Example 37 - Class 12

Transcript

Example 37 If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum. Let ABCD be given trapezium & Given length of three side other than base is 10 ⇒ AD = DC = CB = 10cm. Draw a perpendicular DP & CQ ON AB. Let AP = 𝑥 cm. In ΔAPD & Δ BQC ∠ APD = ∠ BQC AD = BC DP = CQ ∴ ΔAPD ≅ Δ BQC ∴ QB = AP ⇒ QB = 𝑥 cm. Let A be the area of trapezium ABCD A = ﷐1﷮2﷯ (Sum of parallel sides) × (Height) A = ﷐1﷮2﷯ (DC + AB) × DP Now, Since DP & CQ is perpendicular to AB, DPCQ forms a rectangle. ∴ PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 & DC = 10 cm Finding DP In Δ ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 – 𝑥2 DP = ﷐﷮100 −𝑥2﷯ From (1) A = ﷐1﷮2﷯ (DC + AB) DP A = ﷐1﷮2﷯ ﷐10+2𝑥+10﷯ ﷐﷐﷮100−𝑥2﷯﷯ A = ﷐1﷮2﷯ ﷐2𝑥+20﷯ ﷐﷐﷮100−𝑥2﷯﷯ A = ﷐2﷐𝑥 +10﷯ ﷐﷐﷮100 − 𝑥2﷯﷯﷮2﷯ A = ﷐𝑥+10﷯﷐﷮100−𝑥2﷯ We need to find area of trapezium it is maximum i.e. we need to Maximize Area A = ﷐𝑥+10﷯ ﷐﷐﷮100−𝑥2﷯﷯ Diff w.r.t 𝑥 ﷐𝑑𝐴﷮𝑑𝑥﷯= ﷐𝑑﷐﷐𝑥 + 10﷯ ﷐﷐﷮100 − 𝑥2﷯﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐𝑑﷐𝑥 − 10﷯﷮𝑑𝑥﷯ . ﷐﷮100−𝑥2﷯ + ﷐𝑑﷐﷐﷮100 − 𝑥2﷯﷯﷮𝑑𝑥﷯ . ﷐𝑥+10﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐1+0﷯ ﷐﷮100−𝑥2﷯ + ﷐1﷮2﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯ . ﷐𝑑﷐100 − 𝑥2﷯﷮𝑑𝑥﷯ . ﷐𝑥+10﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐﷮100−𝑥2﷯ + ﷐1﷮2﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯ . ﷐0−2𝑥﷯﷐𝑥+10﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐﷮100−𝑥2﷯ – ﷐2𝑥﷐𝑥 + 10﷯﷮2﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐﷐﷐﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯﷮2﷯− 𝑥﷐𝑥 + 10﷯﷮﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐﷐100 − ﷐𝑥﷮2﷯﷯ − 𝑥﷐𝑥 + 10﷯﷮﷐﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯﷯ ﷐𝑑𝐴﷮𝑑𝑥﷯ = ﷐−2𝑥2− 10𝑥 + 100﷮﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯ Putting ﷐𝑑𝐴﷮𝑑𝑥﷯=0 ﷐−2𝑥2− 10𝑥 + 100﷮﷐﷮100 − ﷐𝑥﷮2﷯﷯﷯=0 −2𝑥2− 10𝑥 + 100=0 −2﷐﷐𝑥﷮2﷯+5𝑥−50﷯=0 𝑥2+5𝑥−50=0 𝑥2+10𝑥−5𝑥−50=0 𝑥﷐𝑥+10﷯−5﷐𝑥+10﷯=0 ﷐𝑥−5﷯﷐𝑥+10﷯=0 So, 𝑥=5 & 𝑥=−10 Since 𝑥 represents distance & distance cannot be negative So, 𝑥 = 5 only. Hence value of 𝑥 = 5 Hence, 𝑥 = 5 is point of Maxima ∴ A is Maximum at 𝑥 = 5 Finding maximum area of trapezium A = ﷐𝑥+10﷯ ﷐﷮100−𝑥2﷯ = ﷐5+10﷯ ﷐﷮100−﷐5﷯2﷯ = ﷐15﷯﷐﷮100−25﷯ = 15 ﷐﷮75﷯ = 75﷐﷮3﷯ Hence, maximum area is 75﷐﷮𝟑﷯ cm2