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25. Find the area of the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90° and triangle BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given, v3 = 1.73.] |
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Answer» answer In Δ ABD Using the Pythagoras theorem BD 2 =AB 2 +AD 2
By substituting the values 26 2 =AB 2 +24 2
On further calculation AB 2 =676−576 By subtraction AB 2 =100 AB= 100
So we GET Base = AB = 10cm We KNOW that area of ΔABD= 2 1
×b×h By substituting the values Area of ΔABD= 2 1
×10×24 On further calculation Area of ΔABD=120cm 2
We know that the area of ΔBCD= 3
/4a 2
By substituting the values Area of ΔBCD=(1.73/4)(26) 2
So we get Area of ΔBCD=292.37cm 2
So we get area of quadrilateral ABCD = Area of Δ ABD + Area of Δ BCD By substituting the values Area of quadrilateral ABCD = 120 + 29237 By addition Area of quadrilateral ABCD = 412.37 cm 2
The perimeter of quadrilateral ABCD = AB + BC + CD + DA By substituting the values Perimeter = 10 + 26 + 26 + 24 So we get Perimeter = 86cm Therefore, the area is 412.37 cm 2 and perimeter is |
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