CHAPTER 12 HERON’S FORMULA 12.1 Introduction You have studied in earlier classes about figures of different shapes such as squares, rectangles, triangles and quadrilaterals. You have also calculated perimeters and the areas of some of these figures like rectangle, square etc. For instance, you can find the area and the perimeter of the floor of your classroom. Let us take a walk around the floor along its sides once; the distance we walk is its perimeter. The size of the floor of the room is its area. So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m, i.e., 80 m 2 . Unit of measurement for length or breadth is taken as metre (m) or centimetre (cm) etc. Unit of measurement for area of any plane figure is taken as square metre (m 2 ) or square centimetre (cm 2 ) etc. Suppose that you are sitting in a triangular garden. How would you find its area? From Chapter 9 and from your earlier classes, you know that: Area of a triangle = 1 2 × base × height (I) We see that when the triangle is right angled, we can directly apply the formula by using two sides containing the right angle as base and height. For example, suppose that the sides of a right triangle ABC are 5 cm, 12 cm and 13 cm; we take base as 12 cm and height as 5 cm (see Fig. 12.1). Then the Fig. 12.1
Transcript
CHAPTER 12
HERON’S FORMULA
12.1 IntroductionYou have studied in earlier classes about figures of different shapes such as squares,rectangles, triangles and quadrilaterals. You have also calculated perimeters and theareas of some of these figures like rectangle, square etc. For instance, you can findthe area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is itsperimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeterwould be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m, i.e., 80 m2.
Unit of measurement for length or breadth is taken as metre (m) or centimetre(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m2) orsquare centimetre (cm2) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?From Chapter 9 and from your earlier classes, you know that:
Area of a triangle =12
× base × height (I)
We see that when the triangle is right angled,we can directly apply the formula by using two sidescontaining the right angle as base and height. Forexample, suppose that the sides of a right triangleABC are 5 cm, 12 cm and 13 cm; we take base as12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
198 MATHEMATICS
area of ∆ ABC is given by
12
× base × height = 12
× 12 × 5 cm2, i.e., 30 cm2
Note that we could also take 5 cm as the base and 12 cm as height.Now suppose we want to find the area of an equilateral triangle PQR with side
10cm (see Fig. 12.2). To find its area we need its height. Can you find the height ofthis triangle?
Let us recall how we find its height when weknow its sides. This is possible in an equilateraltriangle. Take the mid-point of QR as M and join it toP. We know that PMQ is a right triangle. Therefore,by using Pythagoras Theorem, we can find the lengthPM as shown below:
PQ2 = PM2 + QM2
i.e., (10)2 = PM2 + (5)2, since QM = MR.Therefore, we have PM2 = 75
i.e., PM = 75 cm = 5 3 cm.
Then area of ∆ PQR = 12
× base × height = 21 10 5 3 cm 25 32× × = cm2.
Let us see now whether we can calculate the area of an isosceles triangle alsowith the help of this formula. For example, we take a triangle XYZ with two equalsides XY and XZ as 5 cm each and unequal side YZ as 8 cm (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from X we drawa perpendicular XP to side YZ. You can see that this perpendicular XP divides thebase YZ of the triangle in two equal parts.
Therefore, YP = PZ = 12
YZ = 4 cm
Then, by using Pythagoras theorem, we getXP2 = XY2 – YP2
= 52 – 42 = 25 – 16 = 9So, XP = 3 cm
Now, area of ∆ XYZ = 12
× base YZ × height XP
= 12
× 8 × 3 cm2 = 12 cm2.
Fig. 12.2
Fig. 12.3
HERON’S FORMULA 199
Now suppose that we know the lengths of the sides of a scalene triangle and notthe height. Can you still find its area? For instance, you have a triangular park whosesides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you wantto apply the formula, you will have to calculate its height. But we do not have a clue tocalculate the height. Try doing so. If you are not able to get it, then go to the nextsection.
12.2 Area of a Triangle — by Heron’s Formula
Heron was born in about 10AD possibly in Alexandria inEgypt. He worked in applied mathematics. His works onmathematical and physical subjects are so numerous andvaried that he is considered to be an encyclopedic writerin these fields. His geometrical works deal largely withproblems on mensuration written in three books. Book Ideals with the area of squares, rectangles, triangles,trapezoids (trapezia), various other specialisedquadrilaterals, the regular polygons, circles, surfaces ofcylinders, cones, spheres etc. In this book, Heron hasderived the famous formula for the area of a triangle interms of its three sides.
The formula given by Heron about the area of a triangle, is also known as Hero’sformula. It is stated as:
Area of a triangle = ( ) ( ) ( )s s a s b s c− − − (II)
where a, b and c are the sides of the triangle, and s = semi-perimeter i.e. half the
perimeter of the triangle = 2
a b c+ + ,
This formula is helpful where it is not possible to find the height of the triangleeasily. Let us apply it to calculate the area of the triangular park ABC, mentionedabove (see Fig. 12.5).Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s = 40 24 32
2+ +
m = 48 m.
Fig. 12.4Heron (10AD – 75 AD)
200 MATHEMATICS
s – a = (48 – 40) m = 8 m,
s – b = (48 – 24) m = 24 m,s – c = (48 – 32) m = 16 m.
Therefore, area of the park ABC
= ( ) ( ) ( )− − −s s a s b s c
= 2 248 8 24 16 m 384m× × × =
We see that 322 + 242 = 1024 + 576 = 1600 = 402. Therefore, the sides of the parkmake a right triangle. The largest side i.e. BC which is 40 m will be the hypotenuseand the angle between the sides AB and AC will be 90°.
By using Formula I, we can check that the area of the park is 12
× 32 × 24 m2
= 384 m2.We find that the area we have got is the same as we found by using Heron’s
formula.Now using Heron’s formula, you verify this fact by finding the areas of other
triangles discussed earlier viz;(i) equilateral triangle with side 10 cm.(ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm.You will see that
For (i), we have s = 10 10 10
2+ +
cm = 15 cm.
Area of triangle = 15(15 10) (15 10) (15 10)− − − cm2
= 2 215 5 5 5 cm 25 3 cm× × × =
For (ii), we have s = 8 5 5 cm 9 cm
2+ +
= .
Area of triangle = 9(9 8) (9 5) (9 5)− − − cm2 = 2 29 1 4 4 cm 12 cm .× × × =
Let us now solve some more examples:
Fig. 12.5
HERON’S FORMULA 201
Example 1 : Find the area of a triangle, two sides of which are 8 cm and 11 cm andthe perimeter is 32 cm (see Fig. 12.6).Solution : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.
Third side c = 32 cm – (8 + 11) cm = 13 cmSo, 2s = 32 i.e. s = 16 cm,
s – a = (16 – 8) cm = 8 cm,s – b = (16 – 11) cm = 5 cm,s – c = (16 – 13) cm = 3 cm.
Therefore, area of the triangle = ( ) ( ) ( )s s a s b s c− − −
= 2 216 8 5 3 cm 8 30 cm× × × =
Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 12.7). Agardener Dhania has to put a fence all around it and also plant grass inside. Howmuch area does she need to plant? Find the cost of fencing it with barbed wire at therate of Rs 20 per metre leaving a space 3m wide for a gate on one side.Solution : For finding area of the park, we have
2s = 50 m + 80 m + 120 m = 250 m.i.e., s = 125 mNow, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m,s – c = (125 – 50) m = 75 m.
Therefore, area of the park = ( ) ( ) ( )s s a s b s c− − −
= 125 5 45 75× × × m2
= 2375 15 m
Also, perimeter of the park = AB + BC + CA = 250 mTherefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)
= 247 mAnd so the cost of fencing = Rs 20 × 247 = Rs 4940
Fig. 12.6
Fig. 12.7
202 MATHEMATICS
Example 3 : The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeteris 300 m. Find its area.Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 12.8).Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)Therefore, 15x = 300, which gives x = 20.So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 mi.e., 60 m, 100 m and 140 m.Can you now find the area [Using Heron’s formula]?
We have s = 60 100 140
2+ +
m = 150 m,
and area will be 150(150 60) (150 100) (150 140)− − − m2
= 150 90 50 10× × × m2
= 21500 3 m
EXERCISE 12.11. 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 is180 cm, what will be the area of the signal board?
2. The triangular side walls of a flyover have been used for advertisements. The sides ofthe walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield anearning of Rs 5000 per m2 per year. A company hired one of its walls for 3 months. Howmuch rent did it pay?
Fig. 12.9
Fig. 12.8
HERON’S FORMULA 203
3. There is a slide in a park. One of its side walls has been painted in some colour with amessage “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of thewall are 15 m, 11 m and 6 m, find the area painted in colour.
Fig. 12.10
4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is42cm.
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find
the area of the triangle.
12.3 Application of Heron’s Formula in Finding Areas of QuadrilateralsSuppose that a farmer has a land to be cultivated and she employs some labourers forthis purpose on the terms of wages calculated by area cultivated per square metre.How will she do this? Many a time, the fields are in the shape of quadrilaterals. Weneed to divide the quadrilateral in triangular parts and then use the formula for area ofthe triangle. Let us look at this problem:
Example 4 : Kamla has a triangular field with sides 240 m, 200 m, 360 m, where shegrew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to theprevious field, she wanted to grow potatoes and onions (see Fig. 12.11). She dividedthe field in two parts by joining the mid-point of the longest side to the opposite vertexand grew patatoes in one part and onions in the other part. How much area (in hectares)has been used for wheat, potatoes and onions? (1 hectare = 10000 m2)Solution : Let ABC be the field where wheat is grown. Also let ACD be the fieldwhich has been divided in two parts by joining C to the mid-point E of AD. For thearea of triangle ABC, we have a = 200 m, b = 240 m, c = 360 m
Therefore, s = 200 240 360
2+ +
m = 400 m.
204 MATHEMATICS
So, area for growing wheat
= 400(400 200) (400 – 240) (400 – 360)− m2
= 2400 200 160 40 m× × ×
= 216000 2 m = 1.6 × 2 hectares
= 2.26 hectares (nearly)Let us now calculate the area of triangle ACD.
Here, we have s = 240 320 400
2+ +
m = 480 m.
So, area of ∆ ACD = 480(480 240) (480 320) (480 400)− − − m2
We notice that the line segment joining the mid-point E of AD to C divides thetriangle ACD in two parts equal in area. Can you give the reason for this? In fact, theyhave the bases AE and ED equal and, of course, they have the same height.
Therefore, area for growing potatoes = area for growing onions= (3.84 ÷ 2) hectares = 1.92 hectares.
Example 5 : Students of a school staged a rally for cleanliness campaign. Theywalked through the lanes in two groups. One group walked through the lanes AB, BCand CA; while the other through AC, CD and DA (see Fig. 12.12). Then they cleanedthe area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 mand ∠ B = 90º, which group cleaned more area and by how much? Find the total areacleaned by the students (neglecting the width of the lanes).Solution : Since AB = 9 m and BC = 40 m, ∠ B = 90°, we have:
AC = 2 29 40+ m
= 81 1600+ m
= 1681 m = 41m
Therefore, the first group has to clean the area of triangle ABC, which is right angled.
Area of ∆ ABC = 12
× base × height
= 12 × 40 × 9 m2 = 180 m2
Fig. 12.11
Fig. 12.12
HERON’S FORMULA 205
The second group has to clean the area of triangle ACD, which is scalene having sides41 m, 15 m and 28 m.
Here, s =41 15 28
2+ +
m = 42 m
Therefore, area of ∆ ACD = ( – ) ( – ) ( – )s s a s b s c
= 42(42 – 41) (42 – 15)(42 – 28) m2
= 42 1 27 14× × × m2 = 126 m2
So first group cleaned 180 m2 which is (180 – 126) m2, i.e., 54 m2 more than the areacleaned by the second group.Total area cleaned by all the students = (180 + 126) m2 = 306 m2.
Example 6 : Sanya has a piece of land which is in the shape of a rhombus(see Fig. 12.13). She wants her one daughter and one son to work on the land andproduce different crops. She divided the land in two equal parts. If the perimeter ofthe land is 400 m and one of the diagonals is 160 m, how much area each of them willget for their crops?Solution : Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of ∆ ABD is given by
s = 100 100 160
2+ +
m = 180 m
Therefore, area of ∆ ABD = 180(180 100) (180 – 100) (180 – 160)−
= 180 80 80 20× × × m2 = 4800 m2
Therefore, each of them will get an area of 4800 m2.
Fig. 12.13
206 MATHEMATICS
Alternative method : Draw CE ⊥ BD (see Fig. 12.14).As BD = 160 m, we have
DE = 160 m ÷ 2 = 80 mAnd, DE2 + CE2 = DC2, which gives
CE = 2 2DC DE−
or, CE = 2 2100 80 m 60 m− =
Therefore, area of ∆ BCD = 1 160 602× × m2 = 4800 m2
EXERCISE 12.21. A park, in the shape of a quadrilateral ABCD, has ∠ C = 90º, AB = 9 m, BC = 12 m,
CD = 5 m and AD = 8 m. How much area does it occupy?2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm,
DA = 5 cm and AC = 5 cm.3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find
the total area of the paper used.
Fig. 12.154. A triangle and a parallelogram have the same base and the same area. If the sides of
the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base28 cm, find the height of the parallelogram.
Fig. 12.14
HERON’S FORMULA 207
5. A rhombus shaped field has green grass for 18 cows to graze. If each side of therhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will eachcow be getting?
6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours(see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of eachcolour is required for the umbrella?
7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base8 cm and sides 6 cm each is to be made of three different shades as shown inFig. 12.17. How much paper of each shade has been used in it?
Fig. 12.16 Fig. 12.17
8. A floral design on a floor is made up of 16 tileswhich are triangular, the sides of the trianglebeing 9 cm, 28 cm and 35 cm (see Fig. 12.18).Find the cost of polishing the tiles at the rateof 50p per cm2.
9. A field is in the shape of a trapezium whose parallelsides are 25 m and 10 m. The non-parallel sidesare 14 m and 13 m. Find the area of the field.
12.4 SummaryIn this chapter, you have studied the following points :1. Area of a triangle with its sides as a, b and c is calculated by using Heron’s formula,
stated asArea of triangle = ( ) ( ) ( )− − −s s a s b s c
where s =2
+ +a b c
2. Area of a quadrilateral whose sides and one diagonal are given, can be calculated bydividing the quadrilateral into two triangles and using the Heron’s formula.
Fig. 12.18
111222 HERON’S FORMULA
EXERCISE 12.1Q.1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral
triangle with side ‘a’. Find the area of the signal board, using Heron’sformula. If its perimeter is 180 cm, what will be the area of the signalboard?
Sol. Each side of the triangle = aPerimeter of the triangle = 3a
a∴ s =
32
∴ Area of the signal board (triangle) = s s a s b s c( ) ( ) ( )− − −
= s s a s a s a( ) ( ) ( )− − − [ a = b = c]
= (s – a) s s a( )− = 32
32
32
aa
a aa−⎛
⎝⎜⎞⎠⎟
−⎛⎝⎜
⎞⎠⎟
= a a a a a2
34 2 2
34
32 2
⋅ = ⋅ =
Hence, area of the signal board = a2
43 sq units Ans.
Now, perimeter = 180 cm
Each side of the triangle = 1803
cm = 60 cm
Area of the triangle = ( )604
2× 3 cm2 = 900 3 cm2 Ans.
Q.2. The triangular side walls of a flyover have been used for advertisements.The sides of the walls are 122 m, 22 m and 120 m (see Fig.). Theadvertisements yield an earning of Rs 5000 per m2 per year. A companyhired one of its walls for 3 months. How much rent did it pay?
Sol. Here, we first find the area of the triangular side walls.a = 122 m, b = 120 m and c = 22 m
∴ s = 122 120 22
2+ +
m = 132 m.
www.tiwariacademy.com Focus on free education
Area of the triangular side wall = s s a s b s c( ) ( ) ( )− − −
= 132 132 122 132 120 132 22−( ) −( ) −( ) m2
= 132 10 12 110× × × m2 = 1320 m2
Rent of 1 m2 of the wall for 1 year = Rs 5000
∴ Rent of 1 m2 of the wall for 1 month = Rs 500012
∴ Rent of the complete wall (1320 m2) for 3 months
= Rs 500012
× 1320 × 3 = Rs 16,50,000 Ans.
Q.3. There is a slide in a park. One of its side walls has been painted in somecolour with a message “KEEP THE PARK GREEN AND CLEAN” (seeFig.). If the sides of the wall are 15 m, 11 m and 6 m, find the area paintedin colour.
Sol. Here a = 15 m, b = 11 m, c = 6 m
∴ s = a b c+ + = + +
215 11 6
2 m = 16 m
Area of the triangle = s s a s b s c( ) ( ) ( )− − −
= 16 16 15 16 11 16 6( ) ( ) ( )− − − m2
= 16 1 5 10× × × m2 = 20 2 m2
Hence, the area painted in colour = 20 2 m2 Ans.Q.4. Find the area of a triangle two sides of which are 18 cm and 10 cm and
the perimeter is 42 cm.Sol. Here a = 18 cm, b = 10 cm, c = ?
Perimeter of the triangle = 42 cm⇒ a + b + c = 42⇒ 18 + 10 + c = 42⇒ c = 42 – 28 = 14
Now, s = a b c+ + =
2422 cm = 21 cm
Area of the triangle = s s a s b s c( ) ( ) ( )− − −
= 21 21 18 21 10 21 14( ) ( ) ( )− − − cm2
= 21 3 11 7× × × cm2 = 7 3 3 11 7× × × × cm2
= 7 × 3 11 cm2 = 21 11 2cm Ans.
P a g e | 3
Q.5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is540 cm. Find its area.
Sol. Let the sides of the triangle be 12x cm 17x cm and 25x cm.Perimeter of the triangle = 540 cm∴ 12x + 17x + 25x = 540
⇒ 54 x = 540
⇒ x = 54054 = 10
∴ Sides of the triangle are (12 × 10) cm, (17 × 10) cm and (25 × 10) cm i.e.,120 cm, 170 cm and 250 cm.Now, suppose a = 120 cm, b = 170 cm, c = 250 cm,
∴ s = a b c+ + =2
5402
cm = 270 cm
Area of the triangle = s s a s b s c( ) ( ) ( )− − −
For the rectangle marked II :Length = 6.5 cm, Breadth = 1 cmArea of the rectangle = 6.5 × 1 cm2 = 6.5 cm2
For the trapezium marked III :Draw AF || DC and AE ⊥ BC.AD = FC = 1 cm, DC = AF = 1 cm∴ BF = BC – FC = (2 – 1) cm = 1 cmHence, ∆ABF is equilateral.Also, E is the mid-point of BF.
Area of the trapezium = 12 (sum of the parallel sides) × distance between them.
= 12 × (BC + AD) × AE =
12 × (2 + 1) × 0.9 cm2 = 1.4 cm2.
For the triangle marked IV :It is a right-triangle
∴ Area of the triangle = 12 × base × height
= 12 × 6 × 1.5 cm cm2 = 4.5 cm2.
For the triangle marked V :This triangle is congruent to the triangle marked IV.
Hence, area of the triangle = 4.5 cm2
Total area of the paper used = (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm2
= 19.4 cm2 Ans.Q.4. A triangle and a parallelogram have the same base and the same area. If
the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogramstands on the base 28 cm, find the height of the parallelogram.
Sol. In the figure, ABCD is a parallelogram and ABE is thetriangle which stands on the base ABFor the triangle ABE, a = 30 cm, b = 28 cm, c = 26 cm.
∴ s = a b c+ + = + +2
30 28 262
cm = 42 cm
∴ Area of the ∆ABE = s s a s b s c( ) ( ) ( )− − −
= 42 42 30 42 28 42 26 2( ) ( ) ( )− − − cm
= 42 12 14 16× × × cm2 = 112896 cm2
= 336 cm2
Now, area of the parallelogram = base × height⇒ 336 = 28 × height [Given, area of the triangle
= area of the parallelogram]
⇒ Height of the parallelogram = 33628 cm = 12 cm Ans.
Q.5. A rhombus shaped field has green grass for 18 cows to graze. If each sideof the rhombus is 30 m and its longer diagonal is 48 m, how much areaof grass field will each cow be getting?
Sol. Clearly, the diagonal AC of the rhombus divides it into two congruenttriangles.For triangle ABC, a = b = 30 m, c = 48 m.
∴ s = a b c+ + = + +2
30 30 482
m = 54 m
∴ Area of the triangle
= s s a s b s c( ) ( ) ( )− − −
= 54 54 30 54 30 54 48( ) ( ) ( )− − − m2
= 54 24 24 6× × × m2 = 432 m2
∴ Area of the rhombus = 2 × 432 m2 = 864 m2
Number of cows = 18Hence, area of the grass field which each cow gets
= 86418 m2 = 48 m2 Ans.
Q.6. An umbrella is made by stitching 10 triangularpieces of cloth of two different colours (see Fig.),each piece measuring 20 cm, 50 cm, and 50 cm.How much cloth of each colour is required for theumbrella?
Sol. First we find the area of one triangular piece.Here, a = b = 50 cm, c = 20 cm
∴ s = a b c+ + = + +
250 50 20
2 cm = 60 m
∴ Area of one triangular piece = s s a s b s c( ) ( ) ( )− − −
= 60 60 50 60 50 60 20( ) ( ) ( )− − − cm2
= 60 10 10 40× × × cm2 = 200 6 cm2
∴ Area of 10 such triangular pieces = 10 × 200 6 cm2
= 2000 6 cm2
Hence, cloth required for each colour = 2000 6
2 cm2 = 1000 6 cm2 Ans.
P a g e | 5
Q.7. A kite in the shape of a square with a diagonal 32 cmand an isosceles triangle of base 8 cm and sides 6 cmeach is to be made of three different shades as shownin figure. How much paper of each shade has beenused in it?
Sol. ABCD is a square.So, AO = OC = OB = ODand ∠AOB = 90° [Diagonals of a square bisect
each other at right angles]
BD = 32 cm (Given) ⇒ OA = 322 cm = 16 cm.
∆ABD is a right triangle.
So, area of ∆ABD = 12
× base × height
= 12
× 32 × 16 cm2 = 256 cm2
Thus, area of ∆BCD = 256 cm2
For triangle CEF, a = b = 6 cm, c = 8 cm.
∴ s = a b c+ + = + +
26 6 8
2 cm = 10 cm
∴ Area of the triangle = s s a s b s c( ) ( ) ( )− − −
= 10 10 6 10 6 10 8( ) ( ) ( )− − − cm2
= 10 4 4 2× × × cm2 = 320 cm2 = 17.92 cm2
Hence, paper needed for shade I = 256 cm2, for shade II= 256 cm2 and for shade III = 17.92 cm2 Ans.
Q.8. A floral design on a floor is made up of 16tiles which are triangular, the sides of thetriangle being 9 cm, 28 cm and 35 cm (seefigure). Find the cost of polishing the tilesat the rate of 50 p per cm2.
Sol. We have lengths of the sides of1 triangular tile are a = 35 cm, b = 28 cm,c = 9 cm.
∴ s = a b c+ +
2 =
35 28 92
+ + cm = 36 cm
∴ Area of 1 triangular tile = s s a s b s c( ) ( ) ( )− − −
= 36 36 35 36 28 36 9( ) ( ) ( )− − − cm2
= 36 1 8 27× × × cm2 = 7776 cm2 = 88.2 cm2
∴ Area of 16 such tiles = 16 × 88.2 cm2
8 cm
P a g e | 6
Cost of polishing 1 cm2 = 50 p = Re 0.50∴ Total cost of polishing the floral design = Rs 16 × 88.2 × 0.50
= Rs 705.60 Ans.
Q.9. A field is in the shape of a trapezium whose parallel sides are 25 m and10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Sol. In the figure ABCD is the field. Draw CF || DA and CG ⊥ AB.
DC = AF = 10 m, AD = FC = 13 mFor ∆BCF, a = 15 m, b = 14 m, c = 13 m
