Question

In the given figure, find the perimeter of $△ABC$.

• A. 40 cm
• B. 42.5 cm
• C. 42 cm
• D. 32.5 cm

Answer 1

The correct option is B 42.5 cm

DE is parallel to BC.
Using basic proportionality theorem, we know that
$\frac{AD}{BD}=\frac{AE}{CE}$
$\frac{AD}{6cm}=\frac{8cm}{12cm}$
$AD=\frac{8cm}{12cm}\times 6cm$
$AD=4cm$

In $△ADE$ and $△ABC$,
$\angle D=\angle B$
$\angle E=\angle C$

By AA criteria
In $△ADE\sim △ABC$
$\frac{AE}{AC}=\frac{DE}{BC}$

$\frac{8cm}{20cm}=\frac{5cm}{BC}$
$BC=\frac{20cm}{8cm}\times 5cm=12.5cm$
$AB=AD+DB=4cm+6cm=10cm$

Perimeter of $△ABC=AB+BC+AC$
$=10cm+12.5cm+20cm=42.5cm$