Question
In a rectangle $\text{ABCD}, AB = 7 \ cm$ and $AD = 25 \ cm.$ Find the height of a triangle whose base is $AB$ and whose area is two times the area of the rectangle $\text{ABCD}.$
✓
Answer
Area of rectangle $\text{ABCD}=A B \times A D=7 \times 25=175 \ cm ^2$
Area of triangle whose base is $A B=\frac{1}{2} \times AB \times$ Height $=\frac{1}{2} \times 7 \times$ Height
Now,
Area of triangle whose base is $A B=2 \times$ Area of rectangle $\text{ABCD}$
$=\frac{1}{2} \times 7 \times$ Height $=175$
$\Rightarrow$ Height
$=\frac{175 \times 2}{7}$
$=50 \ cm .$
Area of triangle whose base is $A B=\frac{1}{2} \times AB \times$ Height $=\frac{1}{2} \times 7 \times$ Height
Now,
Area of triangle whose base is $A B=2 \times$ Area of rectangle $\text{ABCD}$
$=\frac{1}{2} \times 7 \times$ Height $=175$
$\Rightarrow$ Height
$=\frac{175 \times 2}{7}$
$=50 \ cm .$
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