Find the perimeter of triangle PQR. Round your answer to the nearest tenth.

1 answer:

3 0

We are given the coordinate of the 3 vertices. From these coordinates we can find the length of the sides using the distance formula and the sum of all the sides will be equal to the perimeter of the triangle.

Side PQ = $\sqrt{ (6-(-2))^{2}+ (1-1)^{2} }=8$

Side QR = $\sqrt{ (4-6)^{2}+ (-3-1)^{2} }=4.5$

Side RP = $\sqrt{ (2-(-4))^{2}+ (-3-1)^{2} }=7.2$

So, the perimeter of the triangle will be 8 + 4.5 + 7.2 = 19.7

You probably misplaced the digits in option A. The correct answer is 19.7 units.

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Can someone answer these questions ? thank you

Naya [18.7K]

Answer:

Step-by-step explanation:

12a)To rationalize the denominator, multiply the denominator and numerator by √5.

$\frac{15}{\sqrt{5}}=\frac{15*\sqrt{5}}{\sqrt{5}*\sqrt{5}}\\\\=\frac{15\sqrt{5}}{5}\\\\=3\sqrt{5}$

b) (a+b)² = a² + 2ab + b²

(1 +√3)² = 1² + 2*1*√3 + (√3)²

= 1 + 2√3 + 3

= 4 + 2√3

a = 4 ; b =2

13) (a + b)(a - b) = a² - b²

$\frac{(6-\sqrt{5})(6+\sqrt{5})}{\sqrt{31}}=\frac{6^{2}-(\sqrt{5})^{2}}{\sqrt{31}}\\\\ =\frac{36-5}{\sqrt{31}}\\\\=\frac{31}{\sqrt{31}}\\\\=\frac{31*\sqrt{31}}{\sqrt{31}*\sqrt{31}}\\\\=\frac{31\sqrt{31}}{31}\\\\=\sqrt{31}$