Tiêu đề 04-Herons Formula(1).pdf ✅

HERON’S FORMULA 4 CHAPTER ➢ CONTENTS • Some basic terms • Area of triangle • Heron’s formula • Area of quadrilateral • Area of Rhombus • Area of Trapezium ➢ SOME BASIC TERMS  Area : Place, which is covered by base of a body is called area of that body. Area is same from everywhere of the base.  Height : The perpendicular distance is called height and the side having foot of perpendicular, is called base. ❖ EXAMPLES ❖ Ex.1 ABCD is a parallelogram. If DE & BF are perpendiculars from D and B on sides AB & DA respectively then their bases are AB and AD respectively. D C F A E B Ex.2 ABC is an acute angle triangle. AD is height and BC is base B D C A Ex.3 PQR is an obtuse angle triangle at Q. Then height of P from BC is PT but base is QR (not SR). T Q R P S  Area of triangle = 2 1 (base × height) square unit we can use this formula when we can find or given height & base.  Heron’s formula : If we have all sides of triangle and their is no way to find height then we use this formula for area of triangle. Area of  = s(s − a)(s − b)(s − c) Where s is semi perimeter of . 2 1 s = (sum of all sides) = 2 1 (a + b + c) B a C A c b and a, b, c are sides of . Note : Use 2 1 (base) (height) for area of right angle triangle, if any two sides are given.
❖ EXAMPLES ❖ Ex.4 For given figure find the s (s – a). A 5 cm 3 cm B 4 cm C Sol. Perimeter = 2s = 3 + 4 + 5 = 12 cm  semi perimeter = 2 12 = 6 cm  s (s – a) = 6 (6 – 4) = 6 × 2 = 12 cm. Ex.5 If semiperimeter of a triangle is 60 cm & its two sides are 45 cm, 40 cm then find third side. Sol.  Semiperimeter = 60  Perimeter = 2 × 60  Sum of all three sides = 120 (Let third side = x cm)  x + 45 + 40 = 120  x + 85 = 120  x = 120 – 85  x = 35 cm. Ex.6 If perimeter of an equilateral triangle is 96 cm, then find its each side. Sol.  Length of all sides are equal in equilateral  Let length is x cm  x + x + x = 96  3x = 96  x = 32 cm. Ex.7 If one side from two equal sides of a  is 14 cm and semiperimeter is 22.5 cm then find the third side. Sol. Let the third side is x cm  x + 14 + 14 = 2 × 22.5  x = 45 – 28 = 17 cm. Ex.8 Find the length of AD in given figure, if EC = 4 cm and AB = 5 cm. B D C A E 6 cm Sol.  area of ABC = 2 1 (AB × EC) = 2 1 (5 × 4) = 10 square cm. also area of ABC = 2 1 (BC × AD) = 2 1 (6 × AD) = 3AD square cm  3AD = 10  AD = 3.33 3 10 = cm. Note : 2 = 1.41, 3 = 1.73, 5 = 2.23, 6 = 2.45, 7 = 2.64, 8 = 2.82, 11 = 3.31, 15 = 3.87 Ex.9 Find the area of a triangle whose sides are of lengths 52 cm, 56 cm and 60 cm respectively. Sol. Let a = 52 cm, b = 56 cm and c = 60 cm. Perimeter of the triangle = (a + b + c) units = (52 + 56 + 60) cm = 168 cm  168 cm 2 1 (a b c) 2 1 s       = + + =  = 84 cm (s – a) = (84 – 52) cm = 32 cm, (s – b) = (84 – 56) cm = 28 cm and (s – c) = (84 – 60) cm = 24 cm By Heron’s formula, the area of the given triangle is  = s(s − a)(s − b)(s − c) = 2 84 32  28 24cm = 2 14616214264 cm = (14 × 6 × 4 × 2 × 2) cm2 = 1344 cm2 . Ex.10 Using Heron’s formula, find the area of an equilateral triangle of side a units.
Sol. We have : 2 1 s = (a + a + a) = 2 3a  (s – a) = 2 a a 2 3a  =      − , (s – b) = 2 a a 2 3a  =      − and (s – c) = 2 a a 2 3a  =      − So, by Heron’s formula, we have : area = s(s − a)(s − b)(s − c)squnits = sq units 4 3a sq units 2 a 2 a 2 a 2 3a 2            = Hence, area of equilateral triangle of side a is         4 3a 2 sq units. Note :         =            =  2 3a a 2 1 a 4 3 2 =        base height 2 1  height = units 2 3a Ex.11 Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm. Sol. Here, a = 13 cm, b = 13 cm and c = 24 cm.  s = 2 1 (a + b + c) = 2 1 (13 + 13 + 24) cm = 25 cm. (s – a) = (25 – 13) cm = 12 cm, (s – b) = (25 – 13) cm = 12 cm and (s – c) = (25 – 24) cm = 1 cm. So, by Heron’s formula,  = s(s − a)(s − b)(s − c) = 2512 12 1 cm2 = (5×12)cm2 = 60 cm2 . Hence, the area of the given triangle is 60 cm2 . Ex.12 The perimeter of a triangular field is 450 m and its sides are in the ratio 13 : 12 : 5. Find the area of the triangle. Sol. a : b : c = 13 : 12 : 5 a = 13x, b = 12x & c = 5x  Perimeter = 450  13x + 12x + 5x = 450  30x = 450  x = 15. So, the sides of the triangle are a = 13 × 15 = 195 m, b = 12 × 15 = 180 m and c = 5 × 15 = 75 m It is given that perimeter = 450  2s = 450  s = 225  Area = s(s − a)(s − b)(s − c) = 225(225 −195)(225 −180)(225 − 75)  Area = 225 30  45150 = 5 3 3 5 2 3 5 5 2 3 2 2 2 2          Area = 6 6 2 5 3  2 = 53 × 3 3 × 2 = 6750 m2 . Ex.13 Find the percentage increase in the area of a triangle if its each side is doubled. Sol. Let a, b, c be the sides of the old triangle and s be its semi-perimeter. Then, s = 2 1 (a + b + c) The sides of the new triangle are 2a, 2b and 2c. Let s' be its semi-perimeter. Then, 2 1 s' = × (2a + 2b + 2c) = a + b + c = 2s Let  and ' be the areas of the old and new triangles respectively. Then,  = s(s − a)(s − b)(s − c) and ' = s'(s'−2a)(s'−2b)(s'−2c)  ' = 2s(2s − 2a)(2s − 2b)(2s − 2c) [ s' = 2s]  ' = 4 s(s − a)(s − b)(s − c) = 4  Increase in the area of the triangle = ' –  = 4 −  = 3 Hence, percentage increase in area = 100 300% 3  =         Ex.14 The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find (i) the area of the triangle and (ii) the height corresponding to the longest side.
Sol. Perimeter = 144 cm and ratio of sides = 3 : 4 : 5 Sum of ratio terms = (3 + 4 + 5) = 12. Let the lengths of the sides be a, b and c respectively. Then, a =        12 3 144 cm = 36 cm, b =        12 4 144 cm = 48 cm and c =        12 5 144 cm = 60 cm.  2 1 s = (a + b + c) = 2 1 (36 + 48 + 60) cm = 72 cm. (s – a) = (72 – 36) cm = 36 cm, (s – b) = (72 – 48) cm = 24 cm and (s – c) = (72 – 60) cm = 12 cm. (i) By Heron’s formula, the area of the triangle is given by  = s(s − a)(s − b)(s − c) = 2 72362412 cm = 2 3636 24 24 cm = (36 × 24) cm2 = 864 cm2 . Hence, the area of the given triangle is 864 cm2 . (ii) Let base = longest side = 60 cm and the corresponding height = h cm. Then, area =        base height 2 1 sq units =        60  h 2 1 cm2 = (30h) cm2 .  30h = 864  h = 28.8 30 864  =      . Hence, the height corresponding to the longest side is 28.8 cm. Ex.15 Find the area of the shaded region in figure : 12cm 16 cm A B C D 48 cm 52 cm Sol. By Pythagoras theorem, in ADB AB = 2 2 AD + BD = 2 2 12 +16 = 144 + 256 = 400 AB = 20 cm.  area of ABC = s(s − a)(s − b)(s − c)       = + + = 60cm 2 20 48 52 s = 60(60 − 20)(60 − 48)(60 − 52) = 60  40 12 8 = (12 5)(58)12 8 = 2 2 2 5 8 12 = 5 × 8 × 12 = 480 cm2 . also area of ADB = 2 1 (AD) (BD) = 2 1 × 12 × 16 = 96 cm2  Shaded area = ar(ABC) – ar (ADB) = 480 – 96 = 384 cm2 . Ex.16 Find the area of an isosceles triangle of its sides are a cm, a cm and b cm. Sol. Semi perimeter = 2 2a b 2 a a b + = + + cm.   = s(s − a)(s − b)(s − c)  =       − +       − +       − +       + b 2 2a b a 2 2a b a 2 2a b 2 2a b  =       −                   + 2 2a b 2 b 2 b 2 2a b  = (2a b)(2a b) 2 2 b + −  2 2 4a b 4 b  = − square cm. Ex.17 A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board ? [NCERT]