Question

The area of two similar triangles are 49 cm² and 64 cm² respectively.If the difference of the corresponding altitudes is 10 cm, then find the lengths of altitudes (in cm).

Answer 1

Answer:

They are similar triangle:

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

Area 2

Area 1

=(

length 2

length 1

)

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

64

49

=(

x+10

x

)

2

Square root both sides:

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

8

7

=

x+10

x

Cross multiply:

7(x + 10) = 8x7(x+10)=8x

Distribute 7:

7x + 70 = 8x7x+70=8x

Take away 7x from both sides:

x = 70x=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