Parallelogram Base Height Area Perimeter (a) 5 units 3 units 5 × 3 = 15 sq units (b) (c) (d) (e) (f) (g) 212 MATHEMATICS consider the following parallelograms with sides 7 cm and 5 cm (Fig 11.13). Fig 11.13 Find the perimeter and area of each of these parallelograms.Analyse your results. You will find that these parallelograms have different areas but equal perimeters. This shows that to find the area of a parallelogram, you need to know only the base and the corresponding height of the parallelogram. TRY THESE Find the area of following parallelograms: (i) (ii) (iii) In aparallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm. 11.4 AREA OF A TRIANGLE A gardener wants to know the cost of covering the whole of a triangular garden with grass. In this case we need to know the area of the triangular region. Let us find a method to get the area of a triangle. PERIMETER AND AREA 213 Draw a scalene triangle on a piece of paper. Cut out the triangle. B Place this triangle on another piece of paper and cut out another triangle of the same size. So now you have two scalene triangles of the same size. E Are both the triangles congruent? C A Superpose one triangle on the other so that they match. You may have to rotate one of the two triangles. Now place both the triangles such that their corresponding sides are joined (as shown in Fig 11.14). F D Is the figure thus formed a parallelogram? Compare the area of each triangle to the area of the parallelogram. Compare the base and height of the triangles with the base and height of the parallelogram. You will find that the sum of the areas of both the triangles is equal to the area of the parallelogram. The base and the height of the triangle are the same as the base and the height of the parallelogram, respectively. 1 Fig 11.14 Area of each triangle = (Area of parallelogram) 2 1 = (base ×height) (Since area of a parallelogram = base × height) 2 = 1(×1 bh , in short) 2 bh) (or 2 TRY THESE 1. Try the above activity with different types of triangles. 2. Take different parallelograms. Divide each of the parallelograms into two triangles by cutting along any of its diagonals.Are the triangles congruent? In the figure (Fig 11.15) all the triangles are on the baseAB = 6 cm. What can you say about the height of each of the triangles corresponding to the base AB? Can we say all the triangles are equal in area? Yes. Are the triangles congruent also? No. 6 cm We conclude that all the congruent triangles are equal in area but the triangles equal in area need not be congruent. Fig 11.15 214 MATHEMATICS Consider the obtuse-angled triangle ABC of base 6 cm (Fig 11.16). A Its height AD which is perpendicular from the vertex A is outside the triangle. Can you find the area of the triangle? EXAMPLE 6 One of the sides and the corresponding height of a CD B 6cm parallelogram are 4 cm and 3 cm respectively. Find the area of the parallelogram (Fig 11.17). Fig 11.16 SOLUTION Given that length of base (b) = 4 cm, height (h) = 3 cm Area of the parallelogram = b × h = 4 cm × 3 cm = 12 cm2 EXAMPLE 7 Find the height ‘x’ if the area of the parallelogram is 24 cm2 and the base is Fig 11.17 4 cm. SOLUTION Area of parallelogram = b × h Therefore, 24 = 4 × x (Fig 11.18) 24 or = x or x = 6 cm 4 Fig 11.18 So, the height of the parallelogram is 6 cm. EXAMPLE 8 The two sides of the parallelogramABCD are 6 cm and 4 cm. The height corresponding to the base CD is 3 cm (Fig 11.19). Find the (i) area of the parallelogram. (ii) the height corresponding to the baseAD. SOLUTION (i) Area of parallelogram = b × h = 6 cm × 3 cm = 18 cm2 (ii) base (b) = 4 cm, height = x (say), Area = 18 cm2 Area of parallelogram = b × x B A 18 = 4 × x x 18 = x4 D Therefore, x = 4.5 cm 6 cm Thus, the height corresponding to baseAD is 4.5 cm. Fig 11.19 S.No. Base Height Area of the Parallelogram a. 20 cm 246 cm2 b. 15 cm 154.5 cm2 c. 8.4 cm 48.72 cm2 d. 15.6 cm 16.38 cm2 Base Height Area of Triangle 15 cm ______ 87 cm2 _____ 31.4 mm 1256 mm2 22 cm ______ 170.5 cm2 218 MATHEMATICS to decorate these cards. What length of the lace does she require for each? (Fig 11.28) (a) (b) (c) Fig 11.28 You cannot measure the curves with the help of a ruler, as these figures are not “straight”. What can you do? Here is a way to find the length of lace required for shape in Fig 11.28(a). Mark a point on the edge of the card and place the card on the table. Mark the position of the Fig 11.29 point on the table also (Fig 11. 29). Now roll the circular card on the table along a straight line till the marked point again touches the table. Measure the distance along the line. This is the length of the lace required (Fig 11.30). It is also the distance along the edge of the card from the marked point back to the marked point. You can also find the distance by putting a string on the edge Fig 11.30 of the circular object and taking all round it. The distance around a circular region is known as its circumference. DO THIS Take a bottle cap, a bangle or any other circular object and find the circumference. Now, can you find the distance covered by the athlete on the track by this method? Still, it will be very difficult to find the distance around the track or any other circular object by measuring through string. Moreover, the measurement will not be accurate. So, we need some formula for this, as we have for rectilinear figures or shapes. Let us see if there is any relationship between the diameter and the circumference of the circles. Consider the following table: Draw six circles of different radii and find their circumference by using string.Also find the ratio of the circumference to the diameter. Circle Radius Diameter Circumference Ratio of Circumference to Diameter 1. 3.5 cm 7.0 cm 22.0 cm 22 7 = 3.14 2. 7.0 cm 14.0 cm 44.0 cm 44 14 =3.14 3. 10.5 cm 21.0 cm 66.0 cm 66 21 =3.14 4. 21.0 cm 42.0 cm 132.0 cm 132 42 =3.14 5. 5.0 cm 10.0 cm 32.0 cm 32 10 =3.2 6. 15.0 cm 30.0 cm 94.0 cm 94 30 =3.13