Question

the areas of two similar triangles are 81cm​2 and 49cm2 respectively .if the altitude of the bigger triangle is 4.5cm find the corresponding altitude of the similar triangle

Answer 1

Area of big triangle ( A )= 81 cm²



Area of smaller triangle ( a )= 49 cm².



Altitude of bigger triangle ( H ) = 4.5 cm



And,
Let altitude of smaller triangle be X cm.


As we know that the area's of two similar triangles are in the ratio of the squares of the corresponding Altitudes.



So,



Area of bigger triangle / Area of smaller triangle = ( Altitude of bigger triangle / Altitude of smaller triangle)² .




A / a = H² / X²



81 / 49 = (4.5)² / x²



X² = ( 20.25 × 49 / 81 )


X² = 12.25




X = √12.25


X = 3.5 cm.


Hence,



altitude of smaller triangle = 3.5 cm

Answer 2

Answer:

The altitude corresponding to the similar triangle is equal to 3.5cm whose area is 49cm².

Step-by-step explanation:

Consider that two have two similar triangles ΔABC and ΔPQR,

We have given, the area of triangle ΔABC = 81cm²

The area of second triangle ΔPQR = 49cm²

The altitude of triangle  ΔABC is h₁ and the altitude of ΔPQR is h₂.

We have given, the altitude of  ΔABC, h₁  = 4.5cm

For two similar triangles, apply the area of similar triangle theorem:

$\frac{Area\; of \;\triangle ABC}{Area\; of \;\triangle PQR}=\frac{(h_1)^2}{(h_2)^2}$

Substitute the value of area of triangles and  h₂ in above equation;

$\frac{81}{49} =\frac{(4.5)^2}{(h_2)^2}$

$(h_{2})^2=\frac{49}{81}*20.25$

$(h_2)^2=12.25$

$h_2=\sqrt{12.25}$

$h_2=3.5cm$

Therefore, the altitude corresponding to the ΔPQR is equal to 3.5cm.