Question

I dont understand how to do this,if someone could explain it to me I'd really appreciate it.​

Answer 1

Explanation:

9.

m∡BHY = 90°, because is given that BKRY is a kite and we know that diagonals of a kite are perpendicular to each other and ⊥ lines form right angles that measure 90°

10.

m∡RKH = 36°, because is given m∡BKH =36° and we know that the longer diagonal KY bisects ∡ BKR, in two congruent angles ∡BKH ≅∡HKR

11.

m∡HBK = 54°,

because we can use the fact that the sum of angles in ΔBKH is 180°.

m∡HBK = 180 - ∡BKH -∡KHB

m∡HBK = 180 - 36 -90 = 54°

12.

m∡RYB = ?

Not sure about this one.

13.

Perimeter of NITE = NI+IT+TE+EN

Given: EB = 84 feet and BI = BN = 35

Use Pythagorean Theorem from the ΔIBN we have NI = √(BI²+BN²)= √(35²+35²) = √2450 = √(5·5·7·7·2) = 5·7√2 = 35√2

NI≅TI = 35√2, because NI and TI are a pair of consecutive congruent sides.

Use Pythagorean Theorem from the ΔEBN we have NE = √(BN²+BE²)= √(35²+84²) = √8281 = √(7·7·13·13) = 7·13 = 91

NE≅TE = 91, because NE and TE are a pair of consecutive congruent sides.

Perimeter of NITE = NI+IT+TE+EN =

35√2 +35√2+91+91 = 182+70√2 ≈ 281 feet

Answer 2

9. 90°

10. 36°

11. 54°

12. cannot be determined

13. 182 +70√2 ≈ 281 units

14. T, T, T, T, T, F

Explanation:

The diagonals of a kite cross at right angles, so all angles at H in the first figure are 90°. The longer diagonal bisects the vertex angles at either end of it. Of course, the acute angles in each right triangle are complements of each other. The vertex angles at either end of the short diagonal are congruent, and may be acute, right, or obtuse. As in any quadrilateral, the sum of angles is 360°.

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9. ∠BHY = 90°, the angle where the diagonals cross

10. ∠RKH = 36°, congruent to ∠BKH

11. ∠HBK = 54°, complementary to ∠BKH

12. ∠RYB = unknown. None of the angles in ΔRYB can be determined from given information, except to say that ΔRYH ≅ ΔBYH and the angles at H are right angles.

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13. The side lengths can be determined from the Pythagorean theorem (or your knowledge of triangles). ΔEBN has side ratios of 5:12:13, with a scale factor of 7, so EN = 7·13 = 91. ΔIBN has side ratios of 1:1:√2, with a scale factor of 35, so IN = 35√2. The perimeter is twice the sum of these lengths, since the symmetrically opposite sides are the same lengths.

P = 2(91 +35√2) = 182 +70√2 ≈ 281 units

Using the Pythagorean theorem:

EN² = EB² +BN² = 84² +35² ⇒ EN = √8281 = 91

IN² = BI² +BN² = 35² +35² ⇒ IN = √2450 = 35√2 ≈ 49.497

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14.

• True, opposite angles may be supplementary
• True, consecutive angles may be supplementary if the kite is a rhombus (opposite sides are parallel)
• True, opposite angles may be acute
• True, consecutive angles may be obtuse (an example is shown in the figure of problem 13, if the given dimensions are ignored)
• True, opposite angles may be complementary.
• False. Consecutive angles of a kite may not be complementary. (If they were, the figure would be a "dart".)