Question

Consider in the figure below.

The perpendicular bisectors of its sides are , , and . They meet at a single point .
(In other words, is the circumcenter of .)
Suppose , , and .
Find , , and .
Note that the figure is not drawn to scale.

Answer 1

Here BD is perpendicular bisector

So

• UB=BV=74

`\\ \tt\hookrightarrow UV=74+74=148`

Apply Pythagorean theorem

`\\ \tt\hookrightarrow BD^2=UD^2-UB^2=UD^2-VD^2`

• UD=VD=78=TD

`\\ \tt\hookrightarrow TC^2=TD^2-CD^2`

`\\ \tt\hookrightarrow TC^2=78^2-30^2`

`\\ \tt\hookrightarrow TC^2=6084-900=5184`

`\\ \tt\hookrightarrow TC=72`

Answer 2

UV = 148

VD = 78

TC = 72

Explanation:

BD is the perpendicular bisector of side UV.

Therefore, ΔUDV is an isosceles triangle.

This implies that UD = VD and BV = UB so UV = 2 x BV

• Given that UD = 78, and UD = VD, then VD = 78
• Given that BV = 74, and BV = UB, then UV = 2 x 74 = 148

ΔUDC is a right triangle.

Given CD = 30 and UD = 78,

and using Pythagoras' Theorem, we can calculate UC:

UC = √(UD² - CD²)

⇒ UC = √(78² - 30²)

⇒ UC = 72

CD is the perpendicular bisector of side UT.

Therefore, ΔUDT is an isosceles triangle, so UC = TC

Since UC = 72, then TC = 72