
ABCD - параллелограмм. ВН перпендикулярна AD, KD перпендикулярна BC. BK=BH. Угол А равен 45
градусов. Площадь BHDK равна 36. Чему равна площадь ABCD?

Ответы на вопрос

Ответ:
угол ВСD = углу ВАD = 45
угол DKC =90
значит угол KDC =180-90-45=45
т к угол BAD=KDC то треугольник равносторонний и КС=KD=6
S=6*(5+6)=66






I will answer your question in detail. ABCD is a parallelogram. VN is perpendicular to AD, KD is perpendicular to BC. BK=BH. Angle A is 45 degrees. The area of BHDK is 36. What is the area of ABCD?
To find the area of ABCD, we can use the following formula: $$\mathrm {S}=ab \sin \alpha$$ where $a$ and $b$ are the lengths of the adjacent sides and $\alpha$ is the angle between them.
We know that angle A is 45 degrees, so $\alpha = 45^\circ$. We also know that the area of BHDK is 36, so $$36 = \frac{1}{2}bh$$ where $b$ and $h$ are the base and height of the triangle. Since VN is perpendicular to AD, we have $h = a \sin \alpha$. Substituting this into the equation, we get $$36 = \frac{1}{2}b(a \sin \alpha)$$
We can simplify this by dividing both sides by $\sin \alpha$, which is $\frac{\sqrt{2}}{2}$ for 45 degrees. We get $$\frac{36}{\sin \alpha} = ba$$
Now we have an expression for the area of ABCD in terms of $a$ and $b$. To find the values of these sides, we can use the fact that BK=BH and KD is perpendicular to BC.
Let $x$ be the length of BK and $y$ be the length of KD. Then we have $$x = a - y$$ and $$y = b - x$$
Solving for $x$ and $y$, we get $$x = \frac{a + b}{2}$$ and $$y = \frac{a + b}{2}$$
Substituting these into the expression for the area of ABCD, we get $$\mathrm {S}=\frac{(a + b)^2}{4}$$
To find the value of $a + b$, we can use the Pythagorean theorem on triangle BHK. We have $$BH^2 + HK^2 = BK^2$$
Since BH and HK are both equal to $\frac{a + b}{2}$, we get $$(\frac{a + b}{2})^2 + (\frac{a + b}{2})^2 = x^2$$
Simplifying this, we get $$\frac{(a + b)^2}{2} = x^2$$
Taking the square root of both sides, we get $$\sqrt{2}(a + b) = 2x$$
Dividing by $\sqrt{2}$, we get $$a + b = \frac{2x}{\sqrt{2}}$$
Since x is the length of BK, we can use the area of BHDK to find its value. We have $$36 = \frac{1}{2}xy$$
Substituting $y = x$, we get $$36 = \frac{1}{2}x^2$$
Solving for $x$, we get $$x = 6\sqrt{2}$$
Therefore, $$a + b = \frac{12\sqrt{2}}{\sqrt{2}} = 12$$
Finally, substituting this into the expression for the area of ABCD, we get $$\mathrm {S}=\frac{(12)^2}{4} = 36$$
The area of ABCD is 36 square units.
I hope this answer helps you understand how to solve this problem.


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