Question

PLEASE HELP!!

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

Answer 1

AC = 12.7 cm

Explanation:

To find:-

• The length of AC.

Answer :-

We are here given a kite in which AC ⊥ DB and DE = EB = 4cm . We are interested in finding out the length of AC .

`\rule{200}2`

D I A G R A M : -

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`\rule{200}2`

We can see that,

`\sf:\implies AC = AE+EC\\`

We can seperately find AE and EC and then add them up to find AC . We can see that due to the diagonals intersecting each other at right angles , there is formation of 4 right angled triangle, the triangles which we will be using are ∆AED and ∆CED .

We can use Pythagoras theorem here according to which

• The square of hypotenuse (longest side) is equal to the sum of squares of other two sides.

`\sf:\implies h^2 = a^2 + b^2 \\`

In two triangles AED and CED hypotenuse are 7cm and cm respectively.

So that , in ∆AED ,

`\sf:\implies 7^2 = AE^2 + DE^2 \\`

`\sf:\implies AE^2 = 7^2 - DE^2 = 7^2 - 4^2 \\`

`\sf:\implies AE^2 = 49 - 16 \\`

`\sf:\implies AE = √( 33) \\`

`\sf:\implies\red{ AE = 5.74} \\`

Similarly, in ∆CED ,

`\sf:\implies 8^2 = CE^2 + DE^2 \\`

`\sf:\implies CE^2 = 8^2 - DE^2 = 8^2 - 4^2 \\`

`\sf:\implies CE^2 = 64- 16 \\`

`\sf:\implies CE = √( 33) \\`

`\sf:\implies\red{ CE = 6.93} \\`

Now add them up to find AC , as ;

`\sf:\implies AC = AE + CE\\`

`\sf:\implies AC = 5.74 + 6.93 \\`

`\sf:\implies AC = 12.67 \\`

Rounding off to nearest tenth, will give us,

`\sf:\implies \red{ AC = 12.7 \ cm } \\`

Hence the length of AC is 12.7 cm.

Answer 2

• 12.7 cm

----------------------------

Since diagonals of a kite are perpendicular to each other, the triangles AED and CED are right triangles.

Find the length of ED:

• ED = BE = BD/2 = 8 / 2 = 4 cm

Find the length of legs AE and CE using Pythagorean theorem:

• 
  `AE=√(AD^2-ED^2)=√(7^2-4^2)=√(49-16)=√(33)=5.74`
• 
  `CE=√(CD^2-ED^2)=√(8^2-4^2)=√(64-16)=√(48)=6.93`

Find the length of AC:

• AC = AE + CE = 5.74 + 6.93 = 12.67 ≈ 12.7 cm