In the given figure, AETB is a straight line such that AE = TB. If ET = 30 cm, LE = LT = 25 cm and the perimeter of ALAB is 20 cm more than the perimeter of ALET, then the length of AE is
(1 ) 6 cm
(2) 7cm
(3) 8 cm
(4) 9 cm
Answers
In the given figure, AETB is a straight line such that AE = TB. If ET = 30 cm, LE = LT = 25 cm and the perimeter of ALAB is 20 cm more than the perimeter of ALET, then the length of AE is
Consider the attached figure while going through the following steps:
Given,
AETB is a straight line such that AE = TB
ET = 30 cm
LE = LT = 25 cm
The perimeter of ALAB is 20 cm more than the perimeter of ALET
Perimeter of Δ LET = 30 + 25 + 25 = 80 cm
Perimeter of Δ LAB = 80 + 20 = 100 cm
In Δ LET
LE = LT
⇒ ∠ LET = ∠ LTE
∴ 180° - ∠ LET = 180° - ∠ LTE
⇒ ∠ LEA = ∠ LTB
In Δ LEA and Δ LTB
LT = LB = 25 cm
AE = TB
∠ LET = ∠ LTE
∴ Δ LEA ≅ Δ LTB (using SAS criteria)
⇒ AL = BL ( c.p.c.t)
Construction: Draw P ⊥ AB
In Δ LET
P bisects ET
∴ EP = 15 cm
LP = √ (25² - 15²) = 20 cm
Let, AE = TB = x
AL = LB = y
Therefore the perimeter,
2x + 2y + 30 = 100
x + y = 35
In Δ LPM,
20² + (15 + x)² = y²
400 + 15² + x² + 30x = y²
625 + x² + 30x = (35 - x)²
625 + x² + 30x = 1225 + x² - 70x
625 - 1225 = - 70x - 30x
- 600 = - 100 x
x = 6
Therefore, the length of AE is 6 cm.
Option (1) is correct.
Answer:
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Step-by-step explanation:
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