An isosceles triangle has area of 110 ft². If the base is 14 ft, what is the length of the legs? Round your answer to the nearest tenth.
A 21 ft
B 17.2 ft
C 14.8 ft
D 442.9 ft

Question

An isosceles triangle has area of 110 ft². If the base is 14 ft, what is the length of the legs? Round your answer to the nearest tenth.
 A	21 ft
 B	17.2 ft
 C	14.8 ft
 D	442.9 ft

solomonzelman · Answer

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solomonzelman · Answer

the area of this triangle would be:

altitude times 14 divided by 2

solomonzelman · Answer

(I will disregard the units for now)

so, you are given the area to be 110:
(let altitude =a)                                     $\large\color{slate}{   110=a\times14\div2    }$

can you solve for a?

anonymous · Answer

No I'm so confused right now.

solomonzelman · Answer

okay, do you get that:

$\LARGE\color{teal}{   A_{\LARGE \rm ~ isosceles~\triangle}=({\LARGE \rm  altitude} \cdot {\LARGE \rm ~ base}) \div2  }$         ??

anonymous · Answer

yes

solomonzelman · Answer

now, lets plug in the given information.
you know:
base=14
area=110

solomonzelman · Answer

$\LARGE\color{teal}{   A_{\LARGE \rm ~ isosceles~\triangle}=({\LARGE \rm  altitude} \cdot {\LARGE \rm ~ base}) \div2  }$

PLUGGING IN  .....

$\LARGE\color{teal}{   100=({\LARGE \rm  altitude} \cdot {\LARGE \rm ~ 14}) \div2  }$

solomonzelman · Answer

can you solve for altitude?

anonymous · Answer

No, I'm not sure how to get the altitude

triciaal · Answer

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