AREA

Q.1 Find the area of a triangle, two sides of which are 8 cm and 11 cm andthe perimeter is 32 cm ? EXPLANATION : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.Third side c = 32 cm – (8 + 11) cm = 13 cmSo, 2s = 32, i.e., s = 16 cm,s – a = (16 – 8) cm = 8 cm,s – b = (16 – 11) cm = 5 cm,s – c = (16 – 13) cm = 3 cm.Therefore, area of the triangle = s(s ?a) ( s ??b) (s ??c)= 16??8??5?3 cm2 ??8 30 cm2 Q.2 A triangular park ABC has sides 120m, 80m and 50m. Agardener Dhania has to put a fence all around it and also plant grass inside. Howmuch area does she need to plant? Find the cost of fencing it with barbed wire at therate of Rs 20 per metre leaving a space 3m wide for a gate on one side. Solution : For finding area of the park, we have2s = 50 m + 80 m + 120 m = 250 m.i.e., s = 125 mNow, s – a = (125 – 120) m = 5 m,s – b = (125 – 80) m = 45 m,s – c = (125 – 50) m = 75 m.Therefore, area of the park = s(s ?a) (s ??b) (s ??c)= 125 ??5 ??45 ??75 m2= 375 15m2Also, perimeter of the park = AB + BC + CA = 250 mTherefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)= 247 m Q.3 The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area. Suppose that the sides, in metres, are 3x, 5x and 7x .Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)Therefore, 15x = 300, which gives x = 20.So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 mi.e., 60 m, 100 m and 140 m.Can you now find the area [Using Heron’s formula]?We have s =60 100 1402???m = 150 m,and area will be 150(150??60) (150 ??100) (150 ?140) m2= 150 ??90 ?50 ??10 m2= 1500 3m2 Q.4 Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions (see Fig. 12.11). She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew patatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10000 m2) Solution : Let ABC be the field where wheat is grown. Also let ACD be the fieldwhich has been divided in two parts by joining C to the mid-point E of AD. For thearea of triangle ABC, we havea = 200 m, b = 240 m, c = 360 mTherefore, s =200 +240+ 360 / 2m = 400 m. So, area for growing wheat= 400(400 ??200) (400 – 240) (400 – 360) m2= 400 ??200 ??160 ??40 m2= 16000 2m2 = 1.6 × 2 hectares= 2.26 hectares (nearly)Let us now calculate the area of triangle ACD.Here, we have s =240+ 320 +400/2m = 480 m.So, area of ??ACD = 480(480 ??240) (480 ??320) (480 ??400) m2= 480 ??240 ??160 ??80 m2 = 38400 m2 = 3.84 hectaresWe notice that the line segment joining the mid-point E of AD to C divides thetriangle ACD in two parts equal in area. Can you give the reason for this? In fact, theyhave the bases AE and ED equal and, of course, they have the same height.Therefore, area for growing potatoes = area for growing onions= (3.84 ??2) hectares = 1.92 hectares. Q.5 Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA (see Fig. 12.12). Then they cleanedthe area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 mand ??B = 90º, which group cleaned more area and by how much? Find the total areacleaned by the students (neglecting the width of the lanes). Solution : Since AB = 9 m and BC = 40 m, ??B = 90°, we have:AC = 92 ??402 m= 81 ??1600 m= 1681 m = 41mTherefore, the first group has to clean the area of triangle ABC, which is right angled.Area of ??ABC =12 × base × height=12 × 40 × 9 m2 = 180 m2 The second group has to clean the area of triangle ACD, which is scalene having sides41 m, 15 m and 28 m.Here, s =41+ 15 +28 /2m = 42 mTherefore, area of ??ACD = s(s – a) ( s – b) (s – c )= 42(42 – 41) (42 – 15) (42 – 28) m2= 42 ?1 ??27 ??14 m2 = 126 m2So first group cleaned 180 m2 which is (180 – 126) m2, i.e., 54 m2 more than the areacleaned by the second group.Total area cleaned by all the students = (180 + 126) m2 = 306 m2.    

Q.1 Find the area of rhombus whose diagonals are 5cm & 8cm ?A) 20B) 21C) 22D) 23Ans. AArea of rhombus = ½ * product of diagonals½ * 5 * 8 = 20 Q.2 Circumference of circle is 88m. Find its area ?A) 616 m2B) 618 m2C) 620m2D) 622m2Ans. AArea = (Circumference)2/ 4 p(88)2/4p = 88*88*7/4*22 = 616m2 Q.3 Find the area of triangle whose sides measure 4,6 & 8cm ?A) 3v15B) 2v15C) 4v15D) 6v15Ans. AA=4 , b=6 , c=8Therefore,S=a+b+c/2 = 9Now, area = vS(s-a) (s-b) (s-c)Or v9(9-4) (9-6) (9-8)Or v9*5*3*1Or 3v15 Q.4Three metal cubes having volumes 125 cubic cm, 64 cubic cm and 27 cubic cm are melted to form a new cube. Find the edge of the new cube. A) 10 cm B) 8 cm C) 7 cm D) 6 cm Ans. DTotal volume = 216 cm3 Edge = 6 cm Q.5 The radius of the base of a right circular cone is 6 cm and its slant height is 28 cm. Find the curved surface area of the cone. A) 1268 sq.cm B) 865 sq.cm C) 662 sq.cm D) 528 sq.cmAns. DCurved surface area of a cone = prl= p × 6 × 28 = 528 cm2        

Q.1 A gardener increased the area of his rectangular garden by increasing its length by 40% & decreasing its width by 20%.The area of the new garden A) has increased by 20% B) has increased by 12% C) has increased by 8% D) Same as the old area Ans. B Explanation:The area of the new garden==(40-20- 40*20/100)%=(20-80)%=12% Q.2 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot? A) 274 & 100 m B) 275 & 200 m C) 275 & 100 m D) None of the above Ans. C Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.Perimeter of the plot = 2(Length + Breadth) It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.? 100 × Perimeter = 75000100 × 30x = 750003000x = 75000Dividing both sides by 3000, we obtainx = 25Length = 11x m = (11 × 25) m = 275 mBreadth = 4x m = (4 × 25) m = 100 mHence, the dimensions of the plot are 275 m and 100 m respectively. Q.3 A Village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20m x 15m x 6m. The water of this tank will last for A) 2 days B) 3 days C) 4 days D) 5 daysAns. BTotal amount of water for 4000 people=4000*150=6,00,000 ltrs.=600 m3Volume of tank= 20m * 15m * 6m=1800 m3Therefore,required no. of days=1800/600 =3 Q.4 Cost of carpeting a room 8m long & 4m wide with a carpet 2m wide is Rs.50 per metre. Find the cost of carpeting the floor?A) Rs. 650B) Rs. 700C) Rs. 750D) Rs. 800Ans. DLength = L * B/ Width of carpetOr 8 * 4 / 2=16mNow, Cost = 16*50 = Rs.800 Q.5 A garden is 112m long & 78 m wide. It has walking path 25m wide all around it on inside. Find the area of path ?A) 925 m2B) 927 m2C) 929 m2D) 934 m2Ans. AIf path is within the garden then use this formula:2 * B * (L + B – 2 * B)2 * 25 (112 + 78 – 2 * 25)= 5 * 185 = 925 m2