Question

Explore Activity 3: play a Kite!

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Answer 1

$\bold{\huge{\underline{ Solution }}}$

Given :-

• Quadrilateral PLAY is a kite.
• The length of PA = 12 cm and LY = 6 cm

Part 1 :-

• The area of Quadrilateral PLAY is 36 cm²

Part 2 :-

Here, we have,

• The length PA = 12 cm and LY = 6 cm
• Here, PA and LY are the diagonal of quadrilateral PLAY

We know that,

Area of kite

$\bold{ = }{\bold{\dfrac{1}{2}}}{\bold{ {\times} d1 {\times} d2}}$

Subsitute the required values,

$\sf{ = }{\sf{\dfrac{1}{2}}}{\sf{ {\times}12 {\times} 6}}$

$\sf{ = }{\sf{\dfrac{1}{2}}}{\sf{ {\times}72 }}$

$\bold{ = 36 cm^{2}}$

Hence, The area of quadrilateral PLAY is 36 cm²

Part 3 :-

Pythagoras theorem justifies our answers because the diagonals of rhombus are bisects each other at 90°

According to this theorem

• The sum of squares of the base and perpendicular height of the triangle are equal to the square of hypotenuse.

That is ,

$\bold{\red{ (Hypotenuse)^{2}= (Base)^{2}+ (perpendicular )^{2}}}$

By using this theorem in Quadrilateral PLAY

• We can find the length of diagonals of kite and it's area.

[ Note :- Please refer the attachment for the correct diagram ]