Question

If AB = 60 cm, AC = 80 cm and point E is 20 cm away from the point A, then find the length of side DB.​

Answer 1

Answer:

In the triangle ABC, AB= 60, CA = 80 and BC = 100. D is a point on BC such that triangles ADB and ADC have equal perimeters. What is AD?

From the lengths given ABC is a right angled triangle with BC as the hypotenuse.

Let BD =x and CD = 100-x.

Equating the perimeters of triangles ADB and ADC we get 60+x+AD = 80+(100-x)+AD, or

60+x=180-x, or

2x = 120 or x = 60. Hence BD = 60.

In triangle ADB, AD^2 = AB^2+BD^2–2*AB*BD*cos <ABD

AD^2 = 60^2+60^2–2*60*60*(60/100) = 3600+3600–4320 = 2880, or AD = 2880^0.5 = 53.66563146 units.

Check: Perimeter of ABD = 60+60+53.66563146 = 173.6656315 units.

Perimeter of ACD = 80+40+53.66563146 = 173.6656315 units. Correct.

AD = 53.66563146 units.

Step-by-step explanation:

Answer 2

Given:

A ΔABC with,

$AB= 60 cm\\AC= 80 cm\\BC= 100 cm$ ; and

perimeter of ΔADB = perimeter of ΔADC

To find:

Length of AD.

Explanation:

Step 1

Given that $BC= 100 cm$, and D is a point on BC ,

Let's consider the length BD$= x cm$

Hence, CD becomes$(100- x) cm$

We have,

Perimeter of ΔADB = AB + BD + AD

Substituting the values, we get

Perimeter of ΔADB $= 60 + x + AD$

Similarly,

Perimeter of ΔADC $= 80 + (100 -x) + AD$

Now, given that perimeter of ΔADB and ΔADC are equal

Therefore,

$60 + x + AD = 80 + (100 -x) + AD$

or $2x = 120$

    $x = 60 cm$

Hence,$DB = 60 cm$

Step 2

Now, in ΔADB,

$AD^{2} = AB^{2}+BD^{2}-2(AB)(BD)cos\alpha \\AD^{2}=(60)^{2}+(60)^{2}-2(60)(60)\frac{60}{100}\\AD^{2}= 3600+3600-4320\\AD^{2}= 2800 ; or$

$AD = 54 cm$

Final answer :

Hence, the length of $AD = 54 cm$ (approx).

Although your question is incomplete, you might be referring to the question below,

In the triangle ABC, AB= 60 cm , CA= 80 cm and BC= 100 cm. D is a point on BC such that triangles ABC and ADC have equal perimeters, What is AD?