Question

the figure,CP is parallel to BD.if the area of triangle ABD is "a" and the area of triangle BCD is "b".complete the table.
SHAPE AREA
Triangle BPD
Quadrilateral ABCD
Quadrilateral ABPD
Triangle AED

Answer 1

Answer:

ANSWER:

(1) It is given that line AB is tangent to the circle at A.

∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Thus, the measure of ∠CAB is 90º.

(2) Distance of point C from AB = 6 cm (Radius of the circle)

(3) ∆ABC is a right triangle.

CA = 6 cm and AB = 6 cm

Using Pythagoras theorem, we have

BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cm

Thus, d(B, C) = 62–√ cm

(4) In right ∆ABC, AB = CA = 6 cm

∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)

Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)

∴ 2∠ABC = 90º

⇒ ∠ABC = 90°2 = 45º

Thus, the measure of ∠ABC is 45º.

Step-by-step explanation:

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