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In a parallelogram ABCD, the altitude from the vertex B to the side

AD
bisects side
AD
. The perimeter of ABCD is 38 in, and the perimeter of △ABD is 10 in less. Find BD, AB, and AD.

User StathisG
by
4.7k points

1 Answer

7 votes

Answer:

DB = 9 in , AB = 9 in , AD = 10 in

Explanation:

* Lets explain how to solve the problem

- The figure ABCD is a parallelogram

- The altitude from B to the side AD bisects it

- In Δ DAB

∵ BE ⊥ AD where E is the point of intersection between the

altitude from point B to the side AD

∵ The altitude from point B to the side AD bisects it

∴ Point E is the mid-point of AD

- In any triangle if a segment drawn from a vertex perpendicular

and bisects the opposite side, then the triangle is isosceles triangle

∵ BE ⊥ AD and EA = ED

∴ Δ BAD is an isosceles triangle

∴ AB = DB

- The perimeter of the parallelogram is 38 inches

∴ 2 AD + 2 AB = 38 ⇒ divide both sides by 2

∴ AD + AB = 19

- The perimeter of Δ BAD is 10 less

∴ The perimeter of Δ BAD = 38 - 10 = 28

∴ AB + DB + AD = 28

∵ AB + AD = 19

∴ 19 + DB = 28 ⇒ subtract 19 from both sides

DB = 9

∵ AB = DB

AB = 9

∵ AB + AD = 19

∴ 9 + AD = 19 ⇒ subtract 9 from both sides

AD = 10

∴ DB = 9 in , AB = 9 in , AD = 10 in

User Sherelyn
by
4.7k points