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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