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Fittoniya [83]
3 years ago
10

6th grade math help me please ! :))

Mathematics
1 answer:
polet [3.4K]3 years ago
3 0

Cost of 32 ounces yoghurt = $1.62

Cost of 1 ounce = $1.62/32 = $0.05 (approx)

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Calculate the length of sides triangle pqr and determine weather or not triangle is a right angled. P(-4,6) q(6,1) r(2,9)
Tom [10]

\bold{\huge{\underline{ Solution }}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • We have given the coordinates of the triangle PQR that is P(-4,6) , Q(6,1) and R(2,9)

<h3><u>To</u><u> </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>calculate </u><u>the </u><u>length </u><u>of </u><u>the </u><u>sides </u><u>of </u><u>given </u><u>triangle </u><u>and </u><u>also </u><u>we </u><u>have </u><u>to </u><u>determine </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

<u>Here</u><u>, </u><u> </u><u>we </u><u>have </u>

  • Coordinates of P =( x1 = -4 , y1 = 6)
  • Coordinates of Q = ( x2 = 6 , y2 = 1 )
  • Coordinates of R = ( x3 = 2 , y3 = 9 )

<u>By </u><u>using </u><u>distance </u><u>formula </u>

\pink{\bigstar}\boxed{\sf{Distance=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2\;}}}

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

Length of side PQ

\sf{ = }{\sf\sqrt{ (6 - (-4))^{2} + (1 - 6)^{2}}}

\sf{ = }{\sf\sqrt{ (6 + 4 )^{2} + (- 5)^{2}}}

\sf{ = }{\sf\sqrt{ (10)^{2} + (- 5)^{2}}}

\sf{ = }{\sf\sqrt{ 100 + 25 }}

\sf{ = }{\sf\sqrt{ 125 }}

\sf{ = 5 }{\sf\sqrt{ 5 }}

Length of QR

\sf{ = }{\sf\sqrt{(2 - 6)^{2} + (9 - 1)^{2}}}

\sf{ = }{\sf\sqrt{(- 4 )^{2} + (8)^{2}}}

\sf{ = }{\sf\sqrt{16 + 64 }}

\sf{ = }{\sf\sqrt{80 }}

\sf{ = 4 }{\sf\sqrt{5 }}

Length of RP

\sf{ = }{\sf\sqrt{ (-4 - 2 )^{2} + (6 - 9)^{2}}}

\sf{ = }{\sf\sqrt{ (-6 )^{2} + (-3)^{2}}}

\sf{ = }{\sf\sqrt{ 36 + 9 }}

\sf{ = }{\sf\sqrt{ 45 }}

\sf{ = 3}{\sf\sqrt{ 5 }}

<h3><u>Now</u><u>, </u></h3>

We have to determine whether the triangle PQR is right angled triangle

<h3>Therefore, </h3>

<u>By </u><u>using </u><u>Pythagoras </u><u>theorem </u><u>:</u><u>-</u>

  • Pythagoras theorem states that the sum of squares of two sides that is sum of squares of 2 smaller sides of triangle is equal to the square of hypotenuse that is square of longest side of triangle

<u>That </u><u>is</u><u>, </u>

\bold{ PQ^{2} + QR^{2} = PR^{2}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>,</u>

\bold{  125 + 80 = 45 }

\bold{  205  = 45 }

<u>From </u><u>above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

  • The triangle PQR is not a right angled triangle because 205 ≠ 45 .
6 0
2 years ago
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