Question

The areas of two similar triangles are 49 cm sq 64 cm sq respectively. If the difference of the corresponding altitudes is 10 cm, then find the length of their altitudes.

Answer 1

They are similar triangle:

$\dfrac{\text {Area 1}}{\text{Area 2}} = \bigg( \dfrac{\text {length 1}}{\text{length 2}} \bigg)^2$

Given that their difference in altitude is 10 cm:

Let x be the altitude of the smaller triangle:

$\dfrac{49}{64} = \bigg(\dfrac{x}{x + 10} \bigg)^2$

Square root both sides:

$\dfrac{7}{8} = \dfrac{x}{x + 10}$

Cross multiply:

$7(x + 10) = 8x$

Distribute 7:

$7x + 70 = 8x$

Take away 7x from both sides:

$x = 70$

Find the length of the altitudes:

Smaller triangle = x = 70 cm

Big triangle = x + 10 = 70 + 10 = 80 cm

Answer: The length are 70 cm and 80 cm

Answer 2

Answer:

Step-by-step explanation:

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