Solve by substitution y=-2x y=8x-10
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Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to
sq. units.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to