Find the area of this parallelogram<br>8cm<br>10cm<br>14cm

Let ABC be a triangle such that AB=13 BC=14 and CA=15. D is a point on BC such that AD Bisects

Yuri [45]

Answer:

Area of triangle ADC is 54 square unit

Step-by-step explanation:

Here is the complete question:

Let ABC be a triangle such that AB=13, BC=14, and CA=15. D is a point on BC such that AD bisects angle A. Find the area of triangle ADC .

Step-by-step explanation:

Please see the attachment below for an illustrative diagram

Considering the diagram,

BC = BD + DC = 14

Let BD be $x$ ; hence, DC will be $14-x$

and AD be $y$

To, find the area of triangle ADC

Area of triangle ADC = $\frac{1}{2} (DC)(AD)$

= $\frac{1}{2}(14-x)(y)$

We will have to determine $x$ and $y$

First we will find the area of triangle ABC

The area of triangle ABC can be determined using the Heron's formula.

Given a triangle with a,b, and c

$Area =\sqrt{s(s-a)(s-b)(s-c)}$

Where $s = \frac{a+b+c}{2}$

For the given triangle ABC

Let $a$ = AB, $b$ = BC, and $c$ = CA

Hence, $a = 13, b= 14,$ and $c = 15$

∴ $s = \frac{13+14+15}{2} \\s= \frac{42}{2}\\s = 21$

Then,

Area of triangle ABC = $\sqrt{(21)(21-13)(21-14)(21-15)}$

Area of triangle ABC = $\sqrt{(21)(8)(7)(6)}$ = $\sqrt{7056}$

Area of triangle ABC = 84 square unit

Now, considering the diagram

Area of triangle ABC = Area of triangle ADB + Area of triangle ADC

Area of triangle ADB = $\frac{1}{2} (BD)(AD)$

Area of triangle ADB = $\frac{1}{2}(x)(y)$

Hence,

Area of triangle ABC = $\frac{1}{2}(x)(y)$ + $\frac{1}{2}(14-x)(y)$

84 = $\frac{1}{2}(x)(y)$ + $\frac{1}{2}(14-x)(y)$

∴ $84 = \frac{1}{2}(xy) + 7y - \frac{1}{2}(xy)$

$84 = 7y\\y = \frac{84}{7}$

∴ $y = 12$

Hence, $y =$ AD = 12

Now, we can find BD

Considering triangle ADB,

From Pythagorean theorem,

/AB/² = /AD/² + /BD/²

∴13² = 12² + /BD/²

/BD/² = 169 - 144

/BD/ = $\sqrt{25}$

/BD/ = 5

But, BD + DC = 14

Then, DC = 14 - BD = 14 - 5

BD = 9

Now, we can find the area of triangle ADC

Area of triangle ADC = $\frac{1}{2} (DC)(AD)$

Area of triangle ADC = $\frac{1}{2} (9)(12)$

Area of triangle ADC = 9 × 6

Area of triangle ADC = 54 square unit

Hence, Area of triangle ADC is 54 square unit.

5 0

3 years ago

What is seven decreased by 3 times some number

natulia [17]

the answer to the question is 7-3x.

5 0

3 years ago

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Please help, this is due today at 3PM

Agata [3.3K]

It decreases by -.4 over 1

6 0

3 years ago

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Please help look at the picture

k0ka [10]

Answer:

y = - 5x - 1

Step-by-step explanation:

The first thing to do (always) is pay attention to x =0. That at least gives you a partial answer. y = y' - 1. when you put 0 in for whatever y' is you get 0. What about the other numbers? How did they come about.

We have another clue. Whenever you put in a positive number for x, you get a negative answer for y. That's interesting. So y' has a minus sign associated with it.

y = -ax - 1 Make a>0 so there is only 1 minus sign. x cannot have an even power, because an even power would make everything positive except the -1.

So we'll start with y= - ax - 1 We'll also make the assumption that a = 1

y = - x - 1 That won't work. a >1 otherwise x = 2 won't give - 11

y = -ax - 1

let x = 2

let y = - 11 Solve for a

y = - ax - 1

- 11 = -a*-2 - 1 add 1 to both sides

-11 + 1 = - a(-2) - 1 + 1

- 10 = -2a Divide by - 2

- 10/-2 = -2a/-2

a = 5

Answer: y = - 5x - 1

8 0

2 years ago

Whats the slope for this question?

Nastasia [14]

Answer:

Slope: 1

Step-by-step explanation:

A variable is always worth 1 if there is no number in front of it. In this case there is no number, so just draw in a one.

6 0

3 years ago

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Find the area of this parallelogram8cm10cm14cm​

Find the area of this parallelogram8cm10cm14cm