and CGFE are rectangles. M, G, E and N are midpoints of AB, BC, CD and DA respectively.

Calculate the perimeter, in cm, of the coloured region.

Solution:

\begin{array}{l} Using Pythagoras ‘ theorem \\ M G^{2} = M F^{2} + F G^{2} \\ = 5^{2} + 12^{2} \\ = 25 + 144 \\ = 169 \\ M G = 13 cm \\ Perimeter of the coloured region \\ = 13 + 13 + 5 + 12 + 5 + 12 \\ = 60 cm \end{array}

Question 6:
Diagram below shows a trapezium BCDE and a parallelogram ABEF. ABC and FED are straight lines.

The area of ABEF is 72 cm².
Calculate the area, in cm², of trapezium BCDE.

Solution:

\begin{array}{l} Base \times Height = Area of A B F E \\ 9 cm \times Height = 72 {cm}^{2} \\ Height = \frac{72}{9} \\ = 8 cm \\ C D = 8 cm \\ B C = 23 - 9 \\ = 14 cm \\ Area of trapezium B C D E \\ = \frac{1}{2} \times (B C + E D) \times C D \\ = \frac{1}{2} \times (14 + 9) \times 8 \\ = 92 {cm}^{2} \end{array}