Question

Question #9
Revi
In triangle ABC, B = 90° BD=BE=4. BC = BA =10. If the area of triangleAFC is R, then value of 7/2R is​

Answer 1

Answer:

Value of $\frac{7}{2R}$is 0.108426 .

Step-by-step explanation:

Given informations,

ABC is a triangle. Value of Angle B is 90° and length of BC and BA are 10 unit and length of BD and BE is 4 unit.

By Pythagoras theorem (For triangle EBC),

${EC}^{2} = { EB}^{2} + { BC}^{2}$

${EC}^{2} = {4}^{2} + {10}^{2}$

${EC }^{2} = 16 + 100 = 116$

$EC = \sqrt{116} = 10.77$

So,length of EC is 10.77 unit.

Again by Pythagoras theorem (For the triangle ABC),

${AC}^{2} = {AB }^{2} + { BC}^{2}$

$AC = \sqrt{ {10}^{2} + {10}^{2} }$

$AC = \sqrt{100 + 100} = \sqrt{200}$

$AC = 14.14$

So length of AC is 14.14 unit.

Now,in the triangle AFC

F is the midpoint of triangle as it is made by intersection of two perpendicular line.

Now base of triangle AFC = FC

$= \frac{EC}{2} = \frac{10.77}{2} = 5.38$unit.

and height $=AE=AB-BE = 10 - 4 = 6$unit.

So,area of triangle AFC= base x height

$= 5.38 \times 6 = 32.28$square unit.

According to question,

$R = 32.28$

Now,

$\frac{7}{2R} = \frac{7}{2} \times 32.28 = 0.108426$