Question

please help asap in math thank you
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Answer 1

Answer:

7.

this is a right angle triangle.

so,

$x = \sqrt{ {5}^{2} - {3}^{2} } = 4$

8.

total length = 3+7 = 10

Area of largest triangle,

$= \frac{1}{2} \times 10 \times 4 = 20$

9.

y can be get by the congruent property of triangle.

10.

as this is isosceles triangle,

m=n

${m}^{2} + {m}^{2} = 49 \\ 2 {m}^{2} = 49 \\ m = n = \frac{7}{ \sqrt{2} }$

Exercise :

Area of equilateral triangle

$area = \frac{ \sqrt{3} }{4} {a}^{2} = \frac{ \sqrt{3} }{4} \times 400 = 100 \sqrt{3}$

Exercise

lengths of largest triangle in #9

altitude a = 4

base b = 7

hypotenuse c = 8

$c = \sqrt{ {7}^{2} + {4}^{2} } = \sqrt{49 + 16 } \\ = \sqrt{65} = 8$

Answer 2

Answer:

7. 6

8. 21

9. 20

10. m = 6 | n = 7

No Extra Credit