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Maths Geometry Question Help?
I had a go at it and I just want to see if my answer is alright:
In triangle PQR, RP = RQ = 12cm. Let S and T be points on sides RP and RQ respectively, such that RS = RT = 8cm. Let X be the point of intersection of the segments PT and QS. If the area of quadrilateral RSXT is 8 cm^2, what is the area of triangle PQR?
2 Answers
- PopeLv 72 years agoFavorite Answer
Let area(PQR) = u.
∆STR ~ ∆PQR
∴ area(STR) : area(PQR)
= ST² : PQ²
= 8² : 12²
= 4 : 9
area(STR) : area(PQR) = 4 : 9
area(STR) = (4/9)area(PQR) = (4/9)u
area(SPQT) = area(PQR) - area(STR) = u - (4/9)u = (5/9)u
∠STP = ∠TPQ ... (alternate angles, ST || PQ)
Let ∠STP = ∠TPQ = α.
area(STP) : area(TPQ)
= (1/2)(TS)(TP)sin(α) : (1/2)(PT)(PQ)sin(α)
= TS : PQ
= 2 : 3
area(STP) : area(TPQ) = 2 : 3
and area(STP) + area(TPQ) = area(SPQT)
∴ area(STP) = (2/5)area(SPQT) = (2/5)(5/9)u = (2/9)u
∆STX ~ ∆QPX
∴ TX : PX = ST : QP = 2 : 3
TX : PX = 2 : 3
TX : TP = 2 : 5
area(STX) : area(STP) = 2 : 5
area(STX) = (2/5)area(STP) = (2/5)(2/9)u = (4/45)u
area(STR) + area(STX) = area(RSXT)
(4/9)u + (4/45)u = 8 cm
(8/15)u = 8 cm²
u = 15 cm²
