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3 years ago

How to solve the problem

Mathematics

1 answer:

WINSTONCH [101]3 years ago

7 0

As tgose lines are parallel then,

a = 140° ( by alternate interior angles)

then,
tha angle just below "a" on the other paralle line would also be 140° by corresponding angles,

using exterior angle property of triangle

66° + b = 140°

b = 140° -66
= 74°

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4x + 6 < -6<br><br> Show steps used to solve the problem :)

Sergeeva-Olga [200]

Answer:

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7 0

2 years ago

Read 2 more answers

If D is midpoint of side BC of triangle ABC. P and Q are points lying respectively on side AB and AC such that DP is parallel to

LUCKY_DIMON [66]

See proof below

Step-by-step explanation:

Assume triangle ABC to have vertices at;

A(2,-1), B(2,-7) and C(6,-7)

D is midpoint of BC, thus D is at (4,-7)

The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA

Plot the points on a graph tool and join the points to view the sketch.

To prove area of triangle CPQ is 1/4 area of ABC will be;

Find area ABC and CPQ then compare the areas.

Apply the distance formula to find the length of sides of the triangles then find the areas.

The distance formula is;

$d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$

Length of side AB from the sketch is;

$AB=\sqrt{(2-2)^2+(-7--1)^2} =\sqrt{-6^2} =\sqrt{36} =6units$

Length of side BC will be;

$BC=\sqrt{(-7--7)^2+(6-2)^2} =\sqrt{4^2} =\sqrt{16} =4units$

Thus area of triangle ABC will be;

1/2 *base length*height ------because it is a right-triangle

1/2*4*6=12 square units

Find the lengths of all sides of triangle CPQ

Length of side PQ is half that of side BC thus PQ=2 units

Length of side PC is;

$PC=\sqrt{(6-2)^2+(-7--4)^2} =\sqrt{4^2+-3^2} =\sqrt{16+9} =\sqrt{25} =5units$

Length of side QC will be;

$QC=\sqrt{(6-4)^2+(-7--4)^2} =\sqrt{2^2+-3^3} =\sqrt{4+9} =\sqrt{13}$

QC= √13 = 3.6 units

Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;

A=√s(s-a)(s-b)(s-c) where;

A=area of the triangle

s= half the perimeter of the triangle

a=side PQ = 2 units

b=side PC = 5 units

c= side QC = 3.6 units

Finding the perimeter of triangle CPQ will be;

P=sum of all sides

P=2+5+3.6 =10.6 units

s=10.6/2 = 5.3

Area of the triangle CPQ will be;

$A=\sqrt{5.3(5.3-2)(5.3-5)(5.3-3.6)} \\A=\sqrt{5.3(3.3)(0.3)(1.7)} \\A=\sqrt{8.9} =2.98$

A=3.0 (1 decimal place)

Compare the areas;

Area of triangle ABC=12 square units

Area of triangle CPQ = 3 square units

Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4

Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3

Learn More

Area of a triangle ;brainly.com/question/14869984

The Heron's formula : brainly.com/question/10713495

Keywords: midpoint, triangle, sides, parallel, prove , area, equal

#LearnwithBrainly

6 0

2 years ago

I need help I hope it uploaded the pic

jenyasd209 [6]

A number that's bigger than 4.3 is the answer :)

8 0

3 years ago

What is the quotient of -3/8 and -1/3

Inga [223]

Answer:

(-3/8)/(-1/3)

=(-3/8) x (-3)

=9/8

=11/8

Step-by-step explanation:

6 0

3 years ago

Here is a list of incoming and outgoing expenditure for the arena, for one day. Work out the final profit of the arena rounded t

stira [4]

Answer:

Final Profit = £11,000

Step-by-step explanation:

Incoming / Outgoing Profit / Cost Events £22,765

Staff £6,711

Security £3,452

Maintenance £500

Merchandise £967

Final Profit = £22,765 - £6,711 - £3,452 - £500 - £967 = £11,135

6 0

3 years ago