In the figure- AP - 3 cm- AR - 4-5 cm- AQ - 6 cm- AB - 5 cm- and AC - 10 cm- Find the length -in cm- of AD

Question

Toppr Admin · Accepted Answer

-x394-APQ is similar to -x394-ABC-AAA postulate-x2235-APAB-AQACAs AD is line from common vertex-x394-APR-x223C-x394-ABDAPAB-ARAD-x2234-AD-AB-xD7-ARAP-5-xD7-4-53-7-5CM

In the figure, $A P = 3 c m, A R = 4.5 c m, A Q = 6 c m, A B = 5 c m$ , and $A C = 10 c m$ . Find the length (in cm) of $A D$

$Δ A P Q$ is similar to $Δ A B C$
( $A A A$ postulate)
$(∵) \frac{A P}{A B} = \frac{A Q}{A C}$
As $A D$ is line from common vertex,
$Δ A P R \sim Δ A B D$
$\frac{A P}{A B} = \frac{A R}{A D}$
$∴ A D = \frac{A B \times A R}{A P}$
$= \frac{5 \times 4.5}{3}$
$= 7.5 C M$

In the figure, AP=3 cm,AR=4.5 cm,AQ=6 cm,AB=5 cm, and AC=10 cm. Find the length (in cm) of AD

ΔAPQ is similar to ΔABC(AAA postulate)(∵)APAB=AQACAs AD is line from common vertex,ΔAPR∼ΔABDAPAB=ARAD∴AD=AB×ARAP=5×4.53=7.5CM

In the figure, $A P = 3 c m, A R = 4.5 c m, A Q = 6 c m, A B = 5 c m$ , and $A C = 10 c m$ . Find the length (in cm) of $A D$

$Δ A P Q$ is similar to $Δ A B C$
( $A A A$ postulate)
$(∵) \frac{A P}{A B} = \frac{A Q}{A C}$
As $A D$ is line from common vertex,
$Δ A P R \sim Δ A B D$
$\frac{A P}{A B} = \frac{A R}{A D}$
$∴ A D = \frac{A B \times A R}{A P}$
$= \frac{5 \times 4.5}{3}$
$= 7.5 C M$