Perimeter and Area of Quadrilateral

Here we will discuss about the perimeter and area of a quadrilateral and some example problems.

In the quadrilateral PQRS, PR is a diagonal, QM ⊥ PR and SN ⊥ PR.

Then, area (A) of the quadrilateral PQRS = Area of ∆PQR + Area of ∆SPR

                                               = (\(\frac{1}{2}\) × QM × PR) + (\(\frac{1}{2}\) × SN × PR)

                                               = \(\frac{1}{2}\) (QM + SN) × PR

Also, area (A) of the quadrilateral PQRS = Area of ∆PQR + Area of ∆SPR

                     = \(\sqrt{s(s - a)(s - b)(s - e)}\) + \(\sqrt{S(S - c)(S - d)(S - e)}\)

where, s = \(\frac{\textrm{a + b + e}}{2}\) and S = \(\frac{\textrm{c + d + e}}{2}\)

Perimeter (P) = a + b + c + d

Solved example problems on finding the perimeter and area of quadrilateral:

1. PQRS is a quadrilateral whose diagonal QS is perpendicular to the side PQ. If PQ = 4.5 cm, PS = 7.5 cm and the distance of R from QS is 1.5 cm, find the area of the quadrilateral.

Solution:

In the right-angled ∆PQS,

PS\(^{2}\) = PQ\(^{2}\) + QS\(^{2}\)

⟹ (7.5)\(^{2}\) cm\(^{2}\) = (4.5)\(^{2}\) cm\(^{2}\) + QS\(^{2}\)

⟹ QS\(^{2}\) = [(7.5)\(^{2}\) – (4.5)\(^{2}\)] cm\(^{2}\)

⟹ QS\(^{2}\) = (7.5 + 4.5)(7.5 - 4.5)  cm\(^{2}\)

⟹ QS\(^{2}\) = 12 × 3  cm\(^{2}\)

⟹ QS\(^{2}\) = 36  cm\(^{2}\)

⟹ QS = 6  cm.

Therefore, area of the quadrilateral PQRS = Area of the ∆PQS + Area of the ∆QRS

                                                             = \(\frac{1}{2}\) PQ × QS + \(\frac{1}{2}\) RT × QS

                                                             = \(\frac{1}{2}\)(PQ + RT) × QS

                                                             = \(\frac{1}{2}\)(4.5 + 1.5) × 6 cm\(^{2}\)

                                                             = \(\frac{1}{2}\) × 6 × 6 cm\(^{2}\)

                                                             = \(\frac{1}{2}\) × 36 cm\(^{2}\)

                                                             = 18 cm\(^{2}\).

2. PQRS is a quadrilateral in which PQ = 4 cm, QC = 5 cm, RS = 7 cm, SP = 6 cm and the diagonal PR = 8 cm. Find its area.

Solution:

Area of the quadrilateral PQRS = Area of the ∆PQR + Area of the ∆SPR

In the ∆PQR, let a = PQ = 4 cm, b = QR = 5 cm and c = RP = 8 cm.

Therefore, s = \(\frac{1}{2}\)(a + b + c)

= \(\frac{1}{2}\)(4 + 5 + 8) cm

= \(\frac{17}{2}\) cm.

Area of the ∆PQR = \(\sqrt{s(s - a)(s - b)(s - c)}\)

                          = \(\sqrt{\frac{17}{2}(\frac{17}{2} - 4)(\frac{17}{2} - 5)(\frac{17}{2} - 8)}\) cm\(^{2}\)

                         = \(\sqrt{\frac{17}{2} ∙ \frac{9}{2} ∙ \frac{7}{2} ∙ \frac{1}{2}}\) cm\(^{2}\)

                         = \(\sqrt{\frac{17 ∙ 9 ∙ 7 ∙ 1}{16}}\) cm\(^{2}\)

                         = \(\frac{3}{4}\sqrt{119}\) cm\(^{2}\).

In the ∆SPR, let a = PS = 6 cm, b = RS = 7 cm and c = RP = 8 cm.

Therefore, S = \(\frac{1}{2}\)(a + b + c)

= \(\frac{1}{2}\)(6 + 7 + 8) cm

= = \(\frac{21}{2}\) cm.

Area of the ∆SPR = \(\sqrt{S(S - a)(S - b)(S - c)}\)

                          = \(\sqrt{\frac{21}{2}(\frac{21}{2} - 6)(\frac{21}{2} - 7)(\frac{21}{2} - 8)}\) cm\(^{2}\)

                          = \(\sqrt{\frac{21}{2} ∙ \frac{9}{2} ∙ \frac{7}{2} ∙ \frac{5}{2}}\) cm\(^{2}\)

                          = \(\sqrt{\frac{21 ∙ 9 ∙ 7 ∙ 5}{16}}\) cm\(^{2}\)

                          = \(\frac{3}{4}\sqrt{735}\) cm\(^{2}\).

Therefore, area of the quadrilateral PQRS = (\(\frac{3}{4}\sqrt{119}\) + \(\frac{3}{4}\sqrt{735}\)) cm\(^{2}\).

                                                             = \(\frac{3}{4}\)(10.9 + 27.1) cm\(^{2}\)

                                                             = \(\frac{3}{4}\) × 38 cm\(^{2}\)

                                                             = \(\frac{57}{2}\) cm\(^{2}\)

                                                             = 28.5 cm\(^{2}\)

9th Grade Math

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