Question

n a right-angled triangle ABC, AB = 10 3 cm and BC = 20 cm, ∠A = 90°. An equilateral triangle
ABD is constructed with base AB and with vertex D, at a maximum possible distance from C. Find
the length of CD.​

Answer 1

Answer:

Two equilateral triangles can be termed with the base AB.

But vertex D should be at maximum distance from C.

Therefore, triangle ABD will be preferred.

Now,ΔABD is equilateral triangle.

∴AB=BD=AD=10√3 cm

In right ΔABC

(BC) 2=(AB) 2 +(AC) 2

(20) 2=(10 √3 ) 2 +(AC) 2

(AC) 2=400−300

AC=10

In ΔACD,∠CAD=90 0 +60 0=150 0

cos∠CAD= (AD) 2 +(AC) 2 −(CD) 2÷(AD) (AC)

[∵cosC= 2ab ÷a 2 +b 2 −c 2 ]

cos 5π/6=10√3 2+10 2 -CD 2÷2(10√3) (10)

10 √3 2 +10 2−CD 2÷2(AD) (AC)

-√3/2 = 300+100−(CD) 2÷200√3

(CD) 2=700

CD=10√7 cm

Step-by-step explanation:

Hope it helps you.