Q1. a cloth piece of length 6m and 2.5m wide has a 20cm wide border all around inside it. find the cost of painting the border at rs 5 per sq m? Q2. find the area of a right angled iscoleses triangle if the hypotenuse of the triangle is 12cm. Q3.the perimeter of two squares are 848cm and 436cm. find the perimeter of square whose area is equal to the sum of the areas of two squares.
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Q1
Total length including borders = 6 + 2x0.2 = 6.4 m
Total width including borders = 2.5 + 2x0.2 = 2.9 m
Area of border = 6.4 x 2.9 - 6 x 2.5 = 3.56.m²
Cost of painting = 5 x 3.56 = rs.17.8
Q2
The sides of the triangle are xcm, xcm and 12cm
By Pythagoras Theorem x² + x² = 12²
2x² = 144
x² = 72 (Find the square root of both sides)
x = 8.485 cm
Area of triangle = 1/2 x Base x Height
= 1/2 x 8.485 x 8.485 = 36cm²
Q3
Length of side of square 1 = 848/4 = 212 cm
Length of side of square 2 = 436/4 = 109
Sum of the areas of the two squares = 212² + 109² = 56825 cm² This equals the area of the third square.
Length of side of the third square = √56825 = 238.38 cm
Perimeter = 238.38 x 4 = 953.52 cm
Total length including borders = 6 + 2x0.2 = 6.4 m
Total width including borders = 2.5 + 2x0.2 = 2.9 m
Area of border = 6.4 x 2.9 - 6 x 2.5 = 3.56.m²
Cost of painting = 5 x 3.56 = rs.17.8
Q2
The sides of the triangle are xcm, xcm and 12cm
By Pythagoras Theorem x² + x² = 12²
2x² = 144
x² = 72 (Find the square root of both sides)
x = 8.485 cm
Area of triangle = 1/2 x Base x Height
= 1/2 x 8.485 x 8.485 = 36cm²
Q3
Length of side of square 1 = 848/4 = 212 cm
Length of side of square 2 = 436/4 = 109
Sum of the areas of the two squares = 212² + 109² = 56825 cm² This equals the area of the third square.
Length of side of the third square = √56825 = 238.38 cm
Perimeter = 238.38 x 4 = 953.52 cm
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