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

User Redhead
by
3.2k points

2 Answers

13 votes
13 votes

Answer:

2.4

Explanation:

15 votes
15 votes

Answer:

Area of ΔQRS = 2.3 square inches

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;


(q)/(SinQ) =
(r)/(SinR) =
(s)/(SinS)

Using
(r)/(SinR) =
(s)/(SinS)


(r)/(Sin 85) =
(2.3)/(Sin51)

r =
(2.3*Sin85)/(sin51)

= 2.9483

r = 2.9 inches

Also,
(q)/(SinQ) =
(s)/(SinS)


(q)/(Sin44) =
(2.3)/(Sin51)

q =
(2.3*Sin44)/(Sin51)

= 2.0559

q = 2.0 inches

From Herons formula,

Area of a triangle =
√(s(s-q)(s-r)(s-s))

s =
(2.3 + 2.0 + 2.9)/(2)

= 3.6

Area of ΔQRS =
√(3.6(3.6-2.0(3.6-2.9)(3.6-2.3))

= 2.2895

Area of ΔQRS = 2.3 square inches

User Dietrich Ayala
by
2.9k points