HELP!!! Find the lengths of the diagonals of this trapezoid.

Mathematics

1 answer:

Alina [70]2 years ago

5 0

Answer:

√[ (a + b)^2 + c^2 ].

Step-by-step explanation:

They are both equal in length because we have an isosceles trapezoid here. We see form the drawing it is symmetrical about the y axis.

The diagonal = the line joining points (b, c) and (-a, 0)

= √[ (b - -a)^2 + (c - 0)^2 ]

= √[ (a + b)^2 + c^2].

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What is x equal to ? Link down below.

scZoUnD [109]

Answer:

Step-by-step explanation:

Because of supplementary angles that line has to equal to 180

180-75 = 105

so 105 = (2x + 5)

you can subtract 5 from each side and get

100 = (2x)

then divide each side by 2 and you get

x = 50

you can check it by doing (2 x 50 + 5)

2 x 50 = 100 + 5 = 105 + 75 = 180

5 0

2 years ago

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BartSMP [9]

Answer:

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Step-by-step explanation:

$9.( {m}^{3} {n}^{5} )^{ \frac{1}{4} } \\ = m^{\frac{3}{4}}n^{\frac{5}{4}}\\ \\ 10. \sqrt[5]{ \sqrt[4]{x} } \\ = \sqrt[5]{ {x}^{ \frac{1}{4} } } \\ = {( {x}^{ \frac{1}{4} } )}^{ \frac{1}{5} } \\ = {x}^{ \frac{1}{4} \times \frac{1}{5} } \\ = {x}^{ \frac{1}{20} } \\ \\ \sqrt[5]{ \sqrt[3]{ {a}^{2} } } \\ = \sqrt[5]{ {a}^{ \frac{2}{3} } } \\ = {( {a}^{ \frac{2}{3} } )}^{ \frac{1}{5} } \\ = {a}^{ \frac{2}{3} \times \frac{1}{5} } \\ = {a}^{ \frac{2}{15} }$