Question

find the area of the quadrilateral ABCD whose side AB=13 and ACD is a equilateral triangle

Answer 1

heya sakna here

in a quadrilateral ABCD

IN ΔABC

SINCE IT IS A RIGHT ANGLEDΔ     ,LET BC BE X

BY PYTHAGORAS THEOREM

(13)²=(X)²  +(12 )² ,(X)²=25

X=5,BC = 5

AREA TRIANGLE ABC=1/2×12×5 = 30

IN ΔACD SINCE IT IS AN EQUILATERAL Δ

AREA =√3/4 (SIDE)²  =36√3

AREA OF QUADRILATERAL =36√3+30

Answer 2

It is given that

ABCD is a quadrilateral in which ∠BCA = 900 and AB = 13 cm

ABCD is an equilateral triangle in which AC = CD = AD = 12 cm

In right angled △ABC

Using Pythagoras theorem,

AB2 = AC2 + BC2

Substituting the values

132 = 122 + BC2

By further calculation

BC2 = 132 – 122

BC2 = 169 – 144 = 25

So we get

BC = √25 = 5 cm

We know that

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

It can be written as

= ½ × base × height + √3/4 × side2

= ½ × AC × BC + √3/4 × 122

Substituting the values

= ½ × 12 × 5 + √3/4 × 12 × 12

So we get

= 6 × 5 + √3 × 3 × 12

= 30 + 36√3

Substituting the value of √3

= 30 + 36 × 1.732

= 30 + 62.28

= 92.28 cm2

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