Area of dragon is the room enclosed through a kite. A dragon is a quadrilateral in which 2 pairs of nearby sides room equal. The facets of a kite room its 4 angles, that 4 sides, and 2 diagonals. In this article, we will emphasis on the area that a kite and also its formula.

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 1 What Is the Area the a Kite? 2 Area that a kite Formula 3 Derivation of the Area of kite Formula 4 FAQ's on Area that a Kite

The area that a kite have the right to be defined as the quantity of an are enclosed or incorporated by a kite in a two-dimensional plane. Prefer a square, and also a rhombus, a kite does not have actually all 4 sides equal. The area that a kite is constantly expressed in terms of units2 because that example, in2, cm2, m2, etc. Let us learn around the area of a dragon formula in our next section.

The area of a dragon is half the product the the lengths the its diagonals. The formula to identify the area of a dragon is: Area = $$\dfrac12\times d_1\times d_2$$. Right here $$d_1$$ and $$d_2$$ are long and short diagonals of a kite.The area of dragon ABCD given below is ½ × AC × BD. BD = lengthy diagonal and also AC = quick diagonal

Consider a kite ABCD as shown above.

Assume the lengths of the diagonals that ABCD to it is in AC = p, BD = q

We recognize that the much longer diagonal of a kite bisects the shorter diagonal at right angles, i.e., BD bisects AC and also ∠AOB = 90°, ∠BOC = 90°.

Therefore,

AO = OC = AC/2 = p/2

Area of kite ABCD = Area of ΔABD + Area the ΔBCD...(1)

We recognize that,

Area that a triangle = ½ × base × Height

Now, we will calculate the locations of triangles ABD and BCD

Area of ΔABD = ½ × AO × BD = ½ × p/2 × q = (pq)/4

Area of ΔBCD = ½ × OC × BD = ½ × p/2 × q = (pq)/4

Therefore, utilizing (1)

Area of kite ABCD = (pq)/4 + (pq)/4= (pq)/2Substituting the values of p and qArea the a dragon = ½ × AC × BD

Important Notes

A kite has two bag of adjacent equal sides.

Example 1: four friends space flying kites the the very same size in a park. The lengths the diagonals of each kite are 12 in and 15 in. Determine the sum of locations of all the four kites.

Solution:

Lengths of diagonals are:

$$d_1$$ =12in

$$d_2$$ =15in

The area the each kite is:

A = $$\dfrac12\times d_1\times d_2$$= ½ × 12 × 15= 90 in2

Since each kite is of the same size, as such the full area of all the four kites is 4 × 90 = 360in2.Therefore the area that the four kites is 360in2

Example 2: Kate wants to provide a kite-shaped coco box to she friend. She desires to dough a snapshot of herself through her girlfriend to covering the peak of the box. Recognize the area of the peak of package if the diagonals that the lid that the box room 9 in and also 12 in.

See more: Is A Negative Number Squared Negative, Can Negative Numbers Be Negative When Squared

Solution:

$$d_1$$ =9in

$$d_2$$ =12in

Since package is kite-shaped, thus the area the the peak of the box is:

A = $$\dfrac12\times d_1\times d_2$$= ½ × 9 × 12Therefore, the area that the optimal of package is 54in2

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