Class 9 Herons Formula

Transcription

1 ID : in-9-herons-formula [1] Class 9 Herons Formula For more such worksheets visit Answer t he quest ions (1) Find the area of the parallelogram ABCD and the length of the altitude DE in the f igure below: () The sides of a triangle are 11 cm, 13 cm and 0 cm. The altitude to the longest side is : (3) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 6 cm, 8 cm and 4 cm. Find the area of the triangle. (4) A rhombus has all its internal angles equal. If one of the diagonals is 7 cm, f ind the length of other diagonal and the area of the rhombus. (5) Find the percentage increase in the area of a triangle if each side is increased by Y times. (6) Find the area of the unshaded region in the f igure below: (C) 016 Edugain (

2 ID : in-9-herons-formula [] (7) Ankur 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 30 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 to give the shape of a kite. The length of the sides of triangle is a, such that a = 8 3. Find the area of this kite. Choose correct answer(s) f rom given choice (8) If in the f igure below AB = 16cm, BC=6cm, CA = 13cm and BE = 7cm, f ind the area of the trapezium BDCE. a. 8.94cm b cm c cm d cm (9) The adjacent sides of a parallelogram are 8 cm and 11 cm. The ratio of their altitudes is : a. 8:19 b. 64:11 c. 8:11 d. 8:5.5 (10) The perimeter of a triangular f ield is 40 m and the ratio of the sides is 0:15:7. Which of the f ollowing is the area of the f ield in sq m : a. 940 b c. 800 d. 400 (11) Find the area of a trapezium, if its height is 4 cm and the lengths of parallel sides are 13 cm and 17 cm. a. 60 cm b. 884 cm c. 78 cm d. 10 cm (1) Find the area of a quadrilateral whose sides are 7 cm, 4 cm, 17 cm and 6 cm and the angle between f irst two sides is a right angle. a. 04cm b. 88cm c. 850cm d. 84cm (C) 016 Edugain (

3 ID : in-9-herons-formula [3] (13) If in the f igure below AB = 13cm, BC=15cm and CA = 4cm, f ind the area of the rectangle BDCE. a. 108 cm b. 4 cm c cm d. 48 cm (14) In Heron's f ormula,, S is equal to: a. a + b + c abc c. Half of perimeter of the triangle b. a + b + c d. a x b x c (15) The perimeter of a triangular f ield is 36 and the ratio of the sides is 8:5:5. The area of the f ield in sq m is : a. 18 b. 48 c d Edugain ( All Rights Reserved Many more such worksheets can be generated at (C) 016 Edugain (

4 Answers ID : in-9-herons-formula [4] (1) Area : 10 cm Altitude : 0 cm 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)/ = ( )/ = 30 cm. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] = [ 30(30-6) (30-5) (30-9) ] = 60 cm The area of the parallelogram ABCD = Area(ΔABC) = 60 = 10 cm Step 4 The length of the altitude DE = Area(ΔABC) AB = 60 6 = 0 cm. (C) 016 Edugain (

5 () 6.6 cm ID : in-9-herons-formula [5] Let's assume the altitude to the longest side be 'h'. Following picture shows the required triangle, 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)/ = ( )/ = cm. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] = [ ( - 0) ( - 13) ( - 11) ] = 66 cm Step The altitude to the longest side = (The area of the ΔABC) Base 'AB' = 66 0 = 6.6 cm. (C) 016 Edugain (

6 (3) cm ID : in-9-herons-formula [6] 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' 8 = 8x (C) 016 Edugain (

7 Similarly, the area of the triangle ΔBOC = 6x ID : in-9-herons-formula [7] and the area of the triangle ΔAOC = 4x. Step 4 The the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC) = 8x + 6x + 4x = 18x -----() Step 5 By comparing equation (1) and (), we get, 3 (x) = 18x 4 x = 36 3 Step 6 Now, Area(ΔABC) = 3 4 (x) = 3 4 ( 36 3 ) = 34 3 = cm Step 7 Hence, the area of the triangle is cm. (4) 7 cm, 4.5 cm The key thing to note is that all the internal angles of a rhombus add up to 360 So if all the internal angles are equal, then one internal angle is = 90 This is a square with all sides equal. So the other diagonal is also 7 cm in length The area of a rhombus/square is half the product of the diagonals Area = (7 x 7) = 49 = 4.5 cm (C) 016 Edugain (

8 (5) ID : in-9-herons-formula [8] 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 Y times, we get a new triangle XYZ XYZ has sides Ya, Yb and Yc By Heron's f ormula Area of new triangle = Ya + Yb + Yc Where S 1 = = Y x a+b+c = MS Area of XYZ = = = Y x A 1 This means the area increases by (C) 016 Edugain (

9 (6) 90 m ID : in-9-herons-formula [9] 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 (

10 (7) 456 cm ID : in-9-herons-formula [10] 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 (8 3) 4 = 6 cm The diagonal of the square = 30 cm. The area of the square ABCD = ( 1 ) (30) = 450 cm. Step 4 Thus, the area of the kite = Area(ABCD) + Area(BEF) = = 456 cm. (8) d cm (9) c. 8:11 The area of a parallelogram is base x height, where base ref ers to the side, and height is indicated by the altitude Area = b 1 x h 1 = b x h Here, let's take b 1 = 8, and b = 11 8 x h 1 = 11 x h Taking the ration we get h :h 1 = 8:11 (C) 016 Edugain (

11 (10) d. 400 ID : in-9-herons-formula [11] Since we know the perimeter, we can use Heron's f ormula to help us compute the area The f ormula states that the area of a triangle with sides a, b and c, and perimeter S = Step Let us assume the 3 sides are of length a=0x, b=15x and c=7x (we know this because the ratio of the sides is given as 0:15:7) We also know that a+b+c = 40. 0x + 15x + 7x = 40 ( )x = 40 4x = 40 x = 40 = 10 4 Step 4 From this we see that a = 00, b = 150 and c=70. Also S=10 Step 5 Putting these values into Heron's f ormula, Area = = Solving, we f ind the area = 400 (C) 016 Edugain (

12 (11) a. 60 cm ID : in-9-herons-formula [1] T he f ollowing picture shows the trapezium ABCD, Step According to the question, the height of the trapezium ABCD = 4 cm The area of the trapezium ABCD = The height of the trapezium ABCD AB + CD = = 4 30 = 60 cm Thus, the area of the trapezium is 60 cm. (C) 016 Edugain (

13 (1) b. 88cm ID : in-9-herons-formula [13] Let's ABCD is the quadrilateral with AB = 7 cm, BC = 4 cm, CD = 17 cm, DA = 6 cm, and angle ABC = 90, as shown in the f ollowing f igure. Step Let's draw the diagonal AC in the quadrilateral ABCD, The area of the right triangle ABC = (1/) AB BC = 1/ (7) (4) = 84 cm AC = (AB + BC ) = (7 + 4 ) = 5 cm Step 4 The area of the triangle ACD can be calculated using Heron's f ormula. S = (CD + DA + AC)/ = ( )/ = 34 cm The area of the triangle ACD = [ S (S-CD) (S-DA) (S-AC) ] = [ 34 (34-17) (34-6) (34-5) ] = 04 cm Step 5 The area of the quadrilateral ABCD = Area(ABC) + Area(ACD) = = 88 cm (C) 016 Edugain (

14 (13) a. 108 cm ID : in-9-herons-formula [14] The area of the triangle ABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = ( )/ = 16 cm. The area of the ΔABC = [ S(S - AB) (S - BC) (S - CA) ] = [ 16(16-13) (16-15) (16-4) ] = 4 cm Step The height(bd) of the ΔABC = Area(ΔABC) AC = 4 4 = 1 cm In right angled ΔBDC, DC = BC - BD DC = [ BC - BD ] = [ (15) - (1) ] = 9 cm The area of the rectangle BDCE = BD DC = 1 9 = 108 cm (C) 016 Edugain (

15 (14) c. Half of perimeter of the triangle ID : in-9-herons-formula [15] In Heron's f ormula, S represents the semi-perimeter of the triangle. Step Semi-perimeter is def ined as the half of perimeter of the triangle (15) b. 48 Since we know the perimeter, we can use Heron's f ormula to help us compute the area The f ormula states that the area of a triangle with sides a, b and c, and perimeter S = Step Let us assume the 3 sides are of length a=8x, b=5x and c=5x (we know this because the ratio of the sides is given as 8:5:5) We also know that a+b+c = 36. = 36 ( )x = 36 18x = 36 x = 36 = 18 Step 4 From this we see that a = 16 m, b = 10 m and c=10 m. Also S=18 Step 5 Putting these values into Heron's f ormula, Area = Area = 48 m (C) 016 Edugain (