QUESTION 33
The length of the legs of the right triangle are given as,
6 centimeters and 8 centimeters.
The length of the hypotenuse can be found using the Pythagoras Theorem.
[tex] {h}^{2} = {6}^{2} + {8}^{2} [/tex]
[tex] {h}^{2} = 36+ 64[/tex]
[tex] {h}^{2} = 100[/tex]
[tex]h = \sqrt{100} [/tex]
[tex]h = 10cm[/tex]
Answer: C
QUESTION 34
The triangle has a hypotenuse of length, 55 inches and a leg of 33 inches.
The length of the other leg can be found using the Pythagoras Theorem,
[tex] {l}^{2} + {33}^{2} = {55}^{2} [/tex]
[tex] {l}^{2} = {55}^{2} - {33}^{2} [/tex]
[tex] {l}^{2} = 1936[/tex]
[tex]l = \sqrt{1936} [/tex]
[tex]l = 44cm[/tex]
Answer:B
QUESTION 35.
We want to find the distance between,
(2,-1) and (-1,3).
Recall the distance formula,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Substitute the values to get,
[tex]d=\sqrt{( - 1-2)^2+(3- - 1)^2}[/tex]
[tex]d=\sqrt{( - 3)^2+(4)^2}[/tex]
[tex]d=\sqrt{9+16}[/tex]
[tex]d=\sqrt{25}[/tex]
[tex]d = 5[/tex]
Answer: 5 units.
QUESTION 36
We want to find the distance between,
(2,2) and (-3,-3).
We use the distance formula again,
[tex]d=\sqrt{( - 3-2)^2+( - 3- 2)^2}[/tex]
[tex]d=\sqrt{( - 5)^2+( - 5)^2}[/tex]
[tex]d=\sqrt{25+25}[/tex]
[tex]d=\sqrt{50}[/tex]
[tex]d=5\sqrt{2}[/tex]
Answer: D
