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see explanation

Step-by-step explanation:

Using Pythagoras' identity on the right triangles, that is

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let x be the missing side, then

13² = 5² + x²

169 = 25 + x² ( subtract 25 from both sides )

144 = x² ( take the square root of both sides )

x = = 12

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let h be the hypotenuse, then

h² = 24² + 45² = 576 + 2025 = 2601 ( take the square root of both sides )

h = = 51

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Using the converse of Pythagoras' identity, that is

If the square of the longest side equals the sum of the squares on the other 2 sides then the triangle is right.

3, 4, 5

5² = 25 and 3² + 4² = 25 ⇒ right triangle

6, 8, 10

10² = 100 and 6² + 8² = 100 ⇒ right triangle

16, 63, 65

65² = 4225 and 16² + 63² = 4225 ⇒ right triangle

8, 9, 10

10² = 100 and 8² + 9² = 145 ⇒ not a right triangle

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.