Question

The areas of similar triangles are 81 cm2 and 49 cm2 if the altitude of the bigger triangle is 4.5 CM find the corresponding altitude of smaller triangle

Answer 1

altitude of smaller triangle is 2.7

Step-by-step explanation:

area of bigger triangle =1/2base ×height

81cm2= 1/2 base × 4.5cm

81/4.5= 1/2 base

18×2 = base

base = 36 cm

area of smaller triangle =1/2 base × height

49 cm2 =1/2×36 × height

49 cm2 = 18 × height

49/18 = height

height = 2.7 cm

Answer 2

Step-by-step explanation:

gn.that,

ar(∆ABC)= 81cm^

ar(∆PQR)= 49cm^

Let us draw AD perpendicular BC in ∆ABC; Let us draw PL perpendicular to QR....

ar(∆ABC)/ar(∆PQR)= (AD/PL)^

81/49= (4.5/PL)^

(9/7)^=(4.5/PL)^

S.O.B.S

(9/7)=(4.5/PL)

PL= (4.5*7)/9

PL= 0.5*7

PL= 3.5cm....

hope u got the answer buddy