Question

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
ALE 300
B.
AT) 6 cm
(2) 7cm
(3) 8 cm
(4) 9 cm​

Answer 1

Step-by-step explanation:

in the equator answer of the question is 8 cm

Answer 2

Answer:

7 cm.

Step-by-step explanation:

Let us draw a perpendicular on ET from L, which will bisect ET at O.

See the attached diagram.

From ΔEOL, LO²= LE²-EO², ⇒h²= 25²-15² =400, ⇒ h= 20 cm.

Let us assume AE=TB=a cm.

Then from ΔAOL, AL²= h²+(a+15)², ⇒ AL = $\sqrt{400+(a+15)^{2} }$

Given condition, perimeter ALAB=Perimeter ALET +20

Hence, 2 $\sqrt{400+(a+15)^{2} }$+2a+30 =  $\sqrt{400+(a+15)^{2} }$+25+30+20

⇒ $\sqrt{400+(a+15)^{2} }$=45-2a

⇒ 400+a²+30a+225 =2025-180a+4a²

⇒3a²-210a+1400=0

Hence, a=$\frac{210-\sqrt{210^{2}-4*3*1400 } }{6}$ (Taking small value}

⇒ a= $\frac{210-165.22}{6}$ =7.46 ≈7 cm. (Answer)