1. The area of a triangle is \( \color{black} 90 \, cm^2 \). If its base is \( \color{black} 15 \, cm \), find its altitude.
\[ \begin{align*} \text{Area of the triangle} & = 90 \, \text{cm}^2 \\ \text{Base} &= 15 \, \text{cm} \\ \\ \text{Area of a triangle} &= \frac{1}{2}\times \text{base} \times \text{altitude} \\ \\ \frac{1}{2} \times \text{base} \times \text{altitude} & = 90 \, cm^2 \\ \\ \frac{1}{2} \times 15 \, cm \times \text{altitude} & = 90 \, cm^2 \\ \\ \frac{15}{2} \times \text{altitude} & = 90 \\ \\ \text{altitude} & = 90\times \frac{2}{15} \\ \\ \text{altitude} & = \cancelto{6}{90} \times \frac{2}{\cancelto{1}{15} } \\ \\altitude &= \color{green}12 \, cm \end{align*} \]
Answer Altitude \( = \color{red} 12 \, cm \)
2. Find the base of a triangle whose area is \( \color{black} 0.48 \, dm^2 \) and altitude is \( \color{black} 8 \, cm \).
\[ \begin{align*} \text{Area of the triangle} & = 0.48 \, \text{dm}^2 \\ \text{Altitude} & = 8 \, \text{cm} \\ \\ \text{Convert } \implies & dm^2 \text{ to } cm^2 \\ \\ 1 \, \text{dm} &= 10 \, \text{cm} \\ 1 \, \text{dm} \times 1 \, \text{dm} &= 10 \, \text{cm} \times 10 \, \text{cm} \\ 1 \, \text{dm}^2 &= 100 \, \text{cm}^2 \\ \\ \text{Therefore, Area} & = 0.48 \times 100 \\ &= 48 \, \text{cm}^2 \\ \\\text{Area of the triangle} & = 48 \, \text{cm}^2 \\ \\\frac{1}{2} \times \text{base} \times \text{altitude} & = 48 \, \text{cm}^2 \\ \\ \frac{1}{\cancelto{1}{2}} \times \text{base} \times \cancelto{4}{8} \, cm & = 48 \, \text{cm}^2 \\ \\ \text{base} \times 4 \, cm & = 48 \, \text{cm}^2 \\ \\ \text{base} & = \frac{48 \, \text{cm}^2}{4 \, cm } \\ \\ base & = 12\, cm \end{align*} \]
Answer Base \( = \color{red} 12 \, cm \)
3. Find the area of a right angled triangle whose hypotenuse is \( \color{black} 13 \, cm \) and one side is \( \color{black} 5 \, cm \).
\[ \begin{align*} \text{(Hypotenuse) AC} &= 13\text{cm} \\ \text{(One side) AB} &= 5 \, \text{cm} \\ \\ \text{Pythagorean Theorem} \implies & \quad AB^2 + BC^2 = AC^2 \\ \\ BC^2 &= AC^2 - AB^2 \\ BC^2 & = AC^2 - AB^2 \\ & = 13^2 - 5^2 \\ & = 169 - 25 \\ BC^2& = 144 \\ BC^2& = (12)^2 \\ BC & = 12 \, \text{cm} \\ \\ \text{Area of ΔABC} & = \frac{1}{2} \times \text{base} \times \text{height} \\ \\ \text{Area} & = \frac{1}{2} \times BC \times AB \\ \\ & = \frac{1}{2} \times 12 \times 5 \\ \\ \text{Area}& = 30 \, \text{cm}^2 \\ \end{align*} \]
Answer Area \( = \color{red} 30 \, cm^2 \)
4. Find the area of an isosceles right triangle whose equal sides are \( \color{black} 15 \, cm \) each.
\[ \begin{align*} \text{(Base)} BC &= 15 cm \\ \text{(Height)} AB &= 15 cm \\ Area &= \frac{1}{2} \times b \times h \\ \\ & = \frac{1}{2} \times 15 \,cm \times 15 \,cm \\ \\ & = \frac{1}{2} \times 225 \, cm^2 \\ \\ Area & = 112.5 \,\text{cm}^2 \\ \end{align*} \]
Answer Area of the isosceles right angled triangle \( = \color{red} 112.5 \, cm^2 \)
5. Find the area of an isosceles right triangle having the length of each equal side \( \color{black} 5 \, cm \).
\[ \begin{align*} \text{(Base) BC} &= 5 cm \\ \text{(Height) AB} &= 5 cm \\ \text{Area} &= \frac{1}{2} \times b \times h \\ \\ & = \frac{1}{2} \times 5 \,cm \times 5 \, cm \\ \\ & = \frac{1}{2} \times 25 \, cm^2 \\ \\ & = 12.5 \, cm^2 \\ \end{align*} \]
Answer Area of the isosceles right angled triangle \( = \color{red} 12.5 \, cm^2 \)
6. The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36 per hectare is ₹ 486, find its base and height \( \color{black} 1 \, hectare = 10000 \, m^2 \).
\[ \begin{align*} \text{Let Height of the triangular field} & = x \\ \text{Let Base of the triangular field} & = 3x \\ \text{Total cost for cultivation} & = \text{₹} 486 \\ \text{Cost per hectare} & = \text{₹} 36 \\ \text{Area of the triangular field} & = \frac{486}{36} \text{hectare} \\ & = \frac{27}{2} \text{hectare} \\ & = \frac{27}{2} \times 10000 \text{m}^2 \\ & = 135000 \text{m}^2 \\ \text{Area of the triangular field} & = 135000 \text{m}^2 \\ \frac{1}{2} \times \text{base} \times \text{height} & = 135000 \text{m}^2 \\ \frac{1}{2} \times 3x \times x & = 135000 \\ \frac{3x^2}{2} & = 135000 \\ \ x^2 & = \cancelto{45000}{135000} \times \frac{2}{\cancelto{1}{3}} \\ \ x^2 & = 45000 \times \frac{2}{1} \\ \\ \ x^2 & = 90000 \\ \\ \ x^2 & = 300 \times 300 \\ \ x^2 & = 300^2 \\ x & = 300 \, \text{m} \, (\text{Height of the field}) \\ \\ \text{Base} &= 3 \times x \\ &= 3 \times 300 \\ \text{Base} &= 900 \, \text{m} \\ \end{align*} \]
Answer Height \( = \color{red} 300 \, m \), Base \( = \color{red} 900 \, m \)
7. In the given diagram \( ST= \color{black} 5 \, cm \) , \( QR= \color{black} 9 \, cm \). The area of the larger triangle is \( \color{black} 50 \, cm^2 \). What is the area of shaded region?
\[ \begin{align*} \text{Area of larger } \triangle PQR & = 50 \, \text{cm}^2 \\ \\ \text{Consider } & \triangle QSR \\ (Base) QR & = 9 \, \text{cm} \\ (Height) ST & = 5 \, \text{cm} \\ \\ \text{Area of a smaller } \triangle QSR & = \frac{1}{2} \times \text{base} \times \text{height} \\ \\ & = \frac{1}{2} \times 9 \, cm \times 5 \, cm \\ \\ & = \frac{1}{2} \times 45 \ cm^2\\ \text{Area of a smaller } \triangle QSR & = 22.5 \, \text{cm}^2 \\ \\ \text{Area of a shaded region} & = \text{Area of larger } \triangle PQR - \text{Area of smaller } \triangle QSR \\ & = 50 \, \text{cm}^2 - 22.5 \, \text{cm}^2 \\ & = 27.5 \, \text{cm}^2 \\ \end{align*} \]
Answer The area of the shaded region \( = \color{red} 27.5 \, cm^2 \)
8. The area of rhombus is \( \color{black} 98 \, m^2 \). If one of its diagonal is \( \color{black} 14 \, m \), find the other diagonal.
\[ \begin{align*} \text{Area of the rhombus} & = 98 \, \text{m}^2 \\ \text{(One diagonal) } d_1 & = 14 \, \text{m} \\ \text{(Second diagonal) } d_2 & = \text{?} \\ \text{Area of a rhombus} & = \frac{1}{2} \times \text{d}_1 \times \text {d}_2 \\ \frac{1}{\cancelto{1}{2}} \times \cancelto{7}{14} \times d_2 & = 98 \\ 7 \times d_2 & = 98 \\ \\ d_2 & = \frac{\cancelto{14}{98}}{\cancelto{1}{7}} \\ \\ d_2 & = 14 \, \text{m} \\ \end{align*} \]
Answer The other diagonal of the rhombus \( = \color{red} 14 \, m \)
9. The diagonals of rhombus are \( \color{black} 8 \, cm \) and \( \color{black} 6 \, cm \) respectively. Find its area.
\[ \begin{align*} d_1 & = 8 \, \text{cm} \\ d_2 & = 6 \text{cm} \\ \text{Area of a rhombus:} & = \frac{1}{2} \times \text{d}_1 \times \text {d}_2 \\ & = \frac{1}{\cancelto{1}{2}} \times \cancelto{4}{8} \, cm \times 6 \, cm \\ & = 4 \, cm \times 6 \, cm \\ & = 24 \, \text{cm}^2 \\ \end{align*} \]
Answer The are of the rhombus \( = \color{red} 24 \, cm^2 \)
10. Find the area of a rhombus whose side is \( \color{black} 13 \, cm \) and altitude is \( \color{black} 2 \, cm \).
\[ \begin{align*} \text{Side of the rhombus (base)} & = 13 \, \text{cm} \\ \text{Altitude (height)} & = 2 \, \text{cm} \\ \text{Area of a rhombus:} & = \text{base} \times \text{height} \\ \text{Area} & = 13 \, cm \times 2 \, cm \\ & = 26 \, \text{cm}^2 \\ \end{align*} \]
Answer The area of the rhombus \( = \color{red} 26 \, cm^2 \)
11. The area of a rhombus is \( \color{black} 220.5 \, cm^2 \). If its altitude is \( \color{black} 17.5 \, cm \), find the length of each side of rhombus.
\[ \begin{align*} \text{Area of the rhombus} & = 220.5 \, \text{cm}^2 \\ \text{Altitude (height)} & = 17.5 \, \text{cm} \\ \text{Area of a rhombus} & = base \times height \\ base \times height& = 220.5 \, cm^2 \\ \\ base \times 17.5\ cm & = 220.5 \, cm^2 \\ \\ base & = \frac{220.5 \, cm^2}{17.5\ cm} \\ \\ & = \frac{220.5 \times 10}{17.5 \times 10} \\ \\ & = \frac{2205}{175} \\ \\ & = \frac{\cancelto{441}{2205}}{\cancelto{35}{175}} \\ \\ & = \frac{\cancelto{63}{441}}{\cancelto{5}{35}} \\ \\ & = \frac{63}{5} \\ \\ base & = 12.6 \, \text{cm} \\ \end{align*} \]
Answer The length of each side of the rhombus \( = \color{red} 12.6 \, cm \)
12. A diagonal of a quadrilateral is \( \color{black} 40 \, m \) long and the perpendiculars to it from the opposite corners are \( \color{black} 8 \, m \) and \( \color{black} 10 \, m \) respectively. Find its area.
\[ \begin{align*} \color{green}\text{Method - 1} & \\ \text{In triangle } \triangle ADC \\ \text{Base} & = 40 \, \text{m} \\ \text{Height} & = 8 \, \text{m} \\ \text{Area of the } \triangle ADC & = \frac{1}{2} \times \text{base} \times \text{height} \\ & = \frac{1}{\cancelto{1}{2}} \times 40 \times \cancelto{4}{8} \\ & = 40 \times 4 \\ & = 160 \, \text{m}^2 \\ \text{In triangle } \triangle ABC & \\ \text{Base} & = 40 \, \text{m} \\ \text{Height} & = 10 \, \text{m} \\ \text{Area of the } \triangle ABC & = \frac{1}{2} \times \text{base} \times \text{height} \\ & = \frac{1}{\cancelto{1}{2}} \times 40 \times \cancelto{5}{10} \\ & = 20 \times 10 \\ & = 200 \, \text{m}^2 \\ \text{Total area of the quadrilateral} & = \text{Area of the } \triangle ADC + \text{Area of the } \triangle ABC \\ & = 160 \, \text{m}^2 + 200 \, \text{m}^2 \\ \text{Total area of the quadrilateral} & = 360 \, \text{m}^2 \\ \\ \color{green}\text{Method - 2} & \\ \\ \text{(Base) } b & = 40 \, m \\ \text{(Height 1) } h_1 & = 8 \, m \\ \text{(Height 2) } h_2 & = 10 \, m \\ \text{Area} &= \frac{1}{2} \times base \times (h_1 + h_2) \\ &= \frac{1}{2} \times 40 \, m \times (8 \, m + 10 \, m) \\ &= \frac{1}{\cancelto{1}{2}} \times 40 \, m \times \cancelto{9}{18} \, m \\ & = 40 \, m \times 9 \, m \\ Area & = 360 \, m^2\\ \end{align*} \]
Answer The area of the quadrilateral \( = \color{red} 360 \, m^2 \)
13. Find the area of a rhombus whose diagonals are \( \color{black} 15.2 \, cm \) and \( \color{black} 24 \, cm \) respectively.
\[ \begin{align*} \text{Given:} & \\ \text{(First diagonal) } d_1 & = 15.2 \, \text{cm} \\ \text{(Second diagonal) } d_2 & = 24 \, \text{cm} \\ \text{Area of a rhombus} & = \frac{1}{2} \times \text{d}_1 \times \text{d}_2 \\ \text{Area} & = \frac{1}{\cancelto{1}{2}} \times 15.2 \, cm \times \cancelto{12}{24} \, cm \\ & = 15.2 \, cm \times 12 \, cm \\ & = 182.4 \, \text{cm}^2 \\ \end{align*} \]
Answer The area of the rhombus \( = \color{red} 182.4 \, cm^2 \)