Question

ABCD is a kite, so AC⊥DB and DE = EB. Calculate the length of AC, to the
nearest tenth of a centimeter.

Answer 1

Answer:

AC = 8.8cm

Step-by-step explanation:

EC = (Tan x = O/A)

EC = (Tan x = 4/EC)

x = sin^-1 (O/H)

x = sin^-1 (4/7)

x = 34.74°

EC = (Tan 34.75° = 4/EC)

EC = (EC Tan 34.75/tan 34.75 = 4/tan 34.75)

EC = 5.8

EC + AE = AC

AE = (Tan x = O/A)

AE = (Tan x = AE/4)

x = cos^-1 (A/H)

x = sin^-1 (4/5)

x = 36.87°

AE = (Tan 36.87° = AE/4)

AE = (4 Tan 36.87/AE)

3 = AE

EC + AE = AC

5.8 × 3 = 8.8cm

Answer 2

Applying the Pythagorean Theorem, the length of AC to the nearest tenth of a cm, in the given kite, is: 8.7 cm.

From the information given, we know the following:

• DE = EB = 4cm
• AD = 5 cm
• CD = 7 cm
• Triangles AED and CED are right-angled triangles

Thus:

• AC = AE + CE

Find AE and CE using Pythagorean Theorem $c^2 = a^2 + b^2$, where c if the hypotenuse of the right triangle.

Length of AE:

$AE = \sqrt{AD^2 - DE^2}$

• Substitute

$AE = \sqrt{5^2 - 4^2}\\\\AE = \sqrt{25 - 16} \\\\AE = \sqrt{9} \\\\\mathbf{AE = 3}$

Length of CE:

$CE = \sqrt{CD^2 - DE^2}$

• Substitute

$CE = \sqrt{7^2 - 4^2}\\\\CE = \sqrt{49 - 16} \\\\CE = \sqrt{33} \\\\\mathbf{CE = 5.7}$

Length of AC:

$AC = AE + CE$

• Substitute

$AC = 3 + 5.7\\\\\mathbf{AC = 8.7 cm}$

Therefore, the length of AC to the nearest tenth of a cm, in the given kite, is: 8.7 cm.

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