1. The area of a triangle is 90 cm2. If its base is 15 cm, find its altitude.
\[ \begin{align*} \text{Given:} & \quad \text{Area of the triangle} = 90 \, \text{cm}^2, \, \text{Base} = 15 \, \text{cm} \\ \text{Formula:} & \quad \text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{altitude} \\ \text{To find:} & \quad \text{Altitude of the triangle} \\ \text{Rearranging the formula:}\\ \text{Altitude} & = \frac{2 \times \text{Area}}{\text{Base}} \\ & = \frac{2 \times 90}{15} \\ & = 12 \, \text{cm} \\ \text{Anser:} & \quad \text{The altitude of the triangle is 12 cm.} \end{align*} \]
2. Find the base of a triangle whose area is 0.48 dm2 and altitude is 8 cm.
\[ \begin{align*} \text{Given:} & \quad \text{Area of the triangle} = 0.48 \, \text{dm}^2, \, \text{Altitude} = 8 \, \text{cm} \\ \text{Conversion:} & \quad 1 \, \text{dm}^2 = 100 \, \text{cm}^2 \\ & \quad \text{Therefore, Area} = 0.48 \times 100 = 48 \, \text{cm}^2 \\ \text{Formula:} & \quad \text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{altitude} \\ \text{To find:} & \quad \text{Base of the triangle} \\ \text{Rearranging the formula:} & \\ \text{Base} & = \frac{2 \times \text{Area}}{\text{Altitude}} \\ & = \frac{2 \times 48}{8} \\ & = 12 \, \text{cm} \\ \text{Answer:} & \quad \text{The base of the triangle is 12 cm.} \end{align*} \]
3. Find the area of a right angled triangle whose hypotenuse is 13 cm and one side is 5 cm.
\[ \begin{align*} \text{Given:} & \quad \text{(Hypotenuse) AC} = 13 \, \text{cm}, \, \text{(One side) AB} = 5 \, \text{cm} \\ \text{Pythagorean Theorem:} & \quad AC^2 = AB^2 + BC^2 \\ \text{Finding the other side (BC):} & \\ BC & = \sqrt{AC^2 - AB^2} \\ & = \sqrt{13^2 - 5^2} \\ & = \sqrt{169 - 25} \\ & = \sqrt{144} \\ BC & = 12 \, \text{cm} \\ \text{Area of ΔABC:} & \quad \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \\ \text{Area} & = \frac{1}{2} \times AB \times BC \\ & = \frac{1}{2} \times 5 \times 12 \\ & = 30 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the triangle is 30 cm}^2. \end{align*} \]
4. Find the area of an isosceles right triangle whose equal sides are 15 cm each.
\[ \begin{align*} \text{Given:} & \quad \text{(Base) BC = 15 cm} \\ & \quad \text{(Height) AB = 15 cm} \\ \text{Area of ΔABC:} & \quad \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \\ \text{Area} & = \frac{1}{2} \times BC \times AB \\ & = \frac{1}{2} \times 15 \times 15 \\ & = \frac{1}{2} \times 225 \\ & = 112.5 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the isosceles right angled triangle is 112.5 cm}^2. \end{align*} \]
5. Find the area of an isosceles right triangle having the length of each equal side 5cm.
\[ \begin{align*} \text{Given:} & \quad \text{(Base) BC = 5 cm} \\ & \quad \text{(Height) AB = 5 cm} \\ \text{Area of ΔABC:} & \quad \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \\ \text{Area} & = \frac{1}{2} \times BC \times AB \\ & = \frac{1}{2} \times 5 \times 5 \\ & = \frac{1}{2} \times 25 \\ & = 12.5 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the isosceles right angled triangle is 12.5 cm}^2. \end{align*} \]
6. The base of a triangular field is three times its height. If the cost of cultivating the field at Rs. 36 per hectare is Rs. 486, find its base and height (1 hectare = 10000 m2).
\[ \begin{align*} \text{Let Height of the triangular field} & = x \, \text{m} \\ \text{Let Base of the triangular field} & = 3x \, \text{m} \\ \text{Total cost for cultivation:} & = Rs. 486 \\ \text{Cost per hectare:} & = Rs. 36 \\ \text{Area of the triangular field:} & = \frac{486}{36} \text{hectare} \\ & = \frac{27}{2} \text{hectare} \\ & = \frac{27}{2} \times 10000 \text{m}^2 \\ & = 135000 \text{m}^2 \\ \text{Area of the triangular field:} & = 135000 \text{m}^2 \\ \frac{1}{2} \times \text{base} \times \text{height} & = 135000 \text{m}^2 \\ \frac{1}{2} \times 3x \times x & = 135000 \\ \frac{1}{2} \times {3x^2} & = 135000 \\ \ x^2 & = \frac{\cancelto{45000}{135000} \times 2}{\cancelto{1}{3}} \\ \ x^2 & = \frac{45000 \times 2}{1} \\ \ x^2 & = 90000 \\ \ x^2 & = 300 \times 300 \\ \ x^2 & = 300^2 \\ \ x & = \sqrt{300^2} \\ x & = 300 \, \text{m} \, (\text{Height of the field}) \\ \text{Base of the field:} & \quad 3 \times 300 = 900 \, \text{m} \\ \text{Answer:} & \quad \text{The height of the field is 300 m and the base is 900 m.} \end{align*} \]
7. In the given diagram ST = 5 cm, QR = 9 cm. The area of the larger triangle is 50 cm2. What is the area of shaded region?
\[ \begin{align*} \text{Given} & \\ \text{Area of larger triangle } \triangle PQR & = 50 \, \text{cm}^2 \\ \text{Base } QR & = 9 \, \text{cm} \\ \text{Height } ST & = 5 \, \text{cm} \\ \text{Area of a smaller triangle } \triangle QSR & = \frac{1}{2} \times \text{base} \times \text{height} \\ & = \frac{1}{2} \times 9 \times 5 \\ & = 22.5 \, \text{cm}^2 \\ \text{Area of a shaded region} & = \text{Area of larger triangle } \triangle PQR - \text{Area of smaller triangle } \triangle QSR \\ & = 50 \, \text{cm}^2 - 22.5 \, \text{cm}^2 \\ & = 27.5 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the shaded region is 27.5 cm}^2. \end{align*} \]
8. The area of rhombus is 98 m2. If one of its diagonal is 14 m, find the other diagonal.
\[ \begin{align*} \text{Given:} & \\ \text{Area of the rhombus} & = 98 \, \text{m}^2 \\ \text{(One diagonal) } d_1 & = 14 \, \text{m} \\ \text{(Second diagonal) } d_2 & = \text{?} \\ \text{Area of a rhombus:} & = \frac{1}{2} \times \text{d}_1 \times \text {d}_2 \\ 98 & = \frac{1}{\cancelto{1}{2}} \times \cancelto{7}{14} \times d_2 \\ 98 & = 7 \times d_2 \\ d_2 & = \frac{98}{7} \\ d_2 & = 14 \, \text{m} \\ \text{Answer:} & \quad \text{The other diagonal of the rhombus is 14 m.} \end{align*} \]
9. The diagonals of rhombus are 8 cm and 6 cm respectively. Find its area.
\[ \begin{align*} \text{Given:} & \\ \text{(One diagonal) } d_1 & = 8 \, \text{cm} \\ \text{(Second diagonal) } d_2 & = 6 \text{cm} \\ \text{Area of a rhombus:} & = \frac{1}{2} \times \text{d}_1 \times \text {d}_2 \\ & = \frac{1}{\cancelto{1}{2}} \times \cancelto{4}{8} \times 6 \\ & = 8 \times 3 \\ & = 24 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The are of the rhombus is 24 cm}^2 \end{align*} \]
10. Find the area of a rhombus whose side is 13 cm and altitude is 2 cm.
\[ \begin{align*} \text{Given:} & \\ \text{Side of the rhombus (base)} & = 13 \, \text{cm} \\ \text{Altitude (height)} & = 2 \, \text{cm} \\ \text{Area of a rhombus:} & = \text{base} \times \text{height} \\ \text{Area} & = 13 \times 2 \\ & = 26 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the rhombus is 26 cm}^2. \end{align*} \]
11. The area of a rhombus is 220.5 cm2. if Its altitude is 17.5 cm, find the length of each side of rhombus.
\[ \begin{align*} \text{Given} & \\ \text{Area of the rhombus} & = 220.5 \, \text{cm}^2 \\ \text{Altitude (height)} & = 17.5 \, \text{cm} \\ \text{Area of a rhombus:} & \quad \text{Area} = \text{base} \times \text{height} \\ \text{Base} & = \frac{\text{Area}}{\text{Height}} \\ \text{Base (Side length)} & = \frac{220.5}{17.5} \\ & = \frac{220.5 \times 10}{17.5 \times 10} \\ & = \frac{2205}{175} \\ & = \frac{\cancelto{441}{2205}}{\cancelto{35}{175}} \\ & = \frac{\cancelto{63}{441}}{\cancelto{5}{35}} \\ & = \frac{63}{5} \\ & = 12.6 \, \text{cm} \\ \text{Answer:} & \quad \text{The length of each side of the rhombus is 12.6 cm.} \end{align*} \]
12. A diagonal of a quadrilateral is 40 m long and the perpendiculars to it from the opposite corners are 8 m and 10 m respectively. Find its area.
\[ \begin{align*} \text{Given} & \\ \text{In triangle } \triangle ADC \\ \text{Base} & = 40 \, \text{m} \\ \text{Height} & = 8 \, \text{m} \\ \text{Area of the } \triangle ADC & = \frac{1}{2} \times \text{base} \times \text{height} \\ & = \frac{1}{\cancelto{1}{2}} \times 40 \times \cancelto{4}{8} \\ & = 40 \times 4 \\ & = 160 \, \text{m}^2 \\ \text{In triangle } \triangle ABC & \\ \text{Base} & = 40 \, \text{m} \\ \text{Height} & = 10 \, \text{m} \\ \text{Area of the } \triangle ABC & = \frac{1}{2} \times \text{base} \times \text{height} \\ & = \frac{1}{\cancelto{1}{2}} \times 40 \times \cancelto{5}{10} \\ & = 20 \times 10 \\ & = 200 \, \text{m}^2 \\ \text{Total area of the quadrilateral} & = \text{Area of the } \triangle ADC + \text{Area of the } \triangle ABC \\ & = 160 \, \text{m}^2 + 200 \, \text{m}^2 \\ & = 360 \, \text{m}^2 \\ \text{Answer:} & \quad \text{The area of the quadrilateral is 360 m}^2. \end{align*} \]
13. Find the area of a rhombus whose diagonals are 15.2 cm and 24 cm respectively.
\[ \begin{align*} \text{Given:} & \\ \text{(First diagonal) } d_1 & = 15.2 \, \text{cm} \\ \text{(Second diagonal) } d_2 & = 24 \, \text{cm} \\ \text{Area of a rhombus:} & = \frac{1}{2} \times \text{d}_1 \times \text{d}_2 \\ \text{Area} & = \frac{1}{\cancelto{1}{2}} \times 15.2 \times \cancelto{12}{24} \\ & = 15.2 \times 12 \\ & = 182.4 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the rhombus is 182.4 cm}^2. \end{align*} \]