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$Triangle BNR is shown where NR=18.9 centimeters (cm),BN=22.4cm,BR=27.3cm, and NF=15.1cm. Which is the area of /_\BNR ? (A) 142.695cm^(2) (B) 169.12cm^(2) (C) 211.68cm^(2) (D) 206.115cm^(2)$

Triangle $B N R$ is shown where $N R=18.9$ centimeters $(\mathrm{cm}), B N=22.4 \mathrm{~cm}, B R=27.3 \mathrm{~cm}$ , and $N F=15.1 \mathrm{~cm}$ . $\newline$ Which is the area of $\triangle B N R$ ? $\newline$ (A) $142.695 \mathrm{~cm}^{2}$ $\newline$ (B) $169.12 \mathrm{~cm}^{2}$ $\newline$ (C) $211.68 \mathrm{~cm}^{2}$ $\newline$ (D) $206.115 \mathrm{~cm}^{2}$

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Math Problems

Grade 6

Division with decimal divisors

Full solution

Q. Triangle

B N R

is shown where

N R=18.9

centimeters

(\mathrm{cm}), B N=22.4 \mathrm{~cm}, B R=27.3 \mathrm{~cm}

, and

N F=15.1 \mathrm{~cm}

\newline

Which is the area of

\triangle B N R

\newline

(A)

142.695 \mathrm{~cm}^{2}

\newline

(B)

169.12 \mathrm{~cm}^{2}

\newline

(C)

211.68 \mathrm{~cm}^{2}

\newline

(D)

206.115 \mathrm{~cm}^{2}

Calculate Semi-Perimeter: To find the area of triangle BNR, we can use Heron's formula since we know the lengths of all three sides. Heron's formula states that the area of a triangle with sides of lengths $a$ , $b$ , and $c$ is the square root of $s(s-a)(s-b)(s-c)$ , where $s$ is the semi-perimeter of the triangle, or $(a+b+c)/2$ .
Apply Heron's Formula: First, we calculate the semi-perimeter $s$ of triangle BNR. $s = \frac{NR + BN + BR}{2}$ $s = \frac{18.9 \, \text{cm} + 22.4 \, \text{cm} + 27.3 \, \text{cm}}{2}$ $s = \frac{68.6 \, \text{cm}}{2}$ $s = 34.3 \, \text{cm}$
Find Area of Triangle: Now, we apply Heron's formula to find the area $A$ of triangle BNR. $A = \sqrt{s(s - NR)(s - BN)(s - BR)}$ $A = \sqrt{34.3(34.3 - 18.9)(34.3 - 22.4)(34.3 - 27.3)}$ $A = \sqrt{34.3(15.4)(11.9)(7)}$ $A = \sqrt{34.3 \times 15.4 \times 11.9 \times 7}$ $A = \sqrt{93422.274}$ $A \approx 305.65 \, \text{cm}^2$