Grade 9 Herons Formula

ID : ae-9-herons-formula [1] Grade 9 Herons Formula For more such worksheets visit www.edugain.com Answer the questions (1) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. If the lengths of the perpendiculars are a, b and c, find the altitude of the triangle. () Find the area of a quadrilateral whose sides are 6 cm, 8 cm, 9 cm and 17 cm and the angle between first two sides is a right angle. (3) Find the area of the figure below. Also find the altitude to the base of the triangle. (All measurements are in centimeters) (4) If in the figure below AB = 15 cm, BC=0 cm and CA = 7 cm, find the area of the rectangle BDCE. (5) The area of a trapezium is 98 cm and the height is 7 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other. (6) The area of an equilateral triangle with a side of 8 cm is : Copyright 017 www.edugain.com

ID : ae-9-herons-formula [] (7) The perimeter of an isosceles triangle is 18 cm and its unequal side is 63 cm. Find the area of the triangle. (8) Find the area of the unshaded region in the figure below: (9) An umbrella is made by stitching 1 triangular pieces of cloth each piece measuring 30 cm, 17 cm and 17 cm. How much cloth is required for this umbrella. (10) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 4 cm, 5 cm and 3 cm. Find the area of the triangle. (11) A parallelogram has a diagonal of 13 cm. The perpendicular distance of this diagonal from an opposite vertex is 8 cm. Find the area of the parallelogram. (1) The sides of a triangular field are 10 m, 13 m and 13 m. Find the number of rose beds that can be prepared in the field, if each rose bed, on an average needs 600 cm space. (13) The perimeter of a triangular field is 64 meters and the ratio of the sides is 15:13:4. Find the area of the field. (14) Find the percentage increase in the area of a triangle if each side is increased by N times. Choose correct answer(s) from the given choices (15) In Heron's formula,, S is equal to: a. a + b + c b. Half of perimeter of the triangle c. a + b + c abc d. a x b x c 017 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com Copyright 017 www.edugain.com

Answers ID : ae-9-herons-formula [3] (1) a+b+c The following figure shows the required triangle: Step Let's assume the side of the equilateral triangle ΔABC is x. The area of the triangle ΔABC can be calculated using Heron's formula, since all sides of the triangles are known. S = (AB + BC + CA)/ = (x + x + x)/ = 3x/ cm. The area of the ΔABC = 4 x ------(1) The area of the triangle AOB = AB OP = x b Copyright 017 www.edugain.com

= bx ID : ae-9-herons-formula [4] Step 4 Similarly, the area of the triangle ΔBOC = ax and the area of the triangle ΔAOC = cx Step 5 The area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC) = bx + ax + cx = (a + b + c)x -----() Step 6 By comparing the equations (1) and (), we get: x (a + b + c)x = 4 x = Step 7 (a + b + c) ------(3) Now, Area(ΔABC) = Step 8 4 (x) Area(ΔABC) = 'AB' 'Altitude of the triangle ΔABC' 4 (x) = 'x' 'Altitude of the triangle ΔABC' (x) = 'Altitude of the triangle ΔABC' By putting the value of 'x' from the equation (3), we get, Altitude of the triangle ΔABC = (a + b + c) ( ) Altitude of the triangle ΔABC = a+b+c Step 9 Hence, the altitude of the triangle is a+b+c. Copyright 017 www.edugain.com

ID : ae-9-herons-formula [5] () 60 cm Let's ABCD is the quadrilateral with AB = 6 cm, BC = 8 cm, CD = 9 cm, DA = 17 cm, and angle ABC = 90, as shown in the following figure. Step Let's draw the diagonal AC in the quadrilateral ABCD, The area of the right triangle ABC = (1/) AB BC = 1/ (6) (8) = 4 cm AC = (AB + BC ) = (6 + 8 ) = 10 cm Step 4 The area of the triangle ACD can be calculated using Heron's formula. S = (CD + DA + AC)/ = (9 + 17 + 10)/ = 18 cm The area of the triangle ACD = [ S (S-CD) (S-DA) (S-AC) ] = [ 18 (18-9) (18-17) (18-10) ] = 36 cm Step 5 The area of the quadrilateral ABCD = Area(ABC) + Area(ACD) = 4 + 36 = 60 cm Copyright 017 www.edugain.com

(3) Area : 4 cm Altitude : 1 cm ID : ae-9-herons-formula [6] The area of the triangle can be calculated using Heron's formula, since all sides of the triangle are known. S = (4 + 13 + 15)/ = 16 cm. The area of the triangle = [ 16(16-4) (16-13) (16-15) ] = 4 cm Step Let's assume, 'h' is the altitude to the base of the triangle, (The area of the triangle) The altitude to the base of the triangle = The base of the triangle = 4 4 = 1 cm Copyright 017 www.edugain.com

(4) Area : 19 cm ID : ae-9-herons-formula [7] The area of the triangle ABC can be calculated using Heron's formula, since all sides of the triangle are known. S = (7 + 15 + 0)/ = 1 cm. The area of the ΔABC = [ S(S - AB) (S - BC) (S - CA) ] = [ 1(1-15) (1-0) (1-7) ] = 4 cm Step The height(bd) of the ΔABC = Area(ΔABC) AC = 4 7 = 1 cm In right angled ΔBDC, DC = BC - BD DC = [ BC - BD ] = [ (0) - (1) ] = 16 cm The area of the rectangle BDCE = BD DC = 1 16 = 19 cm Copyright 017 www.edugain.com

(5) Lengths : 1 cm and 16 cm. ID : ae-9-herons-formula [8] The following picture shows the trapezium ABCD, Let's assume, CD = x cm and CD AB. According to the question, one side of the trapezium is 4 cm greater than the other. Therefore, AB = x + 4, Area(ABCD) = 98 cm, Height of the trapezium ABCD = 7 cm. Step The area of the trapezium ABCD = The height of the trapezium ABCD AB + CD AB + CD = Area(ABCD) The height of the trapezium ABCD x + 4 + x = 98 7 x = 8-4 x = 4 x = 1 cm. Now, CD = 1 cm, AB = 1 + 4 = 16 cm. Thus, the lengths of its two parallel sides are 1 cm and 16 cm respectively. Copyright 017 www.edugain.com

(6) cm ID : ae-9-herons-formula [9] As per Heron's formula, the area of a triangle with sides a, b and c, and perimeter S = Step Here, a=b=c=8 and S = 3 x a = 1 Therefore Area = Area = Area = cm (7) 5 cm Here we have an isosceles triangle with perimeter 18 cm and the unequal side of 63 cm. The combined size of the other two sides = 18 cm - 63 cm = 65 cm. Since the triangle is isosceles, the other two sides are equal, and the length of each of them will be: 65 = 3.5 cm. Step Now we know that the three sides of the triangle are 3.5, 3.5 and 63 cm respectively. We can now use the Heron's formula to calculate the area of the triangle: Area of a triangle =, where a, b, and c are its sides and S = Let us first calculate the value of S: 63 + 3.5 + 3.5 S = = 18 = 64 Step 4 a+b+c. Area of the triangle = = 5 cm Copyright 017 www.edugain.com

ID : ae-9-herons-formula [10] (8) 66 m If we look at the figure carefully, 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 formula, since all sides of the triangle are known. S = (AB + BC + CA)/ = (39 + 17 + 44)/ = 50 m. The area of the ΔABC = [ S (S - AB) (S - BC) (S - CA) ] = [ 50(50-39) (50-17) (50-44) ] = 330 m Similarly, the area of the triangle ΔACD can be calculated using Heron's formula. S = (AC + CD + DA)/ = (44 + 15 + 37)/ = 48 m. The area of the ΔACD = [ S (S - AC) (S - CD) (S - DA) ] = [ 48(48-44) (48-15) (48-37) ] = 64 m Step 4 Thus, the area of the unshaded region = Area(ABC) - Area(ACD) = 330-64 = 66 m Copyright 017 www.edugain.com

ID : ae-9-herons-formula [11] (9) 1440 cm Following picture shows the triangular piece of cloth, Since all sides of the triangle are known, the area of the triangle can be calculated using Heron's formula. S = (AB + BC + CA)/ = (30 + 17 + 17)/ = 3 cm The area of the ΔABC = [ S (S - AB) (S - BC) (S - AC) ] = [ 3(3-30) (3-17) (3-17) ] = 10 cm Step According to the question, the umbrella is made by stitching 1 triangular pieces of cloth. The cloth required for this umbrella = 1 Area(ABC) = 1 10 = 1440 cm (10) 48 cm Following figure 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 formula, since all sides of the triangles are known. S = (AB + BC + CA)/ Copyright 017 www.edugain.com

= (x + x + x)/ = 3x/ cm. The area of the ΔABC = [S(S - AB)(S - BC)(S - CA) ] = [ 3x (3x/ - x)(3x/ - x)(3x/ - x) ] ID : ae-9-herons-formula [1] = [ 3x = [ 3x (x/)(x/)(x/) ] (x/) 3 ] = [ 3(x/) 4 ] = [ (x/) ] = (x) ------(1) 4 Step The area of the triangle AOB = AB OP = 'x' 5 = 5x Similarly, the area of the triangle ΔBOC = 4x and the area of the triangle ΔAOC = 3x. Step 4 The the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC) = 5x + 4x + 3x = 1x -----() Step 5 By comparing equation (1) and (), we get, (x) = 1x 4 x = 4 Step 6 Copyright 017 www.edugain.com

Now, Area(ΔABC) = 4 (x) ID : ae-9-herons-formula [13] = 4 ( 4 ) = 144 = 48 cm Step 7 Hence, the area of the triangle is 48 cm. (11) 104 cm Consider a parallelogram ABCD as shown in the figure below P is the point where the perpendicular from 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 = 8 cm Area of ACD= 1 x 8 x 13 = 5 Area of parallelogram ABCD = x area of ACD = x 5 = 104 cm Copyright 017 www.edugain.com

(1) 1000 ID : ae-9-herons-formula [14] Following figure shows the triangular field ABC, The area of the triangular field ABC can be calculated using Heron's formula, since all sides of the triangular field are known. S = (AB + BC + CA)/ = (10 + 13 + 13)/ = 18 m. Area(Δ ABC) = [ S (S - AB) (S - BC) (S - CA) ] = [ 18(18-10) (18-13) (18-13) ] = 60 m. = 60 10000 = 600000 cm Step According to the question, the number of rose bed that can be prepared in 600 cm space = 1 rose bed. The number of rose beds that can be prepared in 600000 cm space = 600000 = 1000 = 1000 600 rose beds. Copyright 017 www.edugain.com

(13) 96 ID : ae-9-herons-formula [15] Since we know the perimeter, we can use Heron's formula to help us compute the area The formula 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=15x, b=13x and c=4x (we know this because the ratio of the sides is given as 15:13:4) We also know that a+b+c = 64. = 64 (15 + 13 + 4)x = 64 3x = 64 x = 64 = 3 Step 4 From this we see that a = 30 m, b = 6 m and c=8 m. Also S=3 Step 5 Putting these values into Heron's formula, Area = Area = 96 m Copyright 017 www.edugain.com

(14) ID : ae-9-herons-formula [16] 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 N times, we get a new triangle XYZ XYZ has sides Na, Nb and Nc By Heron's formula Area of new triangle = Na + Nb + Nc Where S 1 = = N x a+b+c = MS Area of XYZ = = = N x A 1 This means the area increases by (15) b. Half of perimeter of the triangle In Heron's formula, S represents the semi-perimeter of the triangle. Step Semi-perimeter is defined as the half of perimeter of the triangle Copyright 017 www.edugain.com