HELLO GUYS
PLZ ANSWER THIS QUESTION
in a trapezium ABCD,
segAB || seg DC
seg BD perpendicular seg AD
seg AC perpendicular seg BC
IF AD = 15 ,BC = 15
and AB = 25
FIND AREA OF ABCD
Answers
Construction: 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.
HOPE IT HELPS YOU..!!!
