Question

in the figure, BC = 7 and BD = 3 then write the ratio of A(triangleADC):A(triangle ABC)​

Answer 1

Step-by-step explanation:

construction , AM perpendicular to BC

hope this helps you

Answer 2

Given:

BC = 7cm

BD = 3cm

To Find:

the ratio of triangle A(triangle ADC): A(triangle ABC)

Solution:

In Δ ABC

To find the measure of DC we will subtract BD from BC and the measurements of BC and BD are given.

DC = BC-BD

     = 7 - 3 [the measurement of BC and BD is given]

     = 4cm

∴ DC = 4cm

Now, constructing AM⊥BC

The formula to find the area of a triangle is 1/2×base×height

Area of ΔADC = 1/2×DC×AM

                        = 1/2×4×AM [substituting the value of DC]

                        = 2AM

Area of ΔABC = 1/2×BC×AM

                       = 1/2×7×AM[substituting the value of BC(given)]

                       = 7AM/2

So, the ratio of ΔADC:ΔABC,

$\frac{ar(ADC)}{ar(ABC} = \frac{2AM}{7AM}*2$

            = 4/7AM

            = 4:7

Therefore, the ratio of the area(ΔADC): area(ΔABC) is 4:7.