SHOWN in the diagram, area of triangle ABC \[=1/2\]area of the parallelogram ABCD. By Heron's formula, area of triangle \[=\sqrt{S\LEFT( S-a \right)\left( S-b \right)\left( S-c \right)}\] Here \[S=\frac{9+8+13}{2}=15\] \[\therefore \] Area of ABC\[=\sqrt{15(6)(2)(7)}\] \[\sqrt{6\times 7\times 30}=6\sqrt{35}\,\text{sq}\text{.}\,CM.\] Hence, area of ABCD \[=2\times 6\sqrt{35}=12\sqrt{35}\,\text{sq}\text{.m}\]

"> SHOWN in the diagram, area of triangle ABC \[=1/2\]area of the parallelogram ABCD. By Heron's formula, area of triangle \[=\sqrt{S\LEFT( S-a \right)\left( S-b \right)\left( S-c \right)}\] Here \[S=\frac{9+8+13}{2}=15\] \[\therefore \] Area of ABC\[=\sqrt{15(6)(2)(7)}\] \[\sqrt{6\times 7\times 30}=6\sqrt{35}\,\text{sq}\text{.}\,CM.\] Hence, area of ABCD \[=2\times 6\sqrt{35}=12\sqrt{35}\,\text{sq}\text{.m}\]

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The adjacent sides of a parallelogram are 8 cm and 9 cm. The diagonal joining the ends of these sides is 13 cm. Its area is

7th Class Mathematics in 7th Class 10 months ago

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As SHOWN in the diagram, area of triangle ABC \[=1/2\]area of the parallelogram ABCD. By Heron's formula, area of triangle \[=\sqrt{S\LEFT( S-a \right)\left( S-b \right)\left( S-c \right)}\] Here \[S=\frac{9+8+13}{2}=15\] \[\therefore \] Area of ABC\[=\sqrt{15(6)(2)(7)}\] \[\sqrt{6\times 7\times 30}=6\sqrt{35}\,\text{sq}\text{.}\,CM.\] Hence, area of ABCD \[=2\times 6\sqrt{35}=12\sqrt{35}\,\text{sq}\text{.m}\]

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