Grade 9 Herons Formula

Transcription

1 ID : ww-9-herons-formula [1] Grade 9 Herons Formula For more such worksheets visit Answer t he quest ions (1) Find the area of the unshaded region in the f igure below: () The perimeter of a rhombus is 164 cm and one of its diagonals is 80 cm. What is the length of other diagonal. (3) The area of a triangle with sides 6 cm, 5 cm and 5 cm is : (4) The area of a trapezium is 75 cm and the height is 5 cm. Find the lengths of its two parallel sides if one side is 6 cm greater than the other. (5) The perimeter of a quadrilateral is 56 cm. If the f irst three sides of a quadrilateral, taken in order are 17 cm, 16 cm and 15 cm respectively, and the angle between f ourth side and the third side is a right angle, f ind the area of the quadrilateral. (6) Find the percentage increase in the area of a triangle if each side is increased by X times. (7) The perimeter of an isosceles triangle is 64 cm. The ratio of the equal side to its base is 5 : 6. Find the area of the triangle. (8) The sides of a quadrilateral, taken in order are 17 cm, 16 cm, 15 cm and 8 cm respectively. The angle contained by the last two sides is a right angle. Find the area of the quadrilateral. (C) 016 Edugain (

2 ID : ww-9-herons-formula [] Choose correct answer(s) f rom given choice (9) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 8 cm, 10 cm and 6 cm. Find the altitude of the triangle. a. 4 cm b. 8 cm c. 4 3 cm d. cm (10) Which triangle has the maximum area f or a given perimeter : a. Equilateral T riangle b. Obtuse Angle T riangle c. Can not be determined d. Isoceles T riangle (11) Find the area of a quadrilateral ABCD where AB = 10 cm, BC = 17 cm, CD = 13 cm, DB = 0 cm and AC = 1 cm. a. 16cm b. 10cm c. 84cm d. 100cm (1) A rhombus has all its internal angles equal. If one of the diagonals is 14 cm, f ind the length of other diagonal and the area of the rhombus. a. 15 cm, 105 cm b. 14 cm, 98 cm c. 98 cm, 14 cm d. 14 cm, 196 cm (13) Sharon makes the kite using two pieces of paper. 1st piece of paper is cut in the shape of square where one diagonal is of the length cm. At one of the vertex of this square a second piece of paper is attached which is of the shape of an equilateral triangle of length 4 cm to give the shape of a kite. Find the area of this kite. a. 490cm b. 45cm c. 4cm d cm (14) A parallelogram has a diagonal of 13 cm. T he perpendicular distance of this diagonal f rom an opposite vertex is 7 cm. Find the area of the parallelogram. a. 91cm b. 0cm c..75cm d. 45.5cm (C) 016 Edugain (

3 (15) The area of the parallelogram ABCD and the length of the altitude DE respectively in the f igure below are ID : ww-9-herons-formula [3] a. 86.4cm, 31.cm b. 7cm, 4cm c. 54cm, 7cm d. 36cm, 4cm 016 Edugain ( All Rights Reserved Many more such worksheets can be generated at (C) 016 Edugain (

4 Answers ID : ww-9-herons-formula [4] (1) 90 m If we look at the f igure caref ully, we notice that, the area of the unshaded region = The area of the triangle ΔABC - The area of the triangle ΔACD. Step The area of the triangle ΔABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = (AB + BC + CA)/ = ( )/ = 4 m. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] = [ 4(4-8) (4-17) (4-39) ] = 10 m Similarly, the area of the triangle ΔACD can be calculated using Heron's f ormula. S = (AC + CD + DA)/ = ( )/ = 40 m. The area of the ΔACD = [ S (S - AC) (S - CD) (S - DA) ] = [ 40(40-39) (40-16) (40-5) ] = 10 m Step 4 Thus, the area of the unshaded region = Area(ABC) - Area(ACD) = = 90 m (C) 016 Edugain (

5 () 18 cm ID : ww-9-herons-formula [5] One way to solve this is as f ollows We know that the a) The sides of a rhombus are equal. Theref ore one side = 1 4 x 164 = 41 b) A diagonal of a rhombus divides the rhombus into equal triangles c) The area of a rhombus is 1 4 (Diagonal1 x Diagonal) Step Taking one of the two triangles f ormed by the diagonal with length 80 cm Area (using Heron's f ormula) = Where S = x = 16 = 81 Area = = 70 (the details of this computation are lef t the the student) From c) above, Area = 70 = 1 4 (Diagonal1 x Diagonal) = 1 4 (80 x Diagonal) Diagonal = x = 18 cm (3) 1 cm The area of a triangle with sides a,b, c is given by Heron's f ormula as Area = Where S is half of the perimeter, i.e S = a + b + c Step Here S = = 16 = 8 Area = = = 1 cm (C) 016 Edugain (

6 (4) Lengths : 1 cm and 18 cm. ID : ww-9-herons-formula [6] T he f ollowing picture shows the trapezium ABCD, Let's assume, CD = x cm and CD AB. According to the question, one side of the trapezium is 6 cm greater than the other. Theref ore, AB = x + 6, Area(ABCD) = 75 cm, Height of the trapezium ABCD = 5 cm. Step The area of the trapezium ABCD = The height of the trapezium ABCD AB + CD AB + CD = Area(ABCD) T he height of the trapezium ABCD x x = 75 5 x = 30-6 x = 4 x = 1 cm. Now, CD = 1 cm, AB = = 18 cm. Thus, the lengths of its two parallel sides are 1 cm and 18 cm respectively. (C) 016 Edugain (

7 (5) Area : 180 cm ID : ww-9-herons-formula [7] Following picture shows the quadrilateral ABCD, The perimeter of the quadrilateral ABCD = 56 cm Theref ore, AB + BC + CD + DA = DA = DA = 56 DA = DA = 8 cm Step Let's draw the line AC. The ΔACD is the right angled triangle. Theref ore, AC = DA + DC AC = [ DA + DC ] = [ ] = 17 cm The area of the right angled triangle ΔACD = DA DC = 8 15 = 60 cm Step 4 Now, we can see that, this quadrilateral consists of the triangles ΔACD and ΔABC. (C) 016 Edugain (

8 ID : ww-9-herons-formula [8] The area of the ΔABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = (AB + BC + CA)/ = ( )/ = 5 cm. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] = [ 5(5-17) (5-16) (5-17) ] = 10 cm Step 5 The area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = = 180 cm (6) Consider a triangle QRS with sides a, b and c. Let S = a+b+c Area of triangle QRS = A 1 Step Increasing the side of each side by X times, we get a new triangle XYZ XYZ has sides Xa, Xb and Xc By Heron's f ormula Area of new triangle = Xa + Xb + Xc Where S 1 = = X x a+b+c = MS Area of XYZ = = = X x A 1 This means the area increases by (C) 016 Edugain (

9 (7) 19 cm ID : ww-9-herons-formula [9] Let's assume, the lengths of the base and the equal sides of the isosceles triangle are b cm and x cm respectively. Following f igure shows the isosceles triangle ABC, The ratio of the equal side to its base is 5 : 6. Theref ore, 5 6 = x b By cross multiplying, we get: b = 6x (1) 5 Step According to the question, the perimeter of the isosceles triangle ABC = 64 cm Theref ore, x + x + b = 64 x + 6x = 64 [From equation (1), b = 6x ] x + 6x = x + 6x = 30 16x = 30 x = 0 cm Putting the value of x in equation (1), we get: b = 10 = 4 cm 5 Step 4 The area of the isosceles triangle ABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = 64/ = 3 cm The area of the isosceles triangle ABC = [ S (S - AB) (S - BC) (S - CA) ] = [ 3(3-4) (3-0) (3-0) ] = 19 cm Step 5 Thus, the area of the triangle is 19 cm. (C) 016 Edugain (

10 ID : ww-9-herons-formula [10] (C) 016 Edugain (

11 (8) Area : 180 cm ID : ww-9-herons-formula [11] Following picture shows the quadrilateral ABCD, Step Let's draw the line AC. The ΔACD is the right angled triangle. Theref ore, AC = AD + DC AC = [ AD + DC ] = [ ] = 17 cm The area of the right angled triangle ΔACD = AD DC = 8 15 = 60 cm Step 4 Now, we can see that, this quadrilateral consists of the triangles ΔACD and ΔABC. The area of the ΔABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = (AB + BC + CA)/ = ( )/ = 5 cm. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] (C) 016 Edugain (

12 = [ 5(5-17) (5-16) (5-17) ] = 10 cm ID : ww-9-herons-formula [1] Step 5 The area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = = 180 cm (C) 016 Edugain (

13 (9) a. 4 cm ID : ww-9-herons-formula [13] Following f igure shows the required triangle, Let's assume the sides of the equilateral triangle ΔABC be x. The area of the triangle ΔABC can be calculated using Heron's f ormula, since all sides of the triangles are known. S = (AB + BC + CA)/ = (x + x + x)/ = 3x/ cm. The area of the ΔABC = [S(S - AB)(S - BC)(S - CA) ] = [ 3x (3x/ - x)(3x/ - x)(3x/ - x) ] = [ 3x = [ 3x (x/)(x/)(x/) ] (x/) 3 ] = [ 3(x/) 4 ] = 3[ (x/) ] = 3 (x) (1) 4 Step The area of the triangle AOB = AB OP = 'x' 10 = 10x (C) 016 Edugain (

14 Similarly, the area of the triangle ΔBOC = 8x ID : ww-9-herons-formula [14] and the area of the triangle ΔAOC = 6x. Step 4 The the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC) = 10x + 8x + 6x = 4x -----() Step 5 By comparing equation (1) and (), we get, 3 (x) = 4x 4 x = (3) Step 6 Let H be altitude of triangle ΔABC. Base Height Area(ΔABC) = 3 4 (x) = xh H = 3 H = 3 Step 7 x ( 48 3 ) = 4 cm Hence, the altitude of the triangle is 4 cm. (10) a. Equilateral T riangle (C) 016 Edugain (

15 (11) b. 10cm ID : ww-9-herons-formula [15] Following picture shows the quadrilateral ABCD, Step Let's draw the diagonal AC in the quadrilateral ABCD, Now, we can see that, this quadrilateral consists of two triangles, i.e. ΔABC and ΔACD, the area of each triangle can be calculated using Heron's f ormula, since all sides of the triangles are known. Step 4 The area of the ΔABC can be calculated using Heron's f ormula. S = (AB + BC + AC)/ = ( )/ = 4 cm. The area of the ΔABC = [ S (S - AB) (S - BC) (S - AC) ] = [ 4(4-10) (4-17) (4-1) ] = 84 cm Step 5 Similarly, the area of the ΔACD can be calculated using Heron's f ormula. S = (AC + CD + AD)/ = ( )/ = 7 cm. The area of the ΔACD = [ S (S - AC) (S - CD) (S - AD) ] = [ 7(7-10) (7-17) (7-1) ] (C) 016 Edugain (

16 = 16 cm ID : ww-9-herons-formula [16] Step 6 The area of the quadrilateral ABCD = The area of the ΔABC + The area of the ΔACD = = 10 cm (C) 016 Edugain (

17 (13) b. 45cm ID : ww-9-herons-formula [17] Following f igure shows the kite, made by two pieces of paper, Step Now, we can see that, this kite consists of a square ABCD and a equilateral triangle BEF. The area of the equilateral triangle ΔBEF can be calculated using Heron's f ormula f or equilateral triangle, Area = 3 a 4 = 3 (4 3) 4 = 3 cm The diagonal of the square = cm. The area of the square ABCD = ( 1 ) () = 4 cm. Step 4 Thus, the area of the kite = Area(ABCD) + Area(BEF) = = 45 cm. (C) 016 Edugain (

18 (14) a. 91cm ID : ww-9-herons-formula [18] Consider a parallelogram ABCD as shown in the f igure below P is the point where the perpendicular f rom point D meets diagonal AC Step From the diagram, we see that ACD is a triangle. The area of ACD is half the area of parallelogram ABCD The area of ACD is 1 x base x height Here base = length of diagonal = 13 cm Height = length of DP = 7 cm Area of ACD= 1 x 7 x 13 = 45.5 Area of parallelogram ABCD = x area of ACD = x 45.5 = 91 cm (C) 016 Edugain (

19 ID : ww-9-herons-formula [19] (15) b. 7cm, 4cm T he diagonal AC divides the parallelogram ABCD into two equal triangles, ΔABC and ΔACD. The area of the parallelogram ABCD = Area(ΔABC). Step The area of the ΔABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = (AB + BC + CA)/ = ( )/ = 7 cm. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] = [ 7(7-3) (7-5) (7-6) ] = 36 cm The area of the parallelogram ABCD = Area(ΔABC) = 36 = 7 cm Step 4 The length of the altitude DE = Area(ΔABC) AB = 36 3 = 4 cm. (C) 016 Edugain (