In isosceles △abc the segment bd (with d∈ ac ) is the median to the base ac . find bd, if the perimeter of △abc is 50 meters, and the perimeter of △abd is 40 meters.

Given

In isosceles ΔABC the segment BD (with D∈ AC ) is the median to the base AC.

perimeter of ΔABC is 50 meters, and the perimeter of ΔABD is 40 meters

Find out the value of the BD.

To proof  

In the isosceles triangle the  median is also become the altitude.

First we  prove that

ΔABD is congurent to Δ CBD

In ΔABD and  Δ CBD

(1) AB = BC  

( As given in the question ΔABC isaisosceles triangle thus two sides of  this triangle are equal i.e AB = BC .)

(2) BD = BD  

( common side of the triangle)

(3)  AD = DC

( BD is a median i.e it divided AC in two equal parts i.e AD = DC )

Thus by using the SSS congurence property.  

Thus

perimeter of ΔABD = perimeter of Δ CBD = 40 meter

( by corresponding part of the congurent triangle)

let the length of the median (BD) = x

Thus the equation becomes

perimeter of ΔABD + perimeter of ΔCBD - 2 ( length of the median BD ) = perimeter of Δ ABC

putting the value in the above equation

we get

40 + 40 - 2x = 50

80 - 2x = 50

30 = 2x

x = 15 meters

thus the length of the median BD is 15 meters.

Hence proved

x = -5

step-by-step explanation:

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629