Pls help (9th honors geo - circles)

• Angles DXC and AXB form a vertical pair, so they are congruent and have the same measure.

• ∆ABD is isosceles, since it's given that AD and BD are congruent. This means the "base angles" BAD and ABD have the same measure; call this measure x.

• The measure of angle ADB can be computed by using the inscribed angle theorem, which says

m∠ADB = 1/2 (100°) = 50°

(that is, it's half the measure of the subtended arc AB whose measure is 100°)

• The interior angle to any triangle sum to 180° in measure. So we have in ∆ABD,

m∠ADB + 2x = 180°

Solve for x :

50° + 2x = 180°

2x = 130°

x = 65°

• Use the inscribed angle theorem again to find the measure of angle BAC. This will be half the measure of the subtended arc BC, so

m∠BAC = 1/2 (50°) = 25°

• Now in ∆ABX, we have

m∠AXB + 25° + 65° = 180°

m∠AXB = 90°

Hence m∠DXC = 90°.

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.