Question

perimeter of a triangle os 32cm and area ofinscribing circle is 38.5sqcm.Let us write the area of the triangle​

Answer 1

Answer:

dear student ,hope your doubt gets clear

from this attached solution

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Let BCD is a triangle of perimeter 32cm.AB, AD and AC are the internal bisectors of the respective angles of the triangle.

The internal bisectors meet at A.

Perpendicular drawn from A on sides BC, CD and DB are AG, AE and AF respectively.

AG=AE=AF

We know that area of a circle=πr^2

= > \frac{22}{7} {r}^{2} = 38.5

= > {r}^{2} = 38.5 \div \frac{22}{7}

= > {r}^{2} = 3.85 \times \frac{7}{22}

= > {r}^{2} = \frac{385 \times 7}{220}

= > {r}^{2} = \frac{539}{44}

= > {r}^{2} = 12.25

= > r = \sqrt{12.25}

= > r = 3.5cm

Now,

Area of ΔBCD= Area of ΔBCA + Area of ΔDCA + Area of ΔBDA

=>Area of ΔBCD= 1/2 × BC × r + 1/2 CD × r + 1/2 BD ×r

=>Area of ΔBCD= 1/2 ×r(BC + CD + BD)

=>Area of ΔBCD= 1/2×3.5(Perimeter of ΔABCD)

=>Area of ΔABCD= 1/2×3.5×3.2

=>Area of ΔABCD=

56 {cm}^{2}