Solution

\[ \begin{align*} a &= 57 \, \text{cm} \\ b &= 33 \, \text{cm} \\ h &= 13 \, \text{cm} \\ \\ \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\ \\ &= \frac{1}{2} \times (57 + 33) \times 13 \\ \\ &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {{\cancel{90}}^{\color{green}45}} \times 13 \\ \\ &= 45 \times 13 \\ \\ \text{Area} &= 585 \, \text{cm}^2 \\ \end{align*} \]

Answer Area of a trapezium \( = \color{red} 585 \, \text{cm}^2 \)

Solution

\[ \begin{align*} \text{Let other } &\text{parallel side be } b \, cm \\ \text{Area} &= 850 \, \text{cm}^2 \\ a &= 64 \, \text{cm} \\ b &= ? \\ h &= 17 \, \text{cm} \\ \\ \color{green} \text{Area of the trapezium} &= 850 \\ \color{green} \frac{1}{2} \times (a + b) \times h &= 850 \\ \\ \frac{1}{2} \times (64 + b) \times 17 &= 850 \\ \\ 64 + b &= \frac{{\cancel{850}}^{\color{green}50} \times 2}{{\cancel{17}}_{\color{green}1}} \\ \\ 64 + b &= 50 \times 2 \\ 64 + b &= 100 \\ b &= 100 - 64 \\ b &= 36 \, \text{cm} \end{align*} \]

Answer The length of the other parallel side \( = \color{red} 36 \, \text{cm} \).

Solution

\[ \begin{align*} \text{Let the distance} &\text{ between the two parallel sides be } h \, cm \\ \text{Area} &= 1,586 \, \text{cm}^2 \\ \text{Sum of parallel sides} &= 122 \, \text{cm} \\ h &= ? \\ \\ \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (\text{Sum of parallel sides}) \times h \\ \\ 1586 &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {{\cancel{122}}^{\color{green}61}} \times h \\ \\ 1586 &= 61 \times h \\ \\ \frac{{\cancel{1586}}^{\color{green}26}}{{\cancel{61}}_{\color{green}1}} & = h\\ \\ h &= 26 \, \text{cm} \end{align*} \]

Answer The distance between the parallel sides \( = \color{red} 26 \, \text{cm} \).

Solution

\[ \begin{align*} dm & \text{ to } m \\ 60 \, \text{dm} &= 6 \, \text{m} \\ \\ \text{Area} &= 28 \, \text{m}^2 \\ a &= 8 \, \text{m} \\ b &= 6 \, \text{m} \\ h &= ? \, \text{m} \\ \\ \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\ \\ 28 &= \frac{1}{2} \times (8 + 6) \times h \\ \\ 28 &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{14}}^{\color{green}7} \times h \\ \\ 28 &= 7 \times h \\ \\ \frac{{\cancel{28}}^{\color{green}4}}{{\cancel{7}}_{\color{green}1}} & = h \\ \\ h &= 4 \, \text{m} \end{align*} \]

Answer The altitude \( = \color{red} 4 \, \text{m} \).

Solution

\[ \begin{align*} a &= 55.6 \, \text{cm} \\ b &= 34.4 \, \text{cm} \\ h &= ? \, \text{cm} \\ \color{green} \text{Area of the trapezium} &= 1080 \, \text{cm}^2 \\ \color{green} \frac{1}{2} \times (a + b) \times h & = 1080 \\ \\ \frac{1}{2} \times (55.6 + 34.4) \times h & = 1080\\ \\ \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{90}}^{\color{green}45} \times h & = 1080\\ \\ 45 \times h & = 1080\\ \\ h &= \frac{{\cancel{1080}}^{\color{green}24}}{{\cancel{45}}_{\color{green}1}} \\ \\ h &= 24 \, \text{cm} \end{align*} \]

Answer The height \( = \color{red} 24 \, \text{cm} \).

Solution

\[ \begin{align*} \text{Area} &= 12 \, \text{m}^2 \\ h &= 3 \, \text{m} \\ a + b &= ? \\ \\ \color{green} \text{Area of the trapezium} &= 12 \, \text{m}^2 \\ \color{green} \frac{1}{2} \times (a + b) \times h &= 12 \\ \\ \frac{1}{2} \times (a + b) \times 3 & = 12 \\ \\ \frac{3}{2} \times (a + b) & = 12 \\ \\ a + b &= \frac{{\cancel{12}}^{\color{green}4} \times 2}{{\cancel{3}}_{\color{green}1}} \\ \\ a + b &= 8 \, \text{m} \end{align*} \]

Answer The sum of the parallel sides \( = \color{red} 8 \, \text{m} \).

Solution

\[ \begin{align*} h &= 8 \, \text{m} \\ \text{Let } a &= x \, (\text{larger side}) \\ \text{Let } b &= x - 4 \, (\text{smaller side}) \\ \\ \color{green} \text{Area of the trapezium} &= 248 \, \text{m}^2 \\ \\ \color{green} \frac{1}{2} \times (a + b) \times h &= 248 \\ \\ \frac{1}{{\cancel{2}}_{\color{green}1}} \times [x + (x - 4)] \times {\cancel{8}}^{\color{green}4} &= 248 \\ 4 \times (x + x - 4) &= 248 \\ 4 \times (2x - 4) &= 248 \\ (2x - 4) &= \frac{{\cancel{248}}^{\color{green}62}}{{\cancel{4}}_{\color{green}1}} \\ 2x - 4 &= 62 \\ 2x &= 62 + 4 \\ 2x &= 66 \\ x &= \frac{66}{2} \\ x &= 33 \, \text{m} \\ \\ \implies a &= 33 \, \text{m} \\ b &= 33 - 4 \\ \implies b &= 29 \, \text{m} \end{align*} \]

Answer The two parallel sides are \( \color{red} 33 \, \text{m} \) and \( \color{red} 29 \, \text{m} \).

Solution

\[ \begin{align*} h &= 10 \, \text{dm} \\ b &= x \, (\text{shorter side}) \\ a &= x + 8 \, (\text{longer side}) \\ \\ 1 \, \text{m}^2 &= 100 \, \text{dm}^2 \\ 1.6 \, \text{m}^2 &= 160 \, \text{dm}^2 \\ \implies \text{Area} &= 160 \, \text{dm}^2 \\ \\ \color{green} \text{Area of the trapezium} &= 160 \, \text{dm}^2 \\ \color{green} \frac{1}{2} \times (a + b) \times h & = 160 \\ \\ \frac{1}{2} \times (x + 8 + x) \times 10 & = 160 \\ \\ \frac{1}{{\cancel{2}}_{\color{green}1}} \times (2x + 8) \times {\cancel{10}}^{\color{green}5} & = 160 \\ \\ (2x + 8) \times 5 & = 160 \\ \\ 2x + 8 & = \frac{\cancel{160}^{32}}{\cancel5_1} \\ \\ 2x + 8 & = 32 \\ 2x & = 32 - 8 \\ 2x & = 24 \\ x & = \frac{24}{2} \\ \\ x &= 12 \, \text{dm} \\ \\ \implies b &= 12 \, \text{dm} \\ a &= 12 + 8 \\ \implies a &= 20 \, \text{dm} \end{align*} \]

Answer The two parallel sides are \( \color{red} 20 \, \text{dm} \) and \( \color{red} 12 \, \text{dm} \).

Solution

\[ \begin{align*} a &= 15 \, \text{m} \\ b &= 8 \, \text{m} \\ h &= ? \\ \\ \color{green} \text{Area of the trapezium} &= 138 \, \text{m}^2 \\ \color{green} \frac{1}{2} \times (a + b) \times h & =138 \\ \\ \frac{1}{2} \times (15 + 8) \times h & = 138 \\ \\ \frac{1}{2} \times 23 \times h & = 138 \\ \\ h &= \frac{{\cancel{138}}^{\color{green}6} \times 2}{{\cancel{23}}_{\color{green}1}} \\ \\ h &= 12 \, \text{m} \end{align*} \]

Answer The depth of the canal \( = \color{red} 12 \, \text{m} \).

Solution

\[ \begin{align*} \text{Draw } & DF \parallel CB \\ \implies DFBC & \text{ is a parallelogram} \\ \text{In } &\triangle ADF \\ DA = DF & = 17 \, cm \\ AB &= 58 \, \text{cm} \\ AF &= 58 - 42 \implies 16 \, \text{cm} \\ DE \perp AF \implies & \text{E is the midpoint of AF} \\ \\ EF = \frac{AF}{2} \\ \\ = \frac{16}{2} \\ \implies EF = 8 \, cm \\ \\ \text{In } \triangle DEF & \text{ (right angled triangle)} \\ (DE)^2 + (EF)^2 &= (DF)^2 \\ (DE)^2 + 8^2 &= 17^2 \\ (DE)^2 + 64 &= 289 \\ (DE)^2 &= 289 - 64 \\ (DE)^2 &= 225 \\ DE &= \sqrt{225} \\ DE &= 15 \, cm \\ \\ \text{In trapezium } & DABC \\ a &= 58 \, \text{cm} \\ b &= 42 \, \text{cm} \\ h &= 15 \, \text{cm} \\ \\ \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\ \\ &= \frac{1}{2} \times (58 + 42) \times 15 \\ \\ &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {{\cancel{100}}^{\color{green}50}} \times 15 \\ \\ &= 50 \times 15 \\ \text{Area} &= 750 \, \text{cm}^2 \end{align*} \]

Answer Area of the trapezium \( = \color{red} 750 \, \text{cm}^2 \).

Solution

\[ \begin{align*} \text{Let shorter side be }(a) &= 4x \\ \text{Let longer side be } (b) &= 7x \\ h &= 14 \, \text{m} \\ \color{green} \text{Area of trapezium} &= 385 \, \text{m}^2 \\ \color{green} \frac{1}{2} \times (a + b) \times h &= 385 \\ \frac{1}{2} \times (4x + 7x) \times 14 &= 385\\ \\ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 11x \times {\cancel{14}}^{\color{green}7} &= 385 \\ 77x &= 385 \\ \\ x &= \frac{{\cancel{385}}^{\color{green}5}}{{\cancel{77}}_{\color{green}1}} \\ \\ x &= 5 \\ \\ \text{Shorter side} &= 4x \\ & = 4 \times 5 \\ \text{Shorter side} &= 20 \, \text{m} \\ \\ \text{Longer side} &= 7x \\ & = 7 \times 5 \\ \text{Longer side} &= 35 \, \text{m} \end{align*} \]

Answer The lengths of the parallel sides are \( \color{red} 20 \, \text{m} \) and \( \color{red} 35 \, \text{m} \).

Solution

\[ \begin{align*} h &= 16 \, m\\ \text{Perimeter} &= 104 \, \text{m} \\ 18 + a + 22 + b & = 104 \\ a + b + 40 & = 104 \\ a + b & = 104 - 40\\ a + b & = 64 \\ \\ \color{green} \text{Area of trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\ \\ & = \frac{1}{\cancel2_1} \times 64 \times \cancel{16}^8 \\ \\ & = 64 \times 8 \\ & = 512 \, \text{m}^2 \end{align*} \]

Answer The area of the trapezium \( = \color{red} 512 \, \text{m}^2 \).