Area ratio questions with answers

Questions

1. Question 1Find an area from two lengths

   Triangles $ABC$ and $PQR$ are similar. $AB = 3$ cm, $PQ = 5$ cm and the area of triangle $PQR$ is $12$ cm². Find the area of triangle $ABC$.

   Show the worked solution

   1. Step 1: Areas of similar figures are in the ratio of the squares of their lengths$\dfrac{A_1}{A_2} = \left(\dfrac{l_1}{l_2}\right)^2$
   2. Step 2: Put the unknown area on top: triangle ABC over triangle PQR$\mathstrut \dfrac{\text{Area of } \triangle ABC}{12} = \left(\dfrac{3}{5}\right)^2$
   3. Step 3: Multiply both sides$\text{Area of } \triangle ABC = \left(\dfrac{3}{5}\right)^2 \times 12$$\text{Area of } \triangle ABC = 4.32$ cm²
2. Question 2Find a length from two areas

   Triangles $ABC$ and $PQR$ are similar, with areas $9$ cm² and $29.16$ cm² respectively. $AB = 5$ cm. Find $PQ$.

   Show the worked solution

   1. Step 1: Lengths are in the ratio of the square roots of the areas$\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}$
   2. Step 2: Put the unknown length on top$\mathstrut \dfrac{PQ}{5} = \sqrt{\dfrac{29.16}{9}}$$\dfrac{PQ}{5} = \dfrac{9}{5}$
   3. Step 3: Multiply both sides$PQ = \dfrac{9}{5} \times 5$$PQ = 9$ cm
3. Question 3A line parallel to a side

   In the diagram, $D$ lies on $AB$ and $E$ lies on $AC$, with $DE$ parallel to $BC$. $AD = 4$ cm, $DB = 5$ cm and the area of triangle $ADE$ is $12$ cm². Find the area of triangle $ABC$.

   Show the worked solution

   1. Step 1: DE ∥ BC, so triangle ADE is similar to triangle ABC$\mathstrut AB = 4 + 5$$AB = 9$ cm
   2. Step 2: Area ratio = (length ratio)²$\dfrac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \left(\dfrac{AD}{AB}\right)^2$$\mathstrut \dfrac{12}{\text{Area of } \triangle ABC} = \left(\dfrac{4}{9}\right)^2$$\text{Area of } \triangle ABC = 60.75$ cm²
4. Question 4Areas on maps

   A map is drawn to a scale of $1 : 500\,000$. A park has an area of $12$ cm² on the map. Find the actual area, in km², of the park.

   Show the worked solution

   1. Step 1: Write the scale as a ratio with units (1 km = 100 000 cm)$\mathstrut 1 : 500\,000$$\mathstrut 1\text{ cm} : 500\,000\text{ cm}$$1\text{ cm} : 5\text{ km}$
   2. Step 2: Square both parts for areas$1^2\text{ cm}^2 : 5^2\text{ km}^2$$1\text{ cm}^2 : 25\text{ km}^2$
   3. Step 3: Multiply both parts by the map area$\mathstrut 12\text{ cm}^2 : 12 \times 25\text{ km}^2$$12\text{ cm}^2 : 300\text{ km}^2$
5. Question 5Crossing lines with parallel sides

   In the diagram, the lines $AD$ and $BC$ meet at $X$, and $AB$ is parallel to $DC$. $AX = 5$ cm, $XD = 10$ cm and the area of triangle $XAB$ is $10$ cm². Find the area of triangle $XDC$.

   Show the worked solution

   1. Step 1: AB ∥ DC, so triangle XDC is similar to triangle XAB$\angle XAB = \angle XDC$ (alternate angles)$\angle AXB = \angle DXC$ (vertically opposite angles)
   2. Step 2: Area ratio = (length ratio)²$\dfrac{\text{Area of } \triangle XDC}{\text{Area of } \triangle XAB} = \left(\dfrac{XD}{XA}\right)^2$$\mathstrut \dfrac{\text{Area of } \triangle XDC}{10} = \left(\dfrac{10}{5}\right)^2$$\text{Area of } \triangle XDC = \left(\dfrac{10}{5}\right)^2 \times 10$$\text{Area of } \triangle XDC = 40$ cm²