**Question 6:**In diagram below,

*ABCH*is a square and

*DEFG*is a rectangle.

*HCDE*is a straight line and

*HC*=

*CD*.

Find the perimeter, in cm, of the whole diagram

*.*

*Solution*:$\begin{array}{l}G{H}^{2}=D{H}^{2}+D{G}^{2}\\ \text{}={24}^{2}+{7}^{2}\\ \text{}=576+49\\ \text{}=625\\ GH=25\text{cm}\\ \\ \text{Perimeterofthewholediagram}\\ =12+12+12+26+7+14+25\\ =108\text{cm}\end{array}$

**Question 7:**In the diagram,

*ABC*and

*EFD*are right-angled triangles.

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

*Solution*:$\begin{array}{l}D{E}^{2}={3}^{2}+{4}^{2}\\ \text{}=9+16\\ \text{}=25\\ DE=\sqrt{25}\\ \text{}=5\text{cm}\\ \\ A{C}^{2}={7}^{2}+{24}^{2}\\ \text{}=49+576\\ \text{}=625\\ AC=\sqrt{625}\\ \text{}=25\text{cm}\\ \\ \text{Perimeteroftheshadedregion}\\ =24+7+\left(25-5\right)+3+4\\ =58\text{cm}\end{array}$

**Question 8:**In the diagram,

*ABD*and

*BCE*are right-angled triangles.

*ABC*and

*DBE*are straight lines. The length of

*AB*is twice the length of

*BC*.

Calculate the length, in cm, of

*CE*.

*Solution*:$\begin{array}{l}A{B}^{2}={25}^{2}-{7}^{2}\\ \text{}=625-49\\ \text{}=576\\ AB=\sqrt{576}\\ \text{}=24\text{cm}\\ BC=24\text{cm}\xf72\\ \text{}=12\text{cm}\\ C{E}^{2}={5}^{2}+{12}^{2}\\ \text{}=25+144\\ \text{}=169\\ CE=\sqrt{169}\\ \text{}=13\text{cm}\end{array}$

**Question 9:**

Diagram below shows two right-angled triangles,

*ABE*and

*CBD*.

*BED*is a straight line.

Find the length, in cm, of

*BC*. Round off the answer to two decimal places.

*Solution*:$\begin{array}{l}{3}^{2}+B{E}^{2}={5}^{2}\\ \text{}B{E}^{2}={5}^{2}-{3}^{2}=16\\ \text{}BE=4\text{cm}\\ \\ B{C}^{2}+{\left(5+4\right)}^{2}={17}^{2}\\ \text{}B{C}^{2}={17}^{2}-{9}^{2}=208\\ \text{}BC=\sqrt{208}\\ \text{}BC=14.42\text{cm}\end{array}$

**Question 10:**

Diagram below shows two right-angled triangles,

*ABC*and

*ADE*.

*ACD*is a straight line.

Find the length, in cm, of

*AE*. Round off the answer to one decimal places.

*Solution*:$\begin{array}{l}A{C}^{2}={12}^{2}+{9}^{2}\\ \text{}=225\\ AC=\sqrt{225}\\ \text{}=15\text{cm}\\ \\ A{E}^{2}={\left(15+9\right)}^{2}+{11.5}^{2}\\ \text{}=576+132.25\\ \text{}=708.25\\ AE=\sqrt{708.25}\\ \text{}=26.6\text{cm}\end{array}$