Question

the areas of two similar triangles are 42cm and 262.5cm. if the altitude of the smaller triangle is 7cm. what is the length of the base of the larger triangle?
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Answer 1

Answer:

75 cm

Step-by-step explanation:

It is given that the triangles are similar. It means that the ratio of their areas are equal to the ratio of their bases and ratio of their altitudes.

Let $A_1$ be the area of 1st triangle.

Let $A_2$ be the area of 2nd triangle.

Let $h_1$ be the altitude of 1st triangle.

Let $h_2$ be the altitude of 2nd triangle.

Let $b_1$ be the base of 1st triangle.

Let $b_2$ be the base of 2nd triangle.

Then $A_1: A_2 = h_1 : h_2 = b_1 : b_2 ...... (1)$

$A_1 = 42 cm^{2} \\A_2 = 262.5 cm^{2}\\h_1 = 7 cm\\b_2 = ?$

We know that area of a triangle is:

$A = \dfrac{1}{2} \times b \times h$

Area of smaller triangle:

$\frac{1}{2} \times b_1 \times 7 = 42\\\Rightarrow b_1 = 12 cm$

Now, using part of equation (1):

$A_1: A_2 = b_1 : b_2 \\\Rightarrow \dfrac{42}{262.5} = \dfrac{12}{b_2}\\\Rightarrow b_2 = 75 cm$

Hence, base of larger triangle = 75 cm