Geometric Calculations And Area/Perimeter Problem Solving

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Geometric Calculations and Area/Perimeter Problem Solving This is a right angle triangle so, area = 1 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ 𝑒𝑖𝑔 ℎ 𝑡 = 1 2 ∗ 3 ∗ 4 = 3 ∗ 2 = 6 . In the same way, if we take 5 as base then x is the height. So, area = 1 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ 𝑒𝑖𝑔 ℎ 𝑡 = > 6 = 1 2 ∗ 5 ∗ 𝑥 12 5 = 𝑥 So, answers are: Area = 6 x = 12 5 . The area of shaded part = 80 ∗ 30 + 1 2 ∗ 80 ∗ 60 + 4 ∗ 50 = 2400 + 2400 + 200
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= 5000 𝑓 𝑡 2 So, we need just 1 bad of lawn food to cover the shaded region. Area of ABCD = 𝐵𝑎𝑠𝑒 × 𝐻𝑒𝑖𝑔 ℎ 𝑡 When we take base as AD then height will be CF. So, Area of ABCD = 𝐴𝐷 × 𝐶𝐹 = 26 × 12 = 312 Now, if we take AB as base then height will be AE. Therefore, Area of ABCD = 𝐴𝐵 × 𝐴𝐸 = > 312 = 𝑥 × 24 = > 𝑥 = 312 24 = > 𝑥 = 13 . Hence, answers are: Area = 312 x = 13.
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We know, area of a rhombus = 𝑏𝑎𝑠𝑒 ∗ ℎ 𝑒𝑖𝑔 ℎ 𝑡 = 5 𝑥 ∗ 4 𝑥 = 20 𝑥 2 So, 20 𝑥 2 = 720 𝑥 2 = 36 𝑥 = 6 So, base = 5x = 5*6 = 30 Height = 4x = 4*6 = 24. So, base is 30 cm and height is 24 cm. Area of trapezoid = 1 2 ( 𝑠𝑢𝑚 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑠𝑖𝑑𝑒𝑠 ) ∗ ℎ 𝑒𝑖𝑔 ℎ 𝑡 = 1 2 ( 4 + 7 ) ∗ 4 = 11 ∗ 2 = 22 . So, area is 22 𝑐 𝑚 2 .
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Area of rhombus = 1 2 ∗ 𝑑 1 ∗ 𝑑 2 = 1 2 ∗ 6 𝑥 ∗ 8 𝑥 = 24 𝑥 2 So, 24 𝑥 2 = 96 𝑥 2 = 4 𝑥 = 2 So, AC = 8x = 8*2 = 16 cm. BD = 6x = 6*2 = 12 cm. Area of a kite = 1 2 ∗ 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑠 = 1 2 ∗ ( 4 + 4 ) ∗ ( 5 + 3 ) = 1 2 ∗ 8 ∗ 8 Area = 4 ∗ 8 = 32 𝑐 𝑚 2 . So, 32 𝑐 𝑚 2 is the answer.

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