# How to calculate triangle MNP ?

## Know S triangle ABC is $24 c {m}^{2}$ In AB, take M, know AM = MB. In BC, take N, know BN = $\frac{1}{3}$ BC In the long stretch of AC, take P, know CP = $\frac{1}{4}$ AC. Calculate S triangle MNP.

Mar 31, 2018

area of $\Delta M N P = 9 {\text{ cm}}^{2}$

#### Explanation:

Let $| X Y Z |$ denote area of $\Delta X Y Z$
let $| M N P | =$ yellow area, and $| N B P | =$ purple area
$| M N P | = | A B P | - | N B P | - | M B N | - | A M P |$
given $| A B C | = 24 {\text{ cm}}^{2}$,
as $C P = \frac{1}{4} A C , \implies | C B P | = \frac{1}{4} | A B C | = \frac{1}{4} \cdot 24 = 6$
as $B N = \frac{1}{3} B C , \implies | N B P | = \frac{1}{3} | C B P | = \frac{1}{3} \cdot 6 = 2$
$\implies | A B P | = | A B C | + | C B P | = 24 + 6 = 30$
as $A M = M B , \implies | A M P | = \frac{1}{2} | A B P | = \frac{1}{2} \cdot 30 = 15$
as $B M = A M , \implies | M B C | = \frac{1}{2} | A B C | = \frac{1}{2} \cdot 24 = 12$
as $B N = \frac{1}{3} B C , \implies | M B N | = \frac{1}{3} | M B C | = \frac{1}{3} \cdot 12 = 4$

Now, $| M N P | = | A B P | - | N B P | - | M B N | - | A M P |$
$= 30 - 2 - 4 - 15 = 9 {\text{ cm}}^{2}$

Apr 4, 2018

Area of $\Delta M N P = 9 {\text{ cm}}^{2}$

#### Explanation:

Solution 2 :

Let $| X Y Z |$ denote area of $\Delta X Y Z$
see Fig 1
given $| A B C | = 24 {\text{ cm}}^{\circ} , \mathmr{and} C P = \frac{1}{4} A C$,
$\implies | C B P | = \frac{1}{4} | A B C | = \frac{1}{4} \cdot 24 = 6 {\text{ cm}}^{2}$
$\implies \textcolor{red}{| A B P |} = | A B C | + | C B P | = 24 + 6 = \textcolor{red}{30} {\text{ cm}}^{2}$
as $B N = \frac{1}{3} B C , \implies \textcolor{red}{| N B P |} = \frac{1}{3} | C B P | = \frac{1}{3} \cdot 6 = \textcolor{red}{2} {\text{ cm}}^{2}$
See Fig 2,
as $M B = \frac{1}{2} A B , \implies | M B C | = \frac{1}{2} | A B C | = \frac{1}{2} \cdot 24 = 12 {\text{ cm}}^{2}$
as $B N = \frac{1}{3} B C , \implies \textcolor{red}{| M B N |} = \frac{1}{3} | M B C | = \frac{1}{3} \cdot 12 = \textcolor{red}{4} {\text{ cm}}^{2}$
See Fig 3,
as $A M = \frac{1}{2} A B , \implies \textcolor{red}{| A M P |} = \frac{1}{2} | A B P | = \frac{1}{2} \cdot 30 = \textcolor{red}{15} {\text{ cm}}^{2}$

Now, $| M N P | = \text{ orange area } = | A B P | - | N B P | - | M B N | - | A M P |$
$= 30 - 2 - 4 - 15 = 9 {\text{ cm}}^{2}$