Solution

\[ \begin{align*} \text{Diagonal }(d) &= 7 \, \text{cm} \\ h_1 &= 4 \, \text{cm} \\ h_2 &= 3 \, \text{cm} \\ \\ \color{green} \text{Area of the quadrilateral } PQRS &= \color{green} \frac{1}{2} \times d \times (h_1 + h_2) \\ \\ &= \frac{1}{2} \times 7 \times (4 + 3) \\ \\ &= \frac{1}{2} \times 7 \times 7 \\ \\ &= \frac{1}{2} \times 49 \\ \\ \text{Area} &= 24.5 \, \text{cm}^2 \\ \end{align*} \]

Answer Area of the quadrilateral PQRS \( = \color{red} 24.5 \, \text{cm}^2 \)

Solution

\[ \begin{align*} \text{Diagonal }(d) &= 6 \, \text{cm} \\ h_1 &= 2 \, \text{cm} \\ h_2 &= 3.5 \, \text{cm} \\ \\ \color{green} \text{Area of the quadrilateral } ABCD &= \color{green} \frac{1}{2} \times d \times (h_1 + h_2) \\ \\ &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{6}}^{\color{green}3} \times (2 + 3.5) \\ \\ &= 3 \times 5.5 \\ \text{Area} &= 16.5 \, \text{cm}^2 \\ \end{align*} \]

Answer Area of the quadrilateral ABCD \( = \color{red} 16.5 \, \text{cm}^2 \)

Solution

\[ \begin{align*} \text{In trapezium } & AEDC \\ a &= 8 \, \text{cm} \\ b &= 12 \, \text{cm} \\ h &= 2 \, \text{cm} \\ \\ \text{In } & \triangle ABC \\ b &= 12 \, \text{cm} \\ h &= 4 \, \text{cm} \\ \\ \color{green} \text{Area of } ABCDE &= \color{green} (\text{Area of trapezium } AEDC) + (\text{Area of } \triangle ABC) \\ \\ &= \color{green} \left[ \frac{1}{2} \times (a + b) \times h \right] + \left[ \frac{1}{2} \times b \times h \right] \\ \\ &= \left[ \frac{1}{2} \times (8 + 12) \times 2 \right] + \left[ \frac{1}{2} \times 12 \times 4 \right] \\ \\ &= \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 20 \times {\cancel{2}}^{\color{green}1} \right] + \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{12}}^{\color{green}6} \times 4 \right] \\ \\ &= 20 + 24 \\ \text{Area} &= 44 \, \text{cm}^2 \\ \end{align*} \]

Answer Area of ABCDE \( = \color{red} 44 \, \text{cm}^2 \)

Solution

\[ \begin{align*} \text{In trapezium } & PTSR \\ \text{TS }(a) &= 6 \, \text{cm} \\ \text{PR }(b) &= 10 \, \text{cm} \\ \text{TY }(h) &= 4 \, \text{cm} \\ \\ \text{In } & \triangle PRQ \\ \text{PR }(b) &= 10 \, \text{cm} \\ \text{QX }(h) &= 4 \, \text{cm} \\ \\ \color{green} \text{Area of } PTSRQ &= \color{green} (\text{Area of trapezium } PTSR) + (\text{Area of } \triangle PRQ) \\ \\ &= \color{green} \left[ \frac{1}{2} \times (a + b) \times h \right] + \left[ \frac{1}{2} \times b \times h \right] \\ \\ &= \left[ \frac{1}{2} \times (6 + 10) \times 4 \right] + \left[ \frac{1}{2} \times 10 \times 5 \right] \\ \\ &= \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 16 \times {\cancel{4}}^{\color{green}2} \right] + \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{10}}^{\color{green}5} \times 5 \right] \\ \\ &= 32 + 25 \\ \text{Area} &= 57 \, \text{cm}^2 \\ \end{align*} \]

Answer Area of polygon PTSRQ \( = \color{red} 57 \, \text{cm}^2 \)

Solution

\[ \begin{align*} \text{In } & \triangle FED \\ \text{FD }(b) &= 5 \, \text{cm} \\ \text{EY }(h) &= 2 \, \text{cm} \\ \\ & \text{In quadrilateral } FDCB \\ \text{FC }(d) &= 7 \, \text{cm} \\ \text{DJ }(h_1) &= 4 \, \text{cm} \\ \text{BX }(h_2) &= 4 \, \text{cm} \\ \\ \text{In } & \triangle FBA \\ \text{FB }(b) &= 6.5 \, \text{cm} \\ \text{AZ }(h) &= 2 \, \text{cm} \\ \\ &\color{green} \text{Area of } ABCDEF \\ &= \color{green} (\text{Area of } \triangle FED) + (\text{Area of quadrilateral } FDCB) +(\text{Area of } \triangle FBA) \\ \\ &= \color{green} \left[ \frac{1}{2} \times b \times h \right] + \left[ \frac{1}{2} \times d \times (h_1 + h_2) \right] + \left[ \frac{1}{2} \times b \times h \right] \\ \\ &= \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{2}}^{\color{green}1} \times 5 \right] + \left[ \frac{1}{2} \times 7 \times (4 + 4) \right] + \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 6.5 \times {\cancel{2}}^{\color{green}1} \right] \\ \\ &= 5 + \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 7 \times {\cancel{8}}^{\color{green}4} \right] + 6.5 \\ \\ &= 5 + 28 + 6.5 \\ &= 33 + 6.5 \\ \text{Area} &= 39.5 \, \text{cm}^2 \\ \end{align*} \]

Answer Area of polygon ABCDEF \( = \color{red} 39.5 \, \text{cm}^2 \)

Solution

\[ \begin{align*} & \text{In trapezium } ABCH \\ \text{AB }(a) &= 5 \, \text{m} \\ \text{HC }(b) &= 11 \, \text{m} \\ \text{AX }(h) &= 4 \, \text{m} \\ \\ & \text{In rectangle } HCDG \\ \text{HC }(l) &= 11 \, \text{m} \\ \text{CD }(b) &= 5 \, \text{m} \\ \\ & \text{In trapezium } GDEF \\ \text{GD }(a) &= 11 \, \text{m} \\ \text{FE }(b) &= 5 \, \text{m} \\ \text{FY }(h) &= 4 \, \text{m} \\ \\ &\color{green} \text{Area of octagon } ABCDEFGH \\ &= \color{green} (\text{Area of trapezium } ABCH) + (\text{Area of rectangle } HCDG) +(\text{Area of trapezium } GDEF) \\ \\ &= \color{green} \left[ \frac{1}{2} \times (a + b) \times h \right] + \left[ l \times b \right] + \left[ \frac{1}{2} \times (a + b) \times h \right] \\ \\ &= \left[ \frac{1}{2} \times (5 + 11) \times 4 \right] + \left[ 11 \times 5 \right] + \left[ \frac{1}{2} \times (11 + 5) \times 4 \right] \\ \\ &= \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 16 \times {\cancel{4}}^{\color{green}2} \right] + \left[ 55 \right] + \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 16 \times {\cancel{4}}^{\color{green}2} \right] \\ \\ &= 32 + 55 + 32 \\ &= 87 + 32 \\ Area &= 119 \, m^2 \end{align*} \]

Answer Area of octagon ABCDEFGH \( = \color{red} 119 \, \text{m}^2 \)

Solution

Neha's Calculation: The given pentagonal park can be divided into two equal trapezium. \[ \begin{align*} & \text{In trapezium } AXED \\ \text{AD }(a) &= 15 \, \text{m} \\ \text{EX }(b) &= 28 \, \text{m} \\ \text{AX }(h) &= 6 \, \text{m} \\ \\ & \text{In trapezium } BXEC \\ \text{EX }(a) &= 28 \, \text{m} \\ \text{CB }(b) &= 15 \, \text{m} \\ \text{BX }(h) &= 6 \, \text{m} \\ \\ &\color{green} \text{Area of pentagonal park} \\ &= \color{green} (\text{Area of trapezium } AXED) + (\text{Area of trapezium } BXEC) \\ \\ &= \color{green} \left[ \frac{1}{2} \times (a + b) \times h \right] + \left[ \frac{1}{2} \times (a + b) \times h \right] \\ \\ &= \left[ \frac{1}{2} \times (15 + 28) \times 6 \right] + \left[ \frac{1}{2} \times (28 + 15 ) \times 6 \right] \\ \\ &= \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 43 \times {\cancel{6}}^{\color{green}3} \right] + \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times 43 \times {\cancel{6}}^{\color{green}3} \right] \\ \\ &= 129 + 129 \\ \color{green} Area &= \color{green} 258 \, m^2 \end{align*} \] Nidhi's Calculation:The given pentagonal park can be divided into a triangle and a rectangle. \[ \begin{align*} & \text{In } \triangle DEC \\ \text{DC }(b) &= 12 \, \text{m} \\ \text{EY }(h) &= 13 \, \text{m} \\ \\ & \text{In rectangle } ABCD \\ \text{AB }(l) &= 12 \, \text{m} \\ \text{BC }(b) &= 15 \, \text{m} \\ \\ &\color{green} \text{Area of pentagonal park} \\ &= \color{green} (\text{Area of } \triangle DEC) + (\text{Area of rectangle } ABCD) \\ \\ &= \color{green} \left[ \frac{1}{2} \times b \times h \right] + \left[ l \times b \right] \\ \\ &= \left[ \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{12}}^{\color{green}6} \times 13 \right] + \left[12 \times 15 \right] \\ \\ &= 78 + 180 \\ Area &= 258 \, m^2 \end{align*} \]

Answer Area of the pentagonal park \( = \color{red} 258 \, \text{m}^2 \)