+0

# GEOMETRY IV

+1
94
4
+249

In the figure, what is the area of triangle ABD? Express your answer as a common fraction.

Nov 15, 2019

#1
+106532
+2

Triangle   ACF is right

Let

A = (0, 4)      B  = (7,0)     E = (0,2)   F(3,0)   C = (0,0)

The line joining   AF  has a slope of  [ -4/3 ]

And the equation of this line  is   y =  (-4/3)x + 4     (1)

The  line joining   BE  has a slope of [ -2/7]

And the equation of this line is  y = (-2/7) x +2    ( 2)

We can find the coordinates of D  by equating (1)  and (2)

(-4/3)x + 4  = (-2/7)x  + 2

4 - 2  = ( 4/3 - 2/7)x

2 =  ( 22/21) x

(2)(21)/22 = x  = 21/11

And y = (-4/3) (21/11) + 4   =  16/11

So....the area of triangle  ACF =  (1/2) (4) (3)  = 6

And the area of triangle DFB  =  (1/2) (16/11) (4)  = 32/11

And the area of triangle ACB = (1/2) (4) (7)  = 14

So....the area of triangle ADB  =

14  - 32/11 - 6    =

8 -32/11

[88 - 32 ] /11  =

56 / 11  units^2

Nov 15, 2019
#2
+23835
+3

In the figure, what is the area of triangle $$ABD$$? Express your answer as a common fraction.

Nov 15, 2019
#3
+118
+4

Here is a more geometric approach. I guess you can call mine Euclidean, whereas our moderator's approach is Cartesian.

Let the line segment HF be perpendicular to CB. Then from the fact that BHF and BEC are similar triangles we can deduce

$$\frac{HF}{2}=\frac{4}{7}$$and therefore $$HF=\frac{8}{7}$$. From the similarity of triangles AED and FHD and the fact that the ratio of the sides is

$$\frac{8}{7}\div2=\frac{4}{7}$$ we can conclude that $$c=(\frac{4}{7})^2\cdot b=\frac{16}{49}\cdot b$$ (lower case letters indicate areas of the respective regions). Furthermore, from the fact that $$(c+e) + a=7\ and \ b+a=6$$ we can conclude that $$(c+e)-b=1$$. We also know that $$e=2\cdot\frac{8}{7}=\frac{16}{7}$$.Therefore,

$$(\frac{16}{49}\cdot b+\frac{16}{7})-b=1$$ and this implies that $$b=\frac{21}{11}$$. Since $$d+b=14-7=7$$, we have $$d=7-\frac{21}{11}=\frac{56}{11}$$, the looked-for area opf the triangle ADB.

Nov 15, 2019
#4
+106532
0