It is known that area of a cyclic quadrilateral is \sqrt{(s-a)(s-b)(s-c)(s-d)} where a, b, c, d are the sides and s= \frac{a+b+c+d}{2} If a circle can also be inscribed in the cyclic quadrilateral then the area of this quadrilateral is

NTSE Stage 2 SAT 2019 Exam Question Paper

Question : 53 of 100

Marks: +1, -0

It is known that area of a cyclic quadrilateral is

√{(s-a)(s-b)(s-c)(s-d)}

where a, b, c, d are the sides and

s=\;\;{a+b+c+d}/{2}

If a circle can also be inscribed in the cyclic quadrilateral then the area of this quadrilateral is

Solution:

since the given quadrilateral is a tangentialquadrilateral.

∴a+c=b+d=x

(say) (Pitot Theorem)

s=\;\;{a+b+c+d}/{2}=\;\;{2x}/{2}=x

Area

=√{(s-a)(s-b)(s-c)(s-d)}

=√{(a+c-a)(b+d-b)(a+c-c)(b+d-d)}

=√{(c)(d)(a)(b)}=√{abcd}

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