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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