Question

orem: $a^2 + b^2 = c^2$. round to the nearest tenth.
5.
right triangle with legs 10 and 24, hypotenuse c
6.
right triangle with legs 35 and 12, hypotenuse c

1. $a = 12$ and $c = 28$
2. $a = 10$ and $b = 14$

Explanation:

Step1: Substitute values into Pythagoras

For problem 5: $a=10$, $b=24$, so $c^2=10^2+24^2$

Step2: Calculate squares and sum

$c^2=100+576=676$

Step3: Solve for c, round to tenth

$c=\sqrt{676}=26.0$

Step4: Substitute values for problem 6

$a=12$, $b=35$, so $c^2=12^2+35^2$

Step5: Calculate squares and sum

$c^2=144+1225=1369$

Step6: Solve for c, round to tenth

$c=\sqrt{1369}=37.0$

Step7: Rearrange formula for problem7

$b^2=c^2-a^2$, $c=28$, $a=12$

Step8: Calculate squares and difference

$b^2=28^2-12^2=784-144=640$

Step9: Solve for b, round to tenth

$b=\sqrt{640}\approx25.3$

Step10: Substitute values for problem8

$a=10$, $b=14$, so $c^2=10^2+14^2$

Step11: Calculate squares and sum

$c^2=100+196=296$

Step12: Solve for c, round to tenth

$c=\sqrt{296}\approx17.2$

Answer:

1. $26.0$
2. $37.0$
3. $25.3$
4. $17.2$