# PERIMETER AND AREA OF PLANE FIGURES

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1 PERIMETER AND AREA OF PLANE FIGURES Q.. Find the area of a triangle whose sides are 8 cm, 4 cm and 30 cm. Also, find the length of altitude corresponding to the largest side of the triangle. Ans. Let ABC be the triangle Let a = 8 cm, b = 4 cm, c = 30 cm a + b + c s = = cm = s = 36 cm Using Hero's Formula Area of ABC = s ( s a) ( s b) ( s c) cm = 36 (36 8) (36 4) (36 30) cm = 36 (8) () (6) cm = cm = cm = 8 cm = 6 cm (ii) Let CL be altitude on the largest side AB of ABC Area of ABC = 6 cm [From (i)] base altitude 6 cm = AB CL 6 cm = 30 CL 6 cm = 5 CL = 6 cm 6 CL = cm = 4.4 cm 5 Hence, altitude on the largest side is 4.4 cm. Q.. Find the area of the triangle whose sides are 3 cm, 4 cm and 5 cm. Also find the height of the triangle corresponding to the longest side. Ans. Let a = 3 cm, b = 4 cm, and c = 5 cm a + b + c s = = = = cm Math Class IX Question Bank

2 Area of triangle = s ( s a) ( s b) ( s c) = ( 3) ( 4) ( 5) = 8 6 cm = 3 3 = 3 = 84 cm Let the height on the longest side be h then, Area of = base height 84 = 5 h Height of triangle, h = = =. cm 5 5 Q.3. Find the area of the triangle whose sides are 30 cm, 4 cm and 8 cm. Also find the length of the altitude corresponding to the smallest side of the triangle. Ans. Let sides of triangle are a = 30 cm, b = 4 cm and c = 8 cm, then a + b + c s = = = = 36 Area of triangle = s ( s a) ( s b) ( s c) = 36 (36 30) (36 4) (36 8) = cm = cm = = 6 cm Let the length of altitude on the smallest side of 8 cm = h. Then area of triangle = base altitude 6 6 = 8 h 9h = 6 h = = 4 cm 9 Altitude of triangle = 4 cm Q.4. The lengths of the sides of triangle are in the ratio 3 : 4 : 5 and its perimeter is 44 cm. Find the area of the triangle. Ans. Ratio of sides = 3 : 4 :5 and perimeter of triangle = 44 cm Let the sides are 3 x, 4 x and 5x 3x + 4x + 5x = 44 x = 44 x = 44 = Sides are 3, 4, 5 i.e. 36 cm, 48 cm and 60 cm. sum of sides 44 s = = = 3 Area of the triangle = s ( s a) ( s b) ( s c) = ( 36) ( 48) ( 60) Math Class IX Question Bank

3 = 36 4 cm = cm = 36 cm = 864 cm Q.5. The base of a triangular field is twice its altitude. If the cost of cultivating the field at Rs 4.50 per 00 m is Rs 5,00, find its base and altitude. Ans. Let altitude of the triangle = x and base of triangle = x Area of triangle = base altitude = x x = x...(i) Total cost of cultivating of the field = Rs 5, 00 Rate of cultivating = Rs 4.50 per 00 m Area of the field = m = = m...(ii) From (i) and (ii), we get x = x = (600) x = 600 Hence, base = 600 = 00 m and altitude = 600 = 600 m Q.6. The perimeter of a right triangle is 60 cm and its hypotenuse is 5 cm. Find the area of the triangle. Ans. In ABC, B = 90 hypotenuse AC = 5 cm and perimeter of triangle = 60 cm. AB + BC = 60 5 = 35 cm Let base BC = x cm then altitude AB = 35 x But AC = AB + BC (Pythagoras theorem) (5) = (35 x) + x 65 = 5 0x + x + x x 0x = 0 x 0x = 0 x 35x = 0 (Dividing both sides by ) x 0x 5x = 0 x ( x 0) 5 ( x 0) = 0 ( x 0) ( x 5) = 0 If x 0 = 0, then x = 0. If x 5 = 0, then x = 5 If x = 0, then base = 0 cm and altitude = 35 0 = 5 cm If x = 5, then base = 5 cm and altitude = 35 5 = 0 cm Hence, sides are 5 cm and 0 cm. Math Class IX 3 Question Bank

4 Q.. Find the length of hypotenuse of an isosceles right angled triangle having an area of 00 cm (Take =.44). Ans. In right angled isosceles triangle ABC, B = 90 and AB = BC Let AB = BC = x cm x Area = BC AB = x x = But area of ABC = 00 cm (Given) x = 00 x = 400 = (0) x = 0 Hypotenuse AC = AB + BC = (0) + (0) = = 800 = 400 = 0 cm = 0.44 cm = 8.80 cm = 8.8 cm Q.8. The lengths of two sides of a right triangle containing the right angle differ by cm. If the area of the triangle is 4 cm, find the perimeter of the triangle. Ans. In ABC, C = 90 Let BC = x cm then AC = x + cm Area of triangle ABC = BC AC 4 = ( ) x x + 48 = x + x x + x 48 = 0 x + 8x 6x 48 = 0 x ( x + 8) 6 ( x + 8) = 0 ( x + 8) ( x 6) = 0 If x + 8 = 0, then x = 8 which is not possible. If x 6 = 0, then x = 6 BC = 6 cm and AC = 6 + = 8 cm But AB = BC + AC (Using Pythagoras Theorem) = (6) + (8) = = 00 = (0) AB = 0 cm Perimeter of the triangle ABC = AB + BC + CA = cm = 4 cm Math Class IX 4 Question Bank

5 Q.9. The sides of a right angled triangle containing the right angle are (5x) cm and (3x ) cm. If its area is 60 cm, find its perimeter. Ans. In ABC, B = 90, Area of triangle = 60 cm (Given) AB = (5 x) cm and BC = (3x ) cm Area of triangle ABC = BC AB 60 = (3x ) 5x 0 = 5x 5x 5x 5x 0 = 0 3x x 4 = 0 3x 9x + 8x 4 = 0 3 x ( x 3) + 8 ( x 3) = 0 ( x 3) (3x + 8) = 0 If x 3 = 0, then x = 3 8 or 3x + 8 = 0, then x =, which is not possible. 3 AB = 5x = 5 3 = 5 cm and BC = 3x = 3 3 = 9 = 8 cm But AC = AB + BC = (Using Pythagoras Theorem) = = 89 AC = 89 = Now, perimeter = ( ) m = 40 m Q.0. Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm. Ans. Perimeter of ABC = 60 cm (Given) 60 Thus each side of triangle = = 0 cm 3 3 Altitude AD = a.3 = 0 =.3 cm 3 Area of ABC = 4 a.3 = (0 ) cm cm = = 4 Math Class IX 5 Question Bank

6 Q.. The area of an equilateral triangle is 36 3 sq. m. Find its perimeter. Ans. Area of equilateral triangle = 36 3 sq.cm. (Given) 3 (side) = 3 (side) = (side) = = Taking square root of both sides, we get (side) = 36 4 (side) = (3 3) (4 4) side = 3 4 = cm Perimeter of equilateral ABC = 3 side = 3 cm = 36 cm. Q.. In the given figure, ABC is an equilateral triangle having each side equal to 0 cm and PBC is right angled at P in which PB = 8 cm. Find the area of the shaded region. Ans. ABC is an equilateral triangle whose each side = 0 cm BPC is right angled triangle in which P = 90 and PB = 8 cm In PBC, BC = PB + PC (Using Pythagoras Theorem) (0) = (8) + PC 00 = 64 + PC PC = = 36 = (6) PC = 6 cm Area of PBC = PB PC cm = = Area of ABC = 3 a =.3 (0) cm = = cm = 43.3 cm 4 Hence, area of shaded portion = Area of ABC Area of PBC = = 9.3 cm Math Class IX 6 Question Bank

7 Q.3. The given figure shows a right angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion. Ans. In right angled ABC, B = 90 AB = 0 cm, BC = 8 cm Area of ABC = AB BC cm = = Side of equilateral BCD = 8 cm Area of equilateral triangle BCD = 3 a =.3 (8) cm 4 4 Hence, area of shaded portion = =. cm = ( ) cm =.88 cm =.9 cm. Q.4. Find the area of an isosceles triangle whose perimeter is 36 cm and base is 6 cm. Ans. ABC is isosceles triangle in which AB = AC. Let, AB = AC = x cm, base BC = 6 cm Perimeter of ABC = 36 cm (Given) AB + AC + BC = 36 cm x + x + 6 = 36 x + 6 = 36 x = 36 6 x = 0 0 x = = 0 AB = x cm = 0 cm and AC = x cm = 0 cm Let a = 6 cm, b = 0 cm and c = 0 cm a + b + c s = = cm = = 8 cm Using Hero s Formula Area of ABC = s ( s a) ( s b) ( s c) = 8 (8 6) (8 0) (8 0) cm = 8 () (8) (8) cm = ( 3 3) () ( ) ( ) cm = 3 3 cm = ( ) ( ) ( ) ( ) (3 3) cm Math Class IX Question Bank

8 = 3 cm = 48 cm Q.5. The base of an isosceles triangle is 4 cm and its area is 9 sq. cm. Find its perimeter. Ans. Let ABC be an isosceles triangle in which AB = AC Let AB = AC = x cm, base BC = 4 cm Let a = 4 cm, b = x cm and c = x cm a + b + c 4 + x + x ( + x) s = = cm = cm = ( + x) cm Area of ABC = 9 cm (Given) s ( s a) ( s b) ( s c) = 9 cm ( + x) ( + x 4) ( + x x) ( + x x) = 9 ( + x) ( x ) () () = 9 ( + x) ( x ) = 9 9 ( + x) ( x ) = = 6 Squaring both sides, we get ( + x) ( x )) = (6) ( + x) ( x ) = 6 6 x 44 + x x = 56 x 44 = 56 x = = 400 Squaring root both sides, we get x = 400 = 0 0 x = 0 a = 4 cm b = x cm b = 0 cm and c = x cm c = 0 cm Perimeter of ABC = a + b + c = ( ) cm = 64 cm Q.6. Each of the equal sides of an isosceles triangle is cm more than its height and the base of the triangle is cm. Find the area of the triangle. Ans. Let height AD = x cm then AB = AC = ( x + ) cm Base BC = cm AD BC BD = DC = = 6 cm In ABD, AB = BD + AD Math Class IX 8 Question Bank

11 Q.0. Find the area and perimeter of quadrilateral ABCD, given below; if, AB = 8 cm, AD = 0 cm, BD = cm, DC = 3 cm and DBC = 90. Ans. In DBC, right angled at B DB = cm, DC = 3 cm DB + BC = DC [Using Pythagoras Theorem] () + BC = (3) 44 + BC = 69 BC = = 5 Taking square root of both sides, we get BC = 5 BC = 5 Area of quadrilateral ABCD = Area of ABD + Area of DBC...(i) To find area of ADB, Let a = cm, b = 0 cm, c = 8 cm a + b + c s = = cm Using Hero s Formula, Area of ABD = s ( s a) ( s b) ( s c) 30 = cm = 5 cm = 5 (5 ) (5 0) (5 8) cm = 5 (3) (5) () cm = (5 5) (3 3) cm = cm = 5 3 cm = = cm Area of DBC = base height = BC DB 5 cm = = 5 6 cm = 30 cm From (i), we get Area of quadrilateral ABCD = Area of ABD + Area of DBC = cm + 30 cm = cm = 69. cm (ii) Perimeter of quadrilateral ABCD = AB + BC + DC + AD = 8 cm + 5 cm + 3 cm + 0 cm = 36 cm Math Class IX Question Bank

14 AC = AC = 36 = (6) AC = 6 cm Area of ABC = AB AC cm = = Area of quadrilateral ABCD = 30 cm Area of ADC = (30 3.5) cm = 6.5 cm Area of ADC = base altitude = AC DL = 6 DL = 3DL 6.5 3DL = 6.5 DL = = 5.5 cm 3 Q.6. The perimeter of a rectangle is 40.5 m and its breadth is 6 m. Find its length and area. Ans. Let ABCD be a rectangle. Let length be l m and breadth be 6 m Perimeter of rectangle = 40.5 m ( l + b) = 40.5 m ( l + 6) = 40.5 m l + = 40.5 l = 40.5 l = l = l = 4.5 m (ii) Area of rectangle ABCD = l b = m = m = 85.5 m. Q.. The perimeter of a rectangular field is 3 5 km. If the length of the field is twice its width; find the area of the rectangle in sq. metres. Ans. Let ABCD be a rectangular field. Let breadth of rectangle = x m and length = x m 3 Perimeter of rectangular field = km 5 3 ( l + b) = 000 m 5 ( x + x) = 600 6x = 600 Math Class IX 4 Question Bank

15 600 x = x = 00 6 l = x m = 00 m = 00 m and b = x m = 00 m Area of rectangular field = l b = m = 0000 m. Q.8. A rectangular plot 85 m long and 60 m broad is to be covered with grass leaving 5 m all around. Find the area to be laid with grass. Ans. Let ABCD be the plot and length of plot = 85 m and breadth = 60 m Let PQRS be the grassy plot Length of grassy lawn = 85 m 5 m = 85 m 0 m = 5 m Breadth of grassy lawn = 60 m 5 m = 60 m 0 m = 50 m Area of grassy lawn = length breadth = (5 50) m = 350 m. Q.9. The length and the breadth of rectangle are 6 cm and 4 cm respectively. Find the height of a triangle whose base is 6 cm and whose area is 3 times that of the rectangle. Ans. Let ABCD be the rectangle length of rectangle = 6 cm, breadth = 4 cm Area of rectangle = length breadth = 6 4 cm = 4 cm In triangle EAB, base of triangle = AB = 6 cm Let height of triangle = EM cm Area of triangle EAB = 3 Area of rectangle ABCD base height 3 4 = AB EM = 6 EM = 3EM = EM = 4 Height of triangle = 4 cm Math Class IX 5 Question Bank

16 Q.30. A foot path of uniform width runs all around inside of a rectangular field 45 m long and 36 m wide. If the area of the path is 34 m. Find the width of the path. Ans. Length of rectangular field ( l ) = 45 m and breadth of the field ( b ) = 36 m Let width of inner path = x m Thus, inner length of rectangular field ( l ) = 45 x = (45 x) m inner breadth of rectangular field ( b ) = 36 x = (36 x) m Area of path = 34 m (Given) l b l b = (45 x) (36 x) = (60 90x x + 4 x ) = (60 6x + 4 x ) = 34 + x x = x x = 34 4x 6x + 34 = 0 x 8x + = 0 x 8x 3x + = 0 x ( x 39) 3 ( x 39) = 0 (x 3) ( x 39) = 0 3 If x 3 = 0 then x = If x 39 = 0, then x = 39 which is not possible. 3 Width of path = m =.5 m Q.3. The adjoining diagram shows two cross paths drawn inside a rectangular field 45 m long and 38 m wide, one parallel to length and the other parallel to breadth. The width of each path is 4 m. Find the cost of gravelling the paths at Rs 5.60 per m. Ans. Length of rectangular field ( l ) = 45 m Breadth of rectangular field ( b ) = 38 m Width of path ( a ) = 4 m Math Class IX 6 Question Bank

17 Area of path = l a + b a a a = ( ) 4 (4) m = ( l + b) a a = = (33 6) m = 36 m Rate of gravelling the path = Rs 5.60 per m Total cost of path = Rs = Rs Q3. A rectangle of area 44 cm has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 5 cm, write down an equation in x and solve it to determine the dimensions of the rectangle. Ans. Area of rectangle = 44 cm and perimeter of rectangle = 5 cm Length of rectangle = x cm Area of rectangle 44 Breadth of rectangle = = cm Length of rectangle x Perimeter Length Breadth = 5 x = = (6 x) cm Area of rectangle = length breadth 44 = x (6 x) x 6x + 44 = 0 44 = 6x x x 8x 8x + 44 = 0 x ( x 8) 8 ( x 8) = 0 ( x 8) ( x 8) = 0 If x 8 = 0, then x = 8 If x 8 = 0, then x = 8 x = 8 or 8 Hence, length of rectangle = 8 cm and breadth of rectangle = 8 cm Q.33. The perimeter of a rectangular plot is 30 m and its area is 000 m. Take the length of the plot as x metres. Use the perimeter to write the value of breadth in terms of x. Use the values of length, breadth and area to write an equation in x. Solve the equation and calculate the length and breadth of the plot. Ans. Area of rectangular field = 000 m and perimeter of rectangular field = 30 m Let length of the field = x m Math Class IX Question Bank

18 30 x Breadth of rectangular field = = (65 x) m Area of rectangular field = l b 000 = x (65 x) 000 = 65x x x 65x = 0 x 40x 5x = 0 x ( x 40) 5 ( x 40) = 0 ( x 40) ( x 5) = 0 If x 40 = 0, then x = 40 If x 5 = 0, then x = 5 Length of the field = 40 m and breadth of the field = 5 m Q.34. If the length of a rectangle is increased by 0 cm and the breadth decreased by 5 cm, the area remains unchanged. If the length is decreased by 5 cm and the breadth is increased by 4 cm, even then the area remains unchanged. Find the dimensions of the rectangle. Ans. Let length of rectangle = x cm and breadth of rectangle = y cm and area of rectangle = l b = xy cm Length of rectangle = ( x + 0) cm and breadth of rectangle = ( y 5) cm ( x + 0) ( y 5) = xy xy 5x + 0y 50 = xy 5x + 0y 50 = 0 x y + 0 = 0 x y = 0...(i) Length of rectangle = x 5 and breadth of rectangle = y + 4 ( x 5) ( y + 4) = xy xy + 4x 5y 0 = xy 4x 5y = 0...(ii) Multiply (i) by 5 and (ii) by, we get 5x 0 y = 50...(iii) 8x 0y = 40...(iv) Subtracting (iii) from (iv), we get 5x 0 y = 50 8x 0 y = 40 3x = 90 x = 30 Putting the value of x in (i), we get Math Class IX 8 Question Bank

19 30 y = 0 y = 0 30 = y = = 0 Hence, length of the rectangle = 30 cm and breadth of rectangle = 0 cm Q.35. A room is 3 m long and 9 m wide. Find the cost of carpeting the room with a carpet 5 cm wide and Rs.50 per meter. Ans. Length of room ( l ) = 3 m and width of room ( b ) = 9 m Area of room = l b = 3 m 9 m = m 5 3 Width of carpet = 5 cm = = m Length of carpet = Area Width = = m = 56 m 4 3 Rate of carpet = Rs.50 per m Total cost = Rs. 56 Rs..50 = Rs Q.36. Find the area and perimeter of a square plot of land, and length of whose diagonal is 5 metres. Give your answer correct to places of decimals. Ans. Let ABCD be the square plot of land. Let diagonal AC = 5 m or side = 5 m 5 5 Area of square = (side) = = m =.5 m 5 Perimeter of square = 4 side = 4 = 30 = 4.43 m Q.3. The area of square plot is 05 m. Find the length of its side and the length of its diagonal. Ans. Let ABCD be the square plot. AC be its diagonal Area of square plot = 05 m Side or length of square plot = Area Diagonal AC = side = 05 m = (5 5) (9 9) m = 5 9 m = 45 m Math Class IX 9 Question Bank

20 = 45 m = 45 m = m = m = m Q.38. The cost of cultivating a square field at the rate of 60 per hectare is Rs 440. Find the cost of putting fence around it at the rate of 5 paise per metre. Ans. Total cost = Rs 440, Rate of cultivation = Rs 60 per hectare 440 Area of the field = = 9 hectare = m 60 Side of square field = = 300 m Perimeter of the square field = 4 side = = 00 m Rate of fencing = 5 paise per metre 00 5 Hence, total cost = Rs = Rs Q.39. The shaded region of the given diagram represents the lawn in the form of a house. On the three sides of the lawn there are flowerbeds having a uniform width of m. (i) Fin d the length and the breadth of the lawn. (ii) Hence, or other wise, find the area of the flowerbeds. Ans. Let BCDE is the lawn (i) Length of lawn BCDE = BC = AD AB CD = 30 m m m = 30 m 4 m = 6 m Breadth of lawn BCDE = BE = AG GH = m m = 0 m (ii) Area of flower-beds = ar. (rectangle ADFG) ar. (lawn BCDE) = 30 m 6 0 m = (360 60) m = 00 m Q.40. A floor which measure 5 m 8 m is to be laid with tiles measuring 50 cm 5 cm. Find the number of tiles required. Further, if a carpet is laid on the floor so that a space of m exist between its edges and the edges of the floor, what fraction of the floor is uncovered. Ans. Let rectangle ABCD be the floor and rectangle PQRS represents the carpet. Math Class IX 0 Question Bank

21 Length of floor = 5 m Breadth of floor = 8 m Area of floor = 5 8 m = 0 m 50 Length of tile = 50 cm = m = m 00 5 breadth of tile = 5 cm = cm = m 00 4 Area of tile = m = m 4 8 Area of floor 0 8 Number of tile required = = = 0 = 960 Area of tile 8 Length of carpet = (5 ) m = (5 ) m = 3 m Breadth of carpet = (8 ) m = (8 ) m = 6 m Area of carpet = l b = 3 6 m = 8 m Area of floor uncovered = Total area of floor Area of carpet = ar. (rectangle ABCD) ar. (rectangle PQRS) = (0 8) m = 4 m Area of floor uncovered 4 Fraction of floor uncovered = = = Total area of floor 0 0 Q.4. Find the area of a parallelogram, if its two adjacent sides are cm and 4 cm and if the diagonal connecting their ends is 8 cm. Ans. Diagonals bisect the parallelogram into two triangles of equal area. Area of parallelogram ABCD = area of ABC Sides of ABC are cm, 4 cm and 8 cm a + b + c s = = = = Area of ABC = s ( s a) ( s b) ( s c) = ( ) ( 4) ( 8) = = 5 = 0 = 0 = 8 0 cm Math Class IX Question Bank

22 Hence, area of parallelogram ABCD = ar. ( ABC) = 8 0 = 6 0 cm = = 6.8 cm = 6.8 cm Q.4. Two adjacent sides of a parallelogram are 0 cm and cm. If one diagonal of it is 6 cm long; find area of the parallelogram. Also, find distance between its shorter sides. Ans. Let ABCD be the parallelogram with sides AD = 0 cm and side AB = cm Diagonal BD = 6 cm Let DM be the distance between the shorter sides AD and BC Area of gm ABCD = ar. ( ABD) Area of ABD, Let a = 6 cm, b = 0 cm and c = cm a + b + c s = = cm 38 = cm = 9 cm Using Hero s Formula Area of ABD = s ( s a) ( s b) ( s c) = 9 (9 6) (9 0) (9 ) cm = 9 (3) (9) () cm = 3 9 cm Area of gm ABCD = ar. ( ABD) = 3 9 (3) () cm = cm = = 59.9 cm = 59.9 cm = 9.8 cm = 9.8 cm (ii) Distance between the shorter sides = DM Area of gm ABCD = 9.8 cm base height = 9.8 BC DM = 9.8 AD DM = DM = DM = =.98 =.98 0 Distance between shorter sides AD and BC =.98 cm Math Class IX Question Bank

23 Q.43. The area of a rhombus is 6 square cm. If its one diagonal is 4 cm; find : (i) length of its other diagonal. (ii) length of its side. (iii) perimeter of the rhombus. Ans. Let ABCD be the rhombus then AB = BC = CD = DA Diagonal BD = 4 cm Diagonals AC and BD bisect each other at O at right angles. 4 DO = OB = cm = cm (i) Area of rhombus = 6 sq. cm one diagonal second diagonal 6 sq. cm = BD AC 6 = 4 AC 6 = 6 AC = cm = 8 cm Hence, other diagonal = 8 cm 8 (ii) AO = OC = cm = 9 cm From triangle AOD in which AOD = 90 Using Pythagoras Theorem, AD = AO + OD AD = (9) + () AD = = 5 Taking square root of both sides, we get AD 8 44 = + AD AD = 5 AD = 5 5 = 5 Side of rhombus = 5 cm (iii) Perimeter of rhombus ABCD = 4 side = 4 5 cm = 60 cm Math Class IX 3 Question Bank

24 Q.44. Find the area of a rhombus, one side of which measures 0 cm and one of whose diagonals is 4 cm. Ans. In rhombus ABCD, each side = 0 cm and One diagonal AC = 4 cm The diagonals of a rhombus bisect each other at right angles. AC and BD bisect each other at O at right angle. Hence, AOB is a right angled triangle in which AB = 0 cm AO = AC = 4 = cm AB = AO + OB [Using Pythagoras Theorem] (0) = () + OB 400 = 44 + BO OB = = 56 = (6) OB = 6 cm Diagonal BD = OB = 6 = 3 cm Product of diagonals 4 3 Area of rhombus = = = 384 cm Q.45. The cross-section of a canal is a trapezium in shape. If the canal is 0 m wide at the top, 6 m wide at the bottom and the area of the cross-section is sq. m; determine its depth. Ans. Let ABCD be the cross-section of canal in the shape of trapezium. AB = 6 m, DC = 0 m Let AL be the depth of canal Area of cross-section = sq. m (sum of parallel sides) depth = (AB DC) AL + = (6 0) AL + = (6) AL = 8 AL = Math Class IX 4 Question Bank

25 AL = m = 9 m 8 Hence, depth of canal = 9 m Q.46. Calculate the area of figure given below: which is not drawn to scale. Ans. From figure, AB = 5 cm, BC = 6 cm, EC = 5 cm, DF = cm Draw BG EC EG = AB = 5 cm GC = EC EG = 5 cm 5 cm = 0 cm In BGC, BGC = 90 Using Pythagoras Theorem BG = BC GC = (6) (0) BG = BG = 56 Taking square root of both sides, we get BG 56 = BG BG = (8 8) (3 3) BG = 8 3 cm = 4 cm AE = BG = 4 cm Area of given figure = Area of trapezium ABCE + Area of ECD = (sum of parallel sides) (height) + base height = (AB + EC) AE + EC DF = (5 + 5) 4 cm + 5 cm = 40 4 cm + 5 cm = 0 4 cm cm = 480 cm + 50 cm = 630 cm Math Class IX 5 Question Bank

26 Q.4. The following diagram shows a pentagonal field ABCDE in which the lengths of AF, FG, GH and HD are 50 m, 40 m, 5 m and 5 m respectively; and the lengths of perpendiculars BF, CH and EG are 50 m, 5 m and 60 m respectively. Determine the area of the field. Ans. ABCDE is a pentagonal field in which AF = 50 m FG = 40 m, GH = 5 m HC = 5 m, BF = 50 m, HD = 5 m GD = 40 m, CH = 5 m EG = 60 m Area of pentagonal field ABCDE = ar. ( ABF) + ar. (trapezium BCHF) + ar. ( CDH) + ar. ( ADE) = base height + (sum of parallel sides) height + base height + base height = AF BF + (BF + CH) FH + HD CH + AD EG = (50 + 5) (FG + GH) (AF + FG + GH + HD) 60 = 5 50 m + 5 (40 + 5) m + 65 m + ( ) 60 m = 50 m m m + (30) 60 m = 50 m + 45 m m m = 50 m m m m = 55.0 m = 55 m Math Class IX 6 Question Bank

27 Q.48. The following diagram shows a trapezium ABCD in which AB DC, D = 90, CD = 4 cm, AC = 6 cm and AB = 8 cm. Find the area of the trapezium. Ans. In trapezium ABCD, D = 90, AB CD, AB = 8 cm, CD = 4 cm and AC = 6 cm In right ADC, AC = CD + AD (Using Pythagoras Theorem) (6) = (4) + AD 66 = 56 + AD AD = = 00 = (0) AD = 0 cm Area of trapezium = (AB + CD) AD (8 4) 0 cm = cm = = Q.49. The two parallel sides of a trapezium are 58 m and 4 m long. The other two sides are equal, each being m. Find its area. Ans. In trapezium ABCD, AB DC and AB = 58 m, CD = 4 m, BC = AD = m From C, draw CE DA and CL AB meeting AB at E such that AE = CD = 4 m and EB = AB AE EB = 58 4 = 6 m CE = DA = m ECB is an isosceles triangle and CL EB CL bisects EB at L EL= EB = 6 = 8 m In CEL, CE = CL + EL (Using Pythagoras Theorem) () = CL + (8) 89 = CL + 64 Math Class IX Question Bank

28 CL = = 5 = (5) CL = 5 m Area of trapezium ABCD = (AB + CD) CL (58 4) 5 m = m 50 m = = Q.50. Find the circumference of the circle whose area is 6 times the area of the circle with diameter.4 m..4 Ans. Radius of the small circle = = 0. m Area of circle = π r = =.54 m and area of bigger circle =.54 6 = 4.64 m Let radius of bigger circle = r m π r = r = 4.64 r = r =. =.84 r =.8 m Circumference of circle = π r =.8 =.6 m Q.5. Calculate the circumference of a circle whose area is equal to the sum of areas of the circles with diameters 4 cm, 3 cm and 96 cm. 4 Ans. Radius of first circle = = cm Area of first circle = π r = cm 3 Similarly, area of second circle whose radius = = 6 cm is = π cm = Math Class IX 8 Question Bank

29 Area of third circle whose radius is cm = = π = cm Area of bigger circle = cm ( ) cm = + + ( ) cm = cm = Let radius of the bigger circle = R cm Area of bigger circle 04 = π R = 04 R = R = 04 R = 04 = 5 cm Circumference of bigger circle = π R = 5 cm 88 = = cm = cm Q.5. The ratio between the areas of two circles is 9 : 5. Find the ratio between : (i) their radii, (ii) their circumference Ans. Ratio between areas of two circles = 9 : 5 Let radius of first circle = r and radius of second circle = r Area of first circle = π r and area of second circle = π r r r r 9 (3) = = r 5 (5) π : π = 9 : 5 π π r : r = 3: 5 Math Class IX 9 Question Bank

30 (ii) Circumference of first circle = π r and circumference of second circle = π r Ratio between their circumferences = π r : π r = r : r = 3: 5 Q.53. The ratio between the circumferences of two circles is 4 : 9, find the ratio between their areas. Ans. Let radius of first circle = r and radius of second circle = r Ratio between their circumferences = π r : π r = 4 : 9 π r 4 r = 4 π r 9 r = 9 r : r = 4 : 9 Area of first circle = π r and Area of second circle Ratio between areas of circles π r : π r r r (9) = π r π r (4) 6 = = = = π r 8 Hence, ratio of areas of circles = 6 : 8 Q.54. The circumference of a circular garden is 5 m. Outside the garden a road, 3.5 m wide runs around it. Calculate the cost of repairing the road at the rate of Rs 35 per 00 sq. m. Ans. Circumference of the garden = 5 m = Circumference of circle 5 Radius ( r) = = 9 m Width of the road outside the garden = 3.5 m Outer radius ( R ) = = 94.5 m and area of the road = π ( R r ) = [(94.5) (9.0) ] m = ( ) ( ) m m = = Cost of repairing the road at the rate of Rs 35 per 00 sq. m = = = Rs (approx.s) 00 Math Class IX 30 Question Bank

31 Q.55. A bucket is raised from a well by means of a rope which is wound round a wheel of diameter cm. Given that the bucket ascents in minute 8 seconds with a uniform speed of.. m/s, calculate the number of complete revolutions the wheel makes in raising the bucket. Ans. Diameter of wheel ( d ) = cm Radius of wheel = cm Circumference of wheel = π r = = 4 cm Time taken by the bucket = minutes 8 seconds = = 88 secnds Speed of rope =. m/s Length of rope = 88. m = 96.8 m Number of revolutions = = = Q.56. The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Given your answer, correct to the nearest km. Ans. Diameter of wheel ( d ) = 84 cm 84 Radius of wheel ( r ) = = 4 cm Circumference of wheel = π r = 4 = 64 cm Distance covered in sec = 5 64 cm = 30 cm Distance covered in one hour = cm = km = 4.5 km/hr Hence, Speed of wheel = 4.5 km/hr. Q.5. A circular running track is the portion bounded by two concentric circles. If the cost of fixing the fence along the outer circumference of the track is Rs 3,696 at Rs 4 per meter and the cost of levelling the track is Rs 6,006 at Rs per sq. m; calculate the width of the track. Ans. Cost of fencing the outer track at the rate of Rs 4 per meter = Rs 3696 Circumference of outer track 3696 = = 54 m 4 Math Class IX 3 Question Bank

32 circumfernec of outer track and radius of outer track = π = = m Cost of levelling the track at the rate of Rs per sq. m = Rs Area of the track = = sq. m Let the radius of inner circle = r Area of the track = π ( R r ) 5005 π ( R r ) = ( R r ) = R r = = r = r = 59.5 r = = 44 = () r = radius of inner circle = m 49 Width of the track = R r = = = 3.5 m Q.58. The given figure shows a rectangle ABCD and a circle in it. Given AB = 0.5 cm and BC = cm. (i) If the area of shaded portion is sq. cm; calculate the radius of the circle. (ii) If the circumference of the circle is 5.4 cm; find the area of the shaded portion. Ans. (i) Length AB of the rectangle = 0.5 cm and breadth of rectangle = cm Area of the rectangle ABCD = 0.5 cm = 3.5 cm Area of shaded portion = cm Area of circle = ( ) cm = 4.64 sq. cm Let radius of the circle = r cm Area of circle = π r 4.64 r = Math Class IX 3 Question Bank

33 r = 4.64 =.84 Hence, r =.84 =.8 cm (ii) Circumference of the circle = π r = 5.4 cm C 5.4 Radius of circle = = =.45 cm π Area of circle = π r = = cm But area of rectangle = 3.5 cm Area of remaining part = ( ) cm = cm Q.59. The sum of the radii of two circles is 40 cm and the difference of their circumferences is 88 cm. Find the ratio between their area. Ans. Difference of circumference of two circles = 88 cm Sum of radii of two circles = 40 cm Let radius of the first circle = r cm Then radius of the second circle = (40 r) cm Circumference of first circle = π r Circumference of second circle = (40 r) π According to the given problem, we get π r (40 r) π = 88 π ( r 40 + r) = 88 π (r 40) = 88 (r 40) = r 40 = = 4 r = = r = = cm Radius of first circle = cm and radius of second circle = (40 ) cm = 63 cm Area of first circle Area of second circle = π r = π () cm = π (63) cm Math Class IX 33 Question Bank

34 π () Ratio of area of circles = = = = π (63) Hence, ratio of areas of circles = : 8. Q.60. Find the area of a circle whose circumference is equal to the sum of the circumferences with diameters 36 cm and 0 cm. Ans. Diameter of the first circle = 36 cm 36 Radius of first circle ( r ) = = 8 cm and circumference of first circle = π r = dπ = 8 = 36 Diameter of the second circle = 0 m Circumference of second circle = π r = dπ = 0 cm Sum of the circumferences of two circles = = 56 = 6 cm Circumference of the third circle = 6 cm Circumference of circle 6 Radius of third circle = = = 8 cm π Area of third circle = π r = 8 8 cm = cm = 464 cm Q.6. A toothed wheel of radius 40 cm is attached to a smaller toothed wheel of diameter 4 cm. How many revolutions will the smaller wheel make when the larger one makes 50 revolutions. Ans. Radius of bigger wheel = 40 cm. 60 Circumference of circle = π r = 40 = cm Diameter of smaller wheel = 4 cm Radius of wheel = cm 58 Circumference of wheel = π r = cm = cm Distance covered by larger wheel in 50 revolutions = cm Math Class IX 34 Question Bank

35 Number of revolutions made by the smaller wheel to cover cm Distance covered = = = 500 revolutions. Circumference of smaller wheel 58 Q.6. The side of a square is 0 cm. Find the areas of the circumscribed and inscribed circles. Ans. Side of the square ( a ) = 0 cm Radius of inscribed circle ( r ) = 0 = 0 cm Area of inscribed circle = π r 0 0 cm = 00 = cm = 34 cm Radius of the circumcircle ( R) = diagonal of the square = ( a) = a = 0 = 0 cm Thus, area of circumcircle = π r 0 0 cm = = = 68 cm Q.63. Find the area of the region bounded by two concentric circles, if the length of the chord of the outer circle, touching the inner circle, is 8 cm. Ans. Let the outer radius = R and inner radius = r and chord AB = 8 cm OC AB, therefore C bisects AB 8 BC = = 4 cm Now in OCB, OB OC = CB (Using Pythagoras Theorem) R r = (4) = 96 Area of the region bounded by these two circles will be = π ( R r ) Math Class IX 35 Question Bank

36 96 cm 8 cm = = = 66 cm Q.64. A sheet is 30 cm long and 0 cm wide. Circular pieces, all of equal diameters are cut from the sheet to prepare discs. Calculate the number of discs prepared if the diameter of each disc is : (i) cm (ii). cm (iii).5 cm (iv).5 cm. Ans. Length of sheet ( l ) = 30 cm and width of sheet ( b ) = 0 cm Area of the sheet = l b = 30 0 = 300 cm (i) Diameter of disc = cm Number of discs to be cut from length = 30 = 30 and number of discs to be cut from width = 0 = 0 Hence, total number of discs = 30 0 = 300 (ii) Diameter of disc =. cm Number of discs to be cut from length = 30. = 5 and number of discs to be cut from width = 0. = 8 Thus total number of discs = 5 8 = 00 (iii) Diameter of disc =.5 cm Number of discs to be cut from length = 30.5 = 0 and number of discs to be cut from width = 0.5 = 6 Total number of discs = 0 6 = 0 (iv) Diameter of disc =.5 cm Number of discs to be cut from length = 30.5 = and number of discs to be cut from width = 0.5 = 4 Total number of discs = 4 = 48 Math Class IX 36 Question Bank

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