ICSE Class 8 Maths Selina Solutions Chapter 20: ICSE Class 8 Maths Selina Solutions Chapter 20 focuses on the concept of calculating the area of trapeziums and polygons. This chapter introduces students to formulas and methods for finding the area of these shapes, emphasizing practical applications in geometry.
It covers step-by-step explanations and examples to help students understand the differences between trapeziums and polygons, and how to apply specific formulas for accurate area calculations. By mastering these techniques, students develop a strong foundation in geometry and problem-solving skills, preparing them for more complex mathematical challenges ahead.Trapezium: A trapezium (in American English) or trapezoid (in British English) is a quadrilateral with at least one pair of parallel sides. The parallel sides are called the bases of the trapezium, and the other two sides are called the legs. The height of a trapezium is the perpendicular distance between its two parallel sides.
Area of Trapezium
The area AA of a trapezium (or trapezoid) can be calculated using the formula: A=12×(a+b)×hA = \frac{1}{2} \times (a + b) \times h where:Polygon: A polygon is a closed geometric figure with straight sides. It is formed by connecting a finite number of line segments in a closed loop. Polygons can have any number of sides, but they must have at least three sides (making them triangles or higher-order polygons). Common examples include triangles, squares, pentagons, hexagons, and so on. Polygons are classified based on the number of sides they have (e.g., triangles have three sides, quadrilaterals have four sides).
Area of Polygon
ICSE Class 8 Maths Selina Solutions Chapter 20 PDF
Question 1.
Find the area of a triangle, whose sides are: (i) 10cm, 24cm and 26cmSolution:-
Sides of Δ are a=10cm b=24cm c=26cmSolution:-
Sides of Δ are a=18mm b=24mm C=30mmSolution:-
Sides of Δ are a=21m b=28m c=35mQuestion 2.
Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm; find. (i) Area of the triangle (ii) Height of the triangle corresponding to 8 cm side.Solution:-
BC=6cm Height AD=4cmQuestion 3.
The sides of a triangle are 16cm, 12cm and 20cm. Find: (i) Area of the triangle; (ii) Height of the triangle, corresponding to the largest side; (iii) Height of the triangle, corresponding to the smallest side.Solution:-
Sides of Δ are a=20cm b=12cm c=16cmQuestion 4.
Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8m side is 6m; find: (i) Area of the triangle; (ii) height of the triangle corresponding to 6.4 m side.Solution:-
ABC is the Δ in which BC=4.8m AC=6.4m and AD=6mQuestion 5.
The base and the height of a triangle are in the ratio 4:5. If the area of the triangle is 40 m2; find its base and height.Solution:-
Question 6.
The base and the height of a triangle are in the ratio 5:3. If the area of the triangle is 67m2.find its base and height.Solution:-
Consider base =5×m Height =3×mQuestion 7.
Question 8.
The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.Solution:-
Consider each side of the equilateral triangle = xQuestion 9.
A field is in the shape of a quadrilateral ABCD in which side AB = 18m, side AD = 24m, side BC = 40m, DC = 50m and angle A = 90°. Find the area of the field.Solution:-
∠A=90° By Pythagoras Theorem, In ∆ABD,Question 10.
The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm . Find its area.Solution:-
Consider the sides of the triangle ABC be 4 x, 5 x and 3x AB=4 x, AC=5 x and BC=3 x Perimeter = 4 x+5x+3x=96 12 x=96Question 11.
One of the equal sides of an isosceles triangle is 13cm and its perimeter is 50cm. Find the area of the triangle.Solution:-
In Isosceles ΔABC AB=AC=13cm But perimeter =50cm BC=50-(13+13)cm =50-26=24cm AD⊥BCQuestion 12.
The altitude and the base of a triangular field are in the ratio 6:5. If its cost is Rs.49, 57,200 at the rate of Rs.36, 720 per hectare and 1 hectare =10,000 sq. m, find (in meter) dimensions of the field,Solution:-
Total cost =349,57,200 Rate = 736,720 per hectare Total area of the triangular fieldQuestion 13.
Find the area of the right-angled triangle with hypotenuse 40cm and one of the other two sides 24cm.Solution:-
In right angled triangle ABC Hypotenuse AC =40cm One side AB=24cmQuestion 14.
Use the information given in the adjoining figure to find: (i) The length of AC. (ii) The area of a ΔABC (iii) The length of BD, correct to one decimal place.Solution:-
AB=24 cm, BC=7 cm ( i ) A C = √ A B 2 + B C 2 = √ 24 2 + 7 2 = √ 576 + 49 = √ 625 = 24 c m (ii)Area of Δ A B C = 1 2 A B × B C = 1 2 × 24 × 7 = 84 c m 2 ( i i i ) B D ⊥ A C Area Δ A B C = 1 2 A C × B D 84 = 1 2 × 25 × B D ⇒ B D = 84 × 2 25 = 168 25 = 6.72 c m =6.7 cmQuestion 15.
Find the length and perimeter of a rectangle, whose area =120cm 2 and breadth =8cmSolution:-
Area of rectangle =120cm 2 Breadth, b=8cm