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Question: Find the area of pentagon\(PQRST\)in which \(QD \bot PR,RE \bot PS\)and \(TF \bot PS\)such that \(PR...

Find the area of pentagonPQRSTPQRSTin which QDPR,REPSQD \bot PR,RE \bot PSand TFPSTF \bot PSsuch that PR=10PR = 10cm, PS=12cm,PS = 12cm, QD=3cm,QD = 3cm, RE=7cmRE = 7cmand TF=5cm.TF = 5cm.

Explanation

Solution

It is better to understand a pentagon is a five-sided polygon in geometry. In the given question, the sides of the pentagon are different so do not use the properties of a regular pentagon.

Complete step by step Solution:
From the given figure,

We can observe that the pentagon can be separated into three triangles namely, ΔPQR,\Delta PQR, ΔPRS\Delta PRSand ΔPTS\Delta PTS
If we find the areas of these three triangles, then the sum of areas of all three angles will give us the area of the given pentagon i.e.
ar(ΔPQR)+ar(ΔPRS)+ar(ΔPTS)=ar(PQRST)ar(\Delta PQR) + ar(\Delta PRS) + ar(\Delta PTS) = ar(PQRST) . . . (1)
We know that, for a triangle with base b and perpendicular h, the area of a triangle is
=12×b×h= \dfrac{1}{2} \times b \times h
In ΔPQR\Delta PQR
It is given that QDPRQD \bot PR
Thus we can write
ar(ΔPQR)=12×PR×QDar(\Delta PQR) = \dfrac{1}{2} \times PR \times QD
Substituting the given values of PR and QD, we get
ar(ΔPQR)=12×10×3ar(\Delta PQR) = \dfrac{1}{2} \times 10 \times 3
ar(ΔPQR)=15cm2\Rightarrow ar(\Delta PQR) = 15c{m^2} . . . (2)
In ΔPRS\Delta PRS
It is given that REPSRE \bot PS
Thus we can write
ar(ΔPRS)=12×RE×PSar(\Delta PRS) = \dfrac{1}{2} \times RE \times PS
Substituting the given values of RE and PS, we get
ar(ΔPRS)=12×7×12ar(\Delta PRS) = \dfrac{1}{2} \times 7 \times 12
ar(ΔPRS)=42cm2\Rightarrow ar(\Delta PRS) = 42c{m^2} . . . (3)
InΔPTS\Delta PTS
It is given that TFPSTF \bot PS
Thus we can write
ar(ΔPTS)=12×PS×TFar(\Delta PTS) = \dfrac{1}{2} \times PS \times TF
Substituting the given values of PS and TF, we get

ar(ΔPTS)=12×12×5ar(\Delta PTS) = \dfrac{1}{2} \times 12 \times 5
ar(ΔPTS)=30cm2\Rightarrow ar(\Delta PTS) = 30c{m^2} . . . (4)
From equation (1)
ar(ΔPQR)+ar(ΔPRS)+ar(ΔPTS)=ar(PQRST)ar(\Delta PQR) + ar(\Delta PRS) + ar(\Delta PTS) = ar(PQRST)
Substituting the values of areas of respective triangles from equation (2), (3) and (4), we get
ar(PQRST)=15+42+30ar(PQRST) = 15 + 42 + 30
ar(PQRST)=87cm2\Rightarrow ar(PQRST) = 87c{m^2}
Hence, Area of pentagonPQRSTPQRSTis 87cm287c{m^2}
Note: This question was made easy by giving all the required values. But it is not always necessary that you will be given all this information. In such cases, to solve the problems on any polygon, you need to know their properties. Like, sum of angles and properties of regular polygons.