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# Pls Help!! Geometry

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In right triangle ABC, M and N are midpoints of legs AB and BC, respectively. Leg AB is 6 units long, and leg BC is 8 units long. How many square units are in the area of triangle APC?

MIRB15  Jul 13, 2017
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#1
+286
0

???????

#2
+18281
+3

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

Centroid of a Triangle: The point where the three medians of the triangle intersect.

So P is the centroid of the triangle ABC.

$$\begin{array}{lcll} \text{Let } AB = 6 \\ \text{Let } BC = 8 \\\\ \text{Let } \vec{B} = \binom {0}{0} \\ \text{Let } \vec{A} = \binom {0}{AB} \\ \text{Let } \vec{C} = \binom {BC}{0} \\ \end{array}$$

$$\vec{P} \text{ the centroid of the triangle ABC}\ =\ ?$$

$$\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 (\vec{A}+\vec{B}+\vec{C}) \\ \vec{P} &=& \frac13 \Big( \binom {0}{AB}+\binom {0}{0}+\binom {BC}{0} \Big) \\ \vec{P} &=& \frac13 \binom {BC}{AB} \quad & | \quad AB = 6 \qquad BC = 8 \\ \vec{P} &=& \frac13 \binom {8}{6} \\ \vec{P} &=& \binom {\frac{8}{3}}{2} \\ \hline \end{array}$$

$$\text{Area}_{\text{APC}} = \ ?$$

$$\begin{array}{|rcll|} \hline \text{Area}_{\text{APC}} &=& \frac12 | (\vec{C}-\vec{P}) \times (\vec{A}-\vec{P}) | \\ &=& \frac12 | \Big(\binom {BC}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{AB}- \binom {\frac{8}{3}}{2} \Big) | \quad & | \quad AB = 6 \qquad BC = 8 \\ &=& \frac12 | \Big(\binom {8}{0}- \binom {\frac{8}{3}}{2} \Big) \times \Big(\binom {0}{6}- \binom {\frac{8}{3}}{2} \Big) | \\ &=& \frac12 | \binom {8-\frac{8}{3}}{0-2} \times \binom {0-\frac{8}{3}}{6-2} | \\ &=& \frac12 | \binom {\frac{16}{3}}{-2} \times \binom {-\frac{8}{3}}{4} | \\ &=& \frac12 \cdot \Big( \frac{16}{3} \cdot 4 - (-2)\cdot (-\frac{8}{3}) \Big) \\ &=& \frac12 \cdot \Big( \frac{64}{3} - \frac{16}{3} \Big) \\ &=& \frac12 \cdot \frac{48}{3} \\ &=& \frac{48}{6} \\ \text{Area}_{\text{APC}} &=& 8 \text{ square units } \\ \hline \end{array}$$

heureka  Jul 14, 2017
#3
+18281
+3

In right triangle ABC, M and N are midpoints of legs AB and BC, respectively.

Leg AB is 6 units long, and

leg BC is 8 units long.

How many square units are in the area of triangle APC?

another approach

Let $$BC = 8$$
Let $$AB = 6$$

$$\text{Let } A = \text{area}_{ABC} = \frac{AB\cdot BC}{2} = \frac{6\cdot 8}{2} = 24$$

$$\text{Let } A_1 = \text{area}_{BPN} = \text{area}_{NPC} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_2 = area_{BCP} = area_{CAP} . \quad ( \text{The base and the height is equal} ) \\ \text{Let } A_3 = area_{APC} = \ ? \\ \text{Let area}_{BAN} = \frac{ AB\cdot \frac{BC}{2} } {2} = \frac{ AB\cdot BC } {4}\\ \text{Let area}_{BMC} = \frac{ BC\cdot \frac{AB}{2} } {2} = \frac{ AB\cdot BC } {4}$$

$$\begin{array}{|rcll|} \hline \text{area}_{BAN} &=& \text{area}_{BMC} = \frac{ AB\cdot BC } {4} \quad & | \quad \text{area}_{BAN} = A_1 +2A_2 \qquad \text{area}_{BMC} = A_2 +2A_1 \\ A_1 +2A_2 &=& A_2 +2A_1 \\ 2A_2-A_2 &=& 2A_1-A_1 \\ A_2 &=& A_1 \\\\ A_1 +2A_2 &=& \frac{ AB\cdot BC }{4} \quad & | \quad A_2 = A_1 \\ 3A_1 &=& \frac{ AB\cdot BC }{4} \\ A_1 &=& \frac{ AB\cdot BC }{12} \quad & | \quad AB = 6 \qquad BC = 8 \\ A_1 &=& \frac{ 6\cdot 8 }{12} \\ \mathbf{A_1} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline A &=& 2A_1+2A_2+A_3 \quad & | \quad A_2 = A_1 \\ A &=& 2A_1+2A_1+A_3 \\ A &=& 4A_1+A_3 \\ A_3 &=& A-4A_1 \quad & | \quad A = 24 \qquad A_1 = 4\\ A_3 &=& 24-4\cdot 4 \\ A_3 &=& 24-16 \\ \mathbf{A_3 } & \mathbf{=} & \mathbf{8} \\ \hline \end{array}$$

The area of triangle APC is 8 square units

heureka  Jul 14, 2017
#4
+75017
+3

Connect MN

Now......MB = 3 and BN = 4...and angle ABC is right....so....triangle MBN is a 3-4-5 Pythagorean Triple right triangle with  MN = 5  and the area = (MB * BN)/2 =  (3*4)/ 2 = 6

And triangle ABC is a 6-8-10 Pythagorean Triple right triangle with AC  = 10

Now....MN divides the sides of triangle ABC proportionally so MN is parallel to AC

And AN is a transversal between two parallels so that angles CAN and MNP  form equal alternate interior angles....and since they are vertical angles, angle APC = angle  MPN.....

So....by angle- angle congruency,  triangle APC is similar to  triangle NPM

And AC can be considered the base of triangle APC and NM can be considered the base of triangle NPM.......but AC = 2NM.....so, by the square of the scale factor, 2....the area of triangle APC = 4 times the area of triangle NPM

Now....... triangle MBC  has an area of (MB * BC) / 2 =  (3 * 8) / 2 =  24/2 = 12

And triangle ANC has an area of  (AB * NC) / 2 =  ( 6 * 4) / 2 = 12

And MBC = NPMB + PNC

And ANC = APC + PNC

So....since MBC = ANC, then

NPMB + PNC = APC + PNC .....subtract PNC from both sides

NPMB  = APC

But NPMB  = area of triangle MBN + area of triangle NPM

So......substituting

6 + NPM  = APC

6 + NPM  = 4 (NPM)

6 = 3 (NPM)

2 = NPM

And APC is 4 times this  =  8  (square units)

CPhill  Jul 14, 2017
edited by CPhill  Jul 14, 2017
edited by CPhill  Jul 14, 2017
edited by CPhill  Jul 14, 2017
#5
+25874
+3

Here's yet another approach:

.

Alan  Jul 14, 2017
#6
+75017
+1

Well.....it kinda' looks like Alan, heureka and I have beaten this horse to death !!!!

CPhill  Jul 14, 2017

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