Suppose ABCD is a trapezium whose sides and h | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the parallel sides be $AB = x$ and $CD = y$,and the height be $h$. The area of the trapezium is given by $\frac{1}{2} 	imes h(x + y) = 12$,wh

Vedclass · Accepted Answer

is $4$

Vedclass · Answer

(B) Let the parallel sides be $AB = x$ and $CD = y$,and the height be $h$. The area of the trapezium is given by $\frac{1}{2} 	imes h(x + y) = 12$,which implies $h(x + y) = 24$.Since $h, x, y$ are integers,$h$ must be a factor of $24$. Also,for a trapezium,the slant side must be greater than the height $h$. Let the non-parallel sides be $s_1$ and $s_2$. By dropping perpendiculars from $C$ and $D$ to $AB$,we have $x = y + p + q$,where $p^2 + h^2 = s_1^2$ and $q^2 + h^2 = s_2^2$.For $h=3$,$x+y=8$. If we assume a right-angled trapezium where one slant side is the height,let $s_1 = h = 3$. Then $s_2 = \sqrt{(x-y)^2 + h^2}$. For $s_2$ to be an integer,$(x-y, h, s_2)$ must be a Pythagorean triple. With $h=3$,the triple is $(4, 3, 5)$,so $x-y=4$. Solving $x+y=8$ and $x-y=4$ gives $2x=12 \Rightarrow x=6$ and $2y=4 \Rightarrow y=2$. The sides are $6, 3, 2, 5$,which are all integers and distinct. Thus,$|AB-CD| = |6-2| = 4$.

Suppose $ABCD$ is a trapezium whose sides and height are integers and $AB$ is parallel to $CD$. If the area of $ABCD$ is $12$ and the sides are distinct,then $|AB-CD|$ is:

Similar Questions

The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

$L \equiv 7x - y + 8 = 0$ is one of the diagonals of a square for which $(-4, 5)$ and $(3, 4)$ are two vertices. Find the coordinates of the two vertices lying on the diagonal $L = 0$.

The circumcentre of the triangle formed by the lines $xy+2x+2y+4=0$ and $x+y+2=0$ is

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

If the orthocenter of the triangle formed by the lines $2x + 3y - 1 = 0$,$x + 2y + 1 = 0$,and $ax + by - 1 = 0$ lies at the origin,then $\frac{1}{a} + \frac{1}{b} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module