Question

A kite is in the shape of a square with diagonal
40 cm and an isosceles triangle of base 6 cm and
sides 4 cm. Find the area of the adjoining figure.
​

Answer 1

Answer:

Given PQRS is a square.

SQ=40cm

Let SR=QR=X

Then, x^2+x^2=(40)^2

Or, 2x^2=1600

Or,x^2=800

Therefore area(A1) of square PQRS=800cm^2

Now area of triangle RVU (A2)= Perimeter of triangle RVU

=(4+4+6)= 14cm

Therefore,area of given figure= A1+A2

=800+14=814cm^2

Answer 2

Given : Diagonal of the square = 40 m

To find : area of the figure

Explanation:

Diagonal of the square = 40 m

thus

$\sqrt{a^2+a^2} = 40\\\\ a= 20 \sqrt{2}$

where a is the side of the square

thus , are of the square  is (side)²

= $(20\sqrt{2})^2 = 800 cm ^2$

now ,

in isosceles triangle

let a = 4 cm , b = 4 cm and c = 6 cm

then using herons formula

$s = \frac{a+b+c}{2}\\\\s = \frac{4+4+6}{2}\\\\ s = 7 cm$

then area of triangle is

$\sqrt{s(s-a)(s-b)(s-c)}\\\\\sqrt{7 (7-4)(7-4)(7-6)}\\\\\sqrt{7 \times3\times3\times1 }\\\\3\sqrt{7}$

= 7.94

now , area of the figure is

area of square + area of triangle

800 + 7.94

= 807.84 cm ²

hence , The are of the figure is 807.84 cm ²

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