Get the area of a quadrilateral?
$\angle A_{1}+\angle C_{3}=30^{\circ}$
$\angle A_{2}+\angle C_{4}=90^{\circ}$
$CD=9, DA=5, BC=8 , AB=4$
2026-05-15 07:22:31.1778829751
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area of a quadrilateral
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This can follow directly from Bretschneider's formula: $$K = \sqrt {(s-a)(s-b)(s-c)(s-d) - abcd \cdot \cos^2 \left(\frac{A+C}{2}\right)}$$
You have $$ \angle A_{1}+\angle C_{3}=30^{\circ}\tag1 $$ and $$ \angle A_{2}+\angle C_{4}=90^{\circ}\tag2 $$ Adding $(1)$ and $(2)$, you will get $$ (\angle A_{1}+\angle A_{2})+(\angle C_{3}+\angle C_{4})=\angle A+\angle C=120^{\circ}.\tag3 $$ Now, divide the quadrilateral into two triangles, $\Delta ABD$ and $\Delta BCD$. Using cosine formula, determine $BD$. From $\Delta ABD$ you will get $$ \begin{align} BD^2&=AB^2+BD^2-2AB\cdot BD\cdot\cos\angle A\\ &=4^2+5^2-2\cdot4\cdot5\cdot\cos\angle A\\ &=41-20\cos\angle A.\tag4 \end{align} $$ From $\Delta BCD$ you will get $$ \begin{align} BD^2&=BC^2+CD^2-2BC\cdot CD\cdot\cos\angle C\\ &=8^2+9^2-2\cdot8\cdot9\cdot\cos(120^\circ-\angle A)\\ &=145-144\cos(120^\circ-\angle A).\tag5 \end{align} $$ Using $(4)$ and $(5)$, you will get $$ 41-20\cos\angle A=145-144\cos(120^\circ-\angle A).\tag6 $$ Solve for $\angle A$ from $(6)$ and use $\angle A$ to obtain $\angle C$ using $(3)$. Thus, you can obtain the area of quadrilateral $ABCD$ by adding area of $\Delta ABD$ and $\Delta BCD$. $$ \begin{align} [ABCD]&=[ABD]+[BCD]\\ &=\frac{1}{2}AB\cdot BD\cdot\sin\angle A+\frac{1}{2}BC\cdot CD\cdot\sin\angle C. \end{align} $$
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