1. A. Tick the correct option.
\( (a) \) The diameter of a circle whose circumference is 22 cm is \( (i) \) \( 3.5 \, cm \) \( (ii) \) \( 7 \, cm \) \( (iii) \) \( 14 \, cm \) \( (iv) \) \( 12 \, cm \)
Answer \( \color{red} 7 \, cm \)
\[ \begin{align*} \text{Circumference} &= 22 \, cm \\ 2 \pi r &= 22 \\ 2 \times \frac{22}{7} \times r &= 22 \\ \\ r &= \frac{\cancelto{1}{22} \times 7}{2 \times \cancelto{1}{22}} \\ \\ r &= \frac{7}{2} \, cm \\ \\ Diameter & = 2 \times radius \\ \\ & = {\cancelto{1}{2}} \times \frac{7}{\cancelto{1}{2}} \\ \\ Diameter & = \color{green} 7 \, cm \\ \end{align*} \]
\( (b) \) Area of a triangle whose base is 3 cm and height is 7 cm is \( (i) \) \( 10.5 \, cm^2 \) \( (ii) \) \( 21 \, cm^2 \) \( (iii) \) \( 10.5 \, cm \) \( (iv) \) \( 105 \, cm^2 \)
Answer \(\color{red} 10.5 \, cm^2 \)
\[ \begin{align*} \text{Area of a triangle} &= \frac{1}{2} \times base \times height \\ \\ &= \frac{1}{2} \times 3 \times 7 \\ \\ &= \frac{1}{2} \times 21 \\ \\ \text{Area of a triangle} &= \color{green} 10.5 \, cm^2 \end{align*} \]
\( (c) \) Area of a field is \( 500000 \, m^2\). Its area in hectare is \( (i) \) 500 hectare \( (ii) \) 50 hectare \( (iii) \) 5 hectare \( (iv) \) 5000 hectare
Answer \( \color{red} 50 \text{ hectare} \)
\[ \begin{align*}\text{Area of the field } &= 50000 \, m^2 \\ \text{1 hectare} &= 10000 \, m^2 \\ \\ \text{Area in hectares} &= \frac{500000}{10000} \\ \\ &= \color{green} 50 \, hectare \\ \end{align*} \]
\( (d) \) Which of the following is the circumference of a circle? \( (i) \) \( 2\pi r^2 \) \( (ii) \) \( \pi d \) \( (iii) \) \( \pi r^2 \) \( (iv) \) \( 2\pi d \)
Answer \( \color{red} \pi d \)
\[ \begin{align*} \text{Circumference of a circle} &= 2\pi r \\ \text{We know} \,( d) &= 2r \\ \text{Circumference} &= \pi d \\ \end{align*} \]
\( (e) \) Area of a right-angled triangle whose sides containing the right angle are 12 dm and 5 dm is– \( (i) \) 60 dm\(^2\) \( (ii) \) 30 dm \( (iii) \) 60 dm \( (iv) \) 30 dm\(^2\)
Answer \( \color{red} 30 \text{ dm}^2 \)
\[ \begin{align*} \text{Area of a right-angled triangle} &= \frac{1}{2} \times base \times height \\ \\ &= \frac{1}{\cancelto{1}{2}} \times \cancelto{6}{12} \, dm \times 5 \, dm \\ \\ &= 6 \, dm \times 5 \, dm \\ \\ \text{Area} &= \color{green} 30 \, dm^2 \\ \end{align*} \]
B. Answer the following questions.
(a) Find the area of a parallelogram whose base is 14 cm and height is 5 cm.
\[ \begin{align*} \text{Area of a parallelogram} &= \text{Base} \times \text{Height} \\ &= 14 \, cm \times 5 \, cm \\ &= 70 \, cm^2 \\ \end{align*} \]
Answer Area of the parallelogram \( = \color{red} 70 \, cm^2 \)
(b) Find the height of a rhombus whose area is \( 168 \, dm^2\) and the corresponding base is 21 dm.
\[ \begin{align*} \text{Area of a rhombus} &= 168 \, dm^2 \\ base &= 21 \, dm \\ \\ \text{Height} &= \frac{Area}{base} \\ \\ &= \frac{168 \, dm^2}{21 \, dm} \\ \\ \text{Height} &= 8 \, dm \\ \end{align*} \]
Answer Height of the rhombus \( = \color{red} 8 \, dm \)
(c) A 44 cm long wire is bent to form a circle. Find the diameter of the circle.
\[ \begin{align*} \text{Circumference of the circle} &= 44 \, cm \\ 2 \pi r &= 44 \\ 2 \times \frac{22}{7} \times r &= 44 \\ \\ r &= \frac{44 \times 7}{2 \times 22} \\ \\ r &= 7 \, cm \\ \\ \text{Diameter} & = 2 \times \text{radius} \\ \\ & = 2 \times 7 \, cm \\ \\ \text{Diameter} & = \color{green} 14 \, cm \\ \end{align*} \]
Answer Diameter of the circle \( = \color{red} 14 \, cm \)
(d) Find the area of a circle whose circumference is 132 cm.
\[ \begin{align*} \text{Circumference of the circle} &= 132 \, cm \\ 2 \pi r &= 132 \\ 2 \times \frac{22}{7} \times r &= 132 \\ \\ r &= \frac{\cancelto{66}{132} \times 7}{\cancelto{1}{2} \times 22} \\ \\ r &= \frac{\cancelto{3}{66} \times 7}{\cancelto{1}{22}} \\ \\ r &= 21 \, cm \\ \\ \text{Area of the circle} &= \pi r^2 \\ \\ \text{Area} &= \frac{22}{\cancelto{1}{7}} \times \cancelto{3}{21} \times 21 \\ \\ &= 22 \times 3 \times 21 \\ \\ &= 66 \times 21 \\ \\ \text{Area} &= \color{green} 1386 \, cm^2 \\ \end{align*} \]
Answer Area of the circle \( = \color{red} 1386 \, cm^2 \)
(e) Find the area of a rhombus whose one side is 8 cm and height is 0.8 dm.
\[ \begin{align*} \text{Side of the rhombus} &= 8 \, cm \\ \text{Height} &= 0.8 \, dm \\ 1 \, dm &= 10 \, cm \\ \text{Height} &= 0.8 \times 10 \, cm \\ &= 8 \, cm \\ \\ \text{Area of the rhombus} &= \text{Side} \times \text{Height} \\ \\ \text{Area} &= 8 \, cm \times 8 \, cm \\ \\ \text{Area} &= \color{green} 64 \, cm^2 \\ \end{align*} \]
Answer Area of the rhombus \( = \color{red} 64 \, cm^2 \)
2. A path \( \color{black} 3.5 \, m \) wide runs inside along the boundary of a square field whose side is \( \color{black} 65 \, m \). Find the area of the path. Also find the cost of manuring the rest of the field at the rate of \( \color{black} \text{₹ } 25 \) per square metre.
Solution
Let ABCD be the square field and EFGH be the internal boundaries of the path.
\[ \begin{align*} \\ (Side ) \, AB & = 65 \, \text{m} \\ \\ \text{Area of the } & \text{square field ABCD} \\ &= \text{side} \times \text{side} \\ &= 65 \, \text{m} \times 65 \, \text{m} \\ &= \color{green} 4225 \, \text{m}^2 \\ \\(Side) \, EF &= 65 - (3.5 + 3.5) \\ &= 65 - 7 \\ &= 58 \, \text{m} \\ \\ \text{Area of the } & \text{inner square EFGH} \\ &= \text{side} \times \text{side} \\ &= 58 \, \text{m} \times 58 \, \text{m} \\ &= \color{green} 3364 \, \text{m}^2 \\ \\ \text{Area of the path } &= (\text{Area of square ABCD}) - (\text{Area of square EFGH}) \\ &= 4225 \, \text{m}^2 - 3364 \, \text{m}^2 \\ &= \color{green} 861 \, \text{m}^2 \\ \end{align*} \]
\[ \begin{align*} \text{Cost of manuring the rest of the field } (EFGH) &= \text{Area of } EFGH \times \text{Rate per } \text{m}^2 \\ &= 3364 \times 25 \\ &= \text{₹} 84100 \\ \end{align*} \]
Answer Area of the path \( = \color{red} 861 \, \text{m}^2 \), Cost of manuring the rest of the field \( = \color{red}\text{₹} 84100 \)
3. A rectangular field has \( \color{black} 80m \) length and \( \color{black} 48m \) breadth. Three roads, each of width \( \color{black} 2m \) pass through the field such that two roads are parallel to the breadth and the third road parallel to the length. Find the area covered by the three roads.
Solution
\[ \begin{align*} \\ (Length) \,AB & = 80m \\ (Breadth) \,BC & = 48m \\ \\\text{Area of } & \text{path EFGH} \\ &= length \times breadth \\ &= 80m \times 2m \\ &= \color{green} 160m^2 \\ \\\text{Area of } & \text{path IJKL} \\ &= length \times breadth \\ &= 2m \times 48m \\ &= \color{green} 96m^2 \\ \\\text{Area of } & \text{path MNOP} \\ &= length \times breadth \\ &= 2m \times 48m \\ &= \color{green} 96m^2 \\ \\\text{Area of } & \text{path STUV } (comman \, \, area) \\ &= side \times side \\ &= 2m \times 24 \\ &= \color{green} 4m^2 \\ \\\text{Area of } & \text{path WXYZ } (comman \, \, area) \\ &= side \times side \\ &= 2m \times 24 \\ &= \color{green} 4m^2 \\ \\\text{Total area covered by the roads} &= EFGH + IJKL + MNOP - STUV - WXYZ \\ & = 160 + 96 + 96 - 4 - 4 \\ & = 352 - 8 \\ & = 344 \, m^2\end{align*} \]
Answer Area covered by the roads \( = \color{red} 344 \, m^2 \)
4. A design on the wall of room consists of \( \color{black} 1000 \) tiles of the shape of parallelogram. If altitude and base of each tile is \( \color{black} 10 \, cm \) and \( \color{black} 4 \, cm \) respectively, find the cost of polishing the design at the rate of \( \color{black} \text{₹ } 9.50 \text{ per } dm^2 \)?.
Solution
\[ \begin{align*} 1 \, cm & = \frac{1}{10} \, dm \\ Base & = 4 \, cm \implies \frac{2}{5}dm \\ Altitude & = 10 \, cm \implies 1 \, dm \\ \\\text{Area of the parallelogram} &= \text{Base} \times \text{Height} \\ &= \frac{2}{5}dm \times 1 \, dm \\ &= \frac{2}{5} \, dm^2 \\ \\\text{Area of 1000 tiles} & = \frac{2}{\cancelto{1}{5}} \times \cancelto{200}{1000} \\ &= 400 \, dm^2 \\ \\\text{Cost of 1000 tiles} & = 400 \times 9.50 \\ &= \color{green} \text{₹ } 3800 \end{align*} \]
Answer Cost of polishing \( = \color{red} \text{₹ } 3800 \)
5. In the diagram, \( \color{black} PQ \parallel SR \), \( \color{black} SP \perp PQ \), \( \color{black} PQ = 60 \, cm \), \( \color{black} SR = 36 \, cm \) and \( \color{black} QR = 25 \, cm \). Find the area of quadrilateral \( \color{black} PQRS \).
Solution
\[ \begin{align*} \text{Draw } & RT \perp PQ \\ PT & = 36 \, cm \\ TQ & = 60 - 36 \\ TQ & = 24 \, cm \\ \\ \triangle RTQ \text{ is a right} & \text{ angled triangle} \\ \text{Using pythagoras } & \text{theorem} \\ (Base)^2 + (Height)^2 &= (Hypotenuse)^2 \\ (24)^2 + (RT)^2 & = (25)^2 \\ 576 + (RT)^2 &= 625 \\ (RT)^2 &= 625 - 576 \\ (RT)^2 &= 49 \\ (RT)^2 &= 7^2 \\ RT &= 7 \, cm \\ \\\text{Area of quadrilateral } PQRS &= (\text{Area of rectangle } PTRS) + (\text{Area of } \triangle RTQ) \\ & = (Length \times Breadth) + \left(\frac{1}{2} \times base \times height \right) \\ & = (36 \times 7) + \left(\frac{1}{\cancelto{1}{2}} \times \cancelto{12}{24} \times 7 \right) \\ & = 252 \, cm^2 + 84 \,cm^2 \\ & = 336 \,cm^2 \\ \end{align*} \]
Answer Area of quadrilateral \( PQRS = \color{red} 336 \,cm^2 \)
6. There is a circular pond and footpath runs along its boundary. A man walks around it, exactly once, keeping close to the edge. If his step is \( \color{black} 44 \, cm \) long and he takes exactly \( \color{black} 600 \) steps to go around the pond, what is the radius of the pond?
Solution
\[ \begin{align*} \text{Length of 1 step } & = 44 \, cm \\ \text{Total steps } &= 600 \\\text{Circumference} &= \text{Distance covered in 600 steps} \\ & = 44 \times 600 \\ Circumference & = 26400 \, cm \\ 2 \pi r & = 26400 \, cm \\ 2 \times \frac{22}{7} \times r &= 26400 \, cm \\ r &= \cancelto{13200}{26400} \times \frac{1}{\cancelto{1}{2}} \times \frac{7}{22} \\ r &= \cancelto{600}{13200} \times \frac{7}{\cancelto{1}{22}} \\ r &= 600 \times 7 \\ r &= 4200 \, cm \\ Convert \implies 1 \, cm &= \frac{1}{100}m \\ \\ r &= \frac{4200}{100} \, m \\ \\ r &= 42 \, m \\ \\ \end{align*} \]
Answer Radius of the pond \( = \color{red} 42 \,m \)
7. From a square metal sheet of side \( \color{black} 28 \, cm \), a circular sheet is cut off. Find the radius of the largest possible circular sheet that can be cut. Also find the area of the remaining sheet.
Solution
\[ \begin{align*} Side & = 28 \, cm \\ Diameter & = 28 \, cm \\ Radius & = 14 \, cm \\ \text{Area of the remaining sheet} & = \text{Area of square} - \text{Area of circle} \\ & = (Side \times Side) - (\pi r^2) \\ & = (28 \times 28) \, cm^2 - \left(\frac{22}{\cancelto{1}{7}} \times \cancelto{2}{14} \times 14 \right) \, cm^2 \\ & = 784 \, cm^2 - (44 \times 14) \, cm^2 \\ & = 784 \, cm^2 - 616 \, cm^2 \\ & = 168 \, cm^2 \\ \end{align*} \]
Answer Area of the remaining sheet \( = \color{red} 168 \,cm^2 \)
8. Triangle ABC is right angled at A. AD is drawn perpendicular to BC. If \( \color{black} AB = 5 \, cm \) and \( \color{black} AC = 12 \, cm \), find the area of \( \color{black} \triangle ABC \). Also find the length of \( \color{black} AD \).
Solution
\[ \begin{align*} \text{Area of } \triangle ABC & = \frac{1}{2} \times Base \times Height \\ & = \frac{1}{2} \times AB \times AC \\ & = \frac{1}{\cancelto{1}{2}} \times 5 \times \cancelto{6}{12} \\ &= 30 \,cm^2 \\ \\ \triangle ABC \text{ is a right} & \text{ angled triangle} \\ \text{Using } & \text{pythagoras theorem} \\ (Hypotenuse)^2 & = (Base)^2 + (Height)^2 \\ (CB)^2 &= (AB)^2 + (AC)^2 \\ (CB)^2 &= (5)^2 + (12)^2 \\ (CB)^2 &= 25 + 144 \\ (CB)^2 &= 169 \\ (CB)^2 &= (13)^2 \\ CB &= 13 \, cm \\ \\ \end{align*} \] \[ \begin{align*} \text{To find} & \text{ length of } AD \\ \text{Consider } & \triangle ABC \\ \text{Base} (CB) & = 13 \, cm \\ \text{Height} (AD) & = \, ? \\ \\\text{Area of } \triangle ABC & = 30 \,cm^2 \\ \frac{1}{2} \times Base \times Height & = 30 \,cm^2 \\ \frac{1}{2} \times 13 \times AD & = 30 \\ \\ AD & = \frac{30 \times 2}{13} \\ \\ AD & = \frac{60}{13} \\ \\ AD & = 4.61 \, cm \\ \end{align*} \]
Answer Area of \( \triangle ABC = \color{red} 30 \,cm^2 \) , Length of \( AD = \color{red} 4.61 \, cm \)