Question

Find the area of quadrilateral ABCD in which BCD is an equilateral triangle each of food sites is 26 CM AD =24cm and angle BAD=90°. Also, find the perimeter of a quadrilateral. (Given,√3=1.73)

Answer 1

SOLUTION:-

Given:

•BCD is an equilateral ∆ each of whose sides is 26cm.

•AD= 24cm.

•angle BAD= 90°

To find:

•The area of the equilateral ABCD.

•The perimeter of a quadrilateral.

Explanation:

Let ABCD is a quadrilateral in which AD=24cm & ∆BCD is an equilateral ∆.

&

In right ∆BAD,

Using Pythagoras Theorem:

=) (BD)² =(AB)² + (AD)²

=) (26)² = (AB)² + (24)²

=) 676 = (AB)² + 576

=) AB² = 676 -576

=) AB² = 100

=) AB= √100

=) AB = 10cm

&

Area of right ∆BAD,

$= > \frac{1}{2} \times b \times h \\ \\ = > \frac{1}{2} \times 10 \times 24 \\ \\ = >( 5 \times 24)cm {}^{2} \\ \\ = > 120 {cm}^{2}$

Now,

In equilateral ∆BCD,

Let a, b and c are the sides of triangle & S is the semi-perimeter.

Therefore,

Using Heron's Formula:

$S = \frac{a + b + c}{2}$

•A= 26cm

•B= 26cm

•C= 26cm

$= >S = \frac{26 + 26 + 26}{2} \\ \\ = > S = \frac{78}{2} \\ \\ = > S = 39cm$

&

Area of the ∆BCD,

$A = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = > \sqrt{39(39 - 26)(39 - 26)(39 - 26)} \\ \\ = > \sqrt{39 \times 13 \times 13 \times 13} \\ \\ = > A = 292.72 {cm}^{2}$

Hence,

Area of the quadrilateral ABCD,

Area of ∆BAD + Area of ∆BCD

=) 120cm² + 292.72cm²

=) 412.72cm²

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Answer 2

Answer:

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