I don't know the area of a parallelogram off the top of my head but you can use the area of the rectangle and the area of the triangle and add those. The area of the triangle is 1/2*base*height and use the pythagorean theorem to find the base. 3² + b² = 5²; 9 + b² = 25; b² = 16; b = 4. The area of the triangle is 1/2*4*3 or 6 in. Since there are two of them, the total area of the triangles are 12 in. The area of the rectangle is length*width or (12 - 4) * 3 which is 8 * 3 or 24in. Add these areas together to get 36in total

4 0

3 years ago

Madison spent 3 times as long on her social project as she did on her other project. If she spent a total of 8 hours on he proje

deff fn [24]

Answer:

6 hours

Step-by-step explanation:

Suppose Madison spent x hours on her social project,

then the number of hours spent on other project will be 1/3 of x=x/3

Total hours spend on all projects will be the addition of hours spent on social project and hour spent on other project, i.e

x + x/3 = 4x/3

It si given that this is 8 hours, therefore

4x/3=8

Multiplying both sides by 3:

4x=24

Dividing across board by 4:

x= 6 hours

5 0

3 years ago

[4]

Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

But

(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

7 0

3 years ago