Question

In ΔQRS, s = 2.3 inches, ∠S=51° and ∠Q=44°. Find the area of ΔQRS, to the nearest 10th of an square inch.

Answer 1

Answer:

Area of ΔQRS = 2.3 square inches

Step-by-step explanation:

From the given information,

<S + <Q + <R = $180^{o}$

51 + 44 + <R = $180^{o}$

95 + <R = $180^{o}$

<R = $180^{o}$ - 95

    = $85^{o}$

<R = $85^{o}$

Applying the Sine rule, we have;

$\frac{q}{SinQ}$ = $\frac{r}{SinR}$ = $\frac{s}{SinS}$

Using $\frac{r}{SinR}$ = $\frac{s}{SinS}$

$\frac{r}{Sin 85}$ = $\frac{2.3}{Sin51}$

r = $\frac{2.3*Sin85}{sin51}$

  = 2.9483

r = 2.9 inches

Also, $\frac{q}{SinQ}$ = $\frac{s}{SinS}$

$\frac{q}{Sin44}$ = $\frac{2.3}{Sin51}$

q = $\frac{2.3*Sin44}{Sin51}$

  = 2.0559

q = 2.0 inches

From Herons formula,

Area of a triangle = $\sqrt{s(s-q)(s-r)(s-s)}$

s = $\frac{2.3 + 2.0 + 2.9}{2}$

  = 3.6

Area of ΔQRS = $\sqrt{3.6(3.6-2.0(3.6-2.9)(3.6-2.3)}$

                        = 2.2895

Area of ΔQRS = 2.3 square inches

Answer 2

Answer:

2.4

Step-by-step explanation: