15. In a trapezium ABCD,
seg AB || seg DC
seg BD I seg AD,
seg AC I seg BC,
If AD = 15, BC = 15
and AB = 25. Find ACO ABCD)
Answers
Answer:
Draw seg DE ⊥ seg AB, A – E – B
and seg CF ⊥ seg AB, A – F- B.
In ∆ ACB, ∠ACB = 90° [Given] ∴ AB2 = AC2 + BC2 [Pythagoras theorem]
∴ 252 = AC2 + 152
∴ AC2 = 625 – 225 = 40
∴ AC = √400 [Taking square root of both sides]
= 20 units Now, A(∆ABC) = 1/2 × BC × AC Also,
A(∆ABC) = 1/2 × AB × CF
∴ BC × AC = AB × CF
∴ 15 × 20 = 25 × CF
∴ CF = (15 x 20)/25 = 12 units In ∆CFB, ∠CFB 90° [Construction]
∴ BC2 = CF2 + FB2 [Pythagoras theorem]
∴ 152 = 122 + FB2
∴ FB2 = 225 – 144
∴ FB2 = 81
∴ FB = √81 [Taking square root of both sides] = 9 units Similarly, we can show that, AE = 9 units Now, AB = AE + EF + FB [A – E – F, E – F – B]
∴ 25 = 9 + EF + 9
∴ EF = 25 – 18 = 7 units In ⟂CDEF, seg EF || seg DC [Given, A – E – F, E – F – B] seg ED || seg FC [Perpendiculars to same line are parallel]
∴ ⟂CDEF is a parallelogram.
∴ DC = EF 7 units [Opposite sides of a parallelogram] A(⟂ABCD) = 1/2 × CF × (AB + CD) = 1/2 × 12 × (25 + 7) = 1/2 × 12 × 32
∴ A(⟂ABCD) = 192 sq.
Step-by-step explanation:
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