Question

in in the given figure triangle ABC triangle D E F G of triangle ABC is equal to 121 CM square and area of triangle d e f is equal to 64 CM square if the median of triangle ABC is 12.1 CM then find the median of triangle d e f ​

Answer 1

Answer:

As, given

ΔABC ~ ΔDEF

Area (ΔABC)=121 cm²

Area (ΔDEF)=64 cm²

Now, median from Vertex A is drawn on side BC is drawn,named F.

BF=F C

As well as median from Vertex D is drawn on side E F , named G.

E G=G F

In Δ ABF and Δ DEG

∠B=∠E →→ ∵ ΔABC ~ ΔDEF

As, ΔABC ~ ΔDEF, so corresponding sides will be proportional.

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

So, $\frac{AB}{DE}=\frac{BF}{EG}$

Δ ABF ~ Δ DEG→→[SAS]

∴ $\frac{AB}{DE}=\frac{BF}{EG}=\frac{AF}{DG}$

As, we know when two triangles are similar,  ratio of areas of triangle is equal to square of ratio of sides as well as medians.

So, ratio of areas of ΔABC and  ΔDEF

$=[\frac{AF}{DG}]^2$

$\frac{121}{64}=\frac{(12.1)^2}{DG^2}\\\\  \frac{11}{8}=\frac{12.1}{DG}\\\\ DG=\frac{12.1 \times 8}{11}\\\\ DG=1.1 \times 8=8.8$

So, length of median of  ΔDEF= 8.8 cm

Answer 2

Answer:

Step-by-step explanation:

Answer:

As, given

ΔABC ~ ΔDEF

Area (ΔABC)=121 cm²

Area (ΔDEF)=64 cm²

Now, median from Vertex A is drawn on side BC is drawn,named F.

BF=F C

As well as median from Vertex D is drawn on side E F , named G.

E G=G F

In Δ ABF and Δ DEG

∠B=∠E →→ ∵ ΔABC ~ ΔDEF

As, ΔABC ~ ΔDEF, so corresponding sides will be proportional.

So,

Δ ABF ~ Δ DEG→→[SAS]

∴

As, we know when two triangles are similar,  ratio of areas of triangle is equal to square of ratio of sides as well as medians.

So, ratio of areas of ΔABC and  ΔDEF

So, length of median of  ΔDEF= 8.8 cm