Question

In triangle , side and the perpendicular bisector of meet in point , and bisects . If and , what is the area of triangle

Answer 1

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, 145–√5  is the area of a triangle.

I supposed here that [ABD] is the perimeter of ▲ ABD.

As  BD  is a bisector of  ∠ABC ,

ABBC=ADDC=97

Let  ∠B=2α

Then in isosceles  △DBC

∠C=α

BC=2∗DC∗cosα=14cosα

Thus  AB=18cosα

The Sum of angles in  △ABC  is  π  so

∠A=π−3α

Let's look at  AC=AD+DC=16 :

AC=BCcosC+ABcosA

16=14cos2α+18cosαcos(π−3α)

[1]8=7cos2α−9cosαcos(3α)

cos(3α)=cos(α+2α)=cosαcos(2α)−sinαsin(2α)=cosα(2cos2α−1)−2cosαsin2α=cosα(4cos2α−3)

With  [1]

8=cos2α(7−9(4cos2α−3))

18cos4−17cos2α+4=0

cos2α={12,49}

First root lead to  α=π4  and  ∠BDC=π−∠DBC−∠C=π−2α=π2 . In such case  ∠A=π−∠ABD−∠ADB=π4, and  △ABD  is isosceles with  AD=BD. As  △DBC  is also isosceles with  BD=DC=7,  AD=7≠9.

Thus first root  cos2α=12  cannot be chosen and we have to stick with the second root  cos2α=49. This gives  cosα=23  and  sinα=5√3.

The area of a triangle ABD=12h∗AD  where h  is the distance from  B  to  AC.

h=BCsinC=14cosαsinα

Area of  triangle ABD=145–√5

= 145–√5.

Incomplete question please read below for the proper question.

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, what is the area of triangle ABD?

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