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Geometry

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Let ABC be a triangle, and let its angle bisectors be AD, BE, and CF which intersect at I. If DI=3, BD=4 and BI=6 then compute the area of triangle BID.

Jun 10, 2024

#1
+1252
+1

While it may not seem the most important, let's calculate the semiperimeter. We get $$\frac{3+4+6}{2}=\frac{13}{2}$$

The reaon why we must find the semiperimeter is because we can use Heron's Formula.

It states that the area is $$\sqrt{s(s-a)(s-b)(s-c)}$$ where s is the semiperimeter and a,b, and c are the sidelengths.

Applying this, we get $$Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}$$

$$Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}$$

This rounds to about $$5.3327$$

Thanks! :)

Jun 10, 2024

#1
+1252
+1

While it may not seem the most important, let's calculate the semiperimeter. We get $$\frac{3+4+6}{2}=\frac{13}{2}$$

The reaon why we must find the semiperimeter is because we can use Heron's Formula.

It states that the area is $$\sqrt{s(s-a)(s-b)(s-c)}$$ where s is the semiperimeter and a,b, and c are the sidelengths.

Applying this, we get $$Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}$$

$$Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}$$

This rounds to about $$5.3327$$

Thanks! :)

NotThatSmart Jun 10, 2024