In the given figure,AB = 12 , cm,CD = 8 , cm, | vedclass

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

Question

(A) Let $\angle BAE = 	heta$. Since $	riangle ABE$ is a right-angled triangle at $B$,$\angle AEB = 90^{\circ} - 	heta$.Given $\angle AEC = 90^{\cir

Vedclass · Accepted Answer

$x$ has two possible values whose difference is $4$.

Vedclass · Answer

(A) Let $\angle BAE = 	heta$. Since $	riangle ABE$ is a right-angled triangle at $B$,$\angle AEB = 90^{\circ} - 	heta$.Given $\angle AEC = 90^{\circ}$,we have $\angle CED = 180^{\circ} - (90^{\circ} - 	heta) - 90^{\circ} = 	heta$.In $	riangle CDE$,$\angle CED = 	heta$ and $\angle CDE = 90^{\circ}$,so $\angle ECD = 90^{\circ} - 	heta$.Thus,$	riangle ABE \sim 	riangle ECD$ by $AA$ similarity.Therefore,$\frac{AB}{BE} = \frac{ED}{CD}$.Given $AB = 12$,$CD = 8$,$BD = 20$,and $BE = x$,we have $ED = 20 - x$.Substituting these values: $\frac{12}{x} = \frac{20 - x}{8}$.$96 = 20x - x^2$.$x^2 - 20x + 96 = 0$.$(x - 12)(x - 8) = 0$.So,$x = 8$ or $x = 12$.The difference between the two values is $|12 - 8| = 4$.

In the given figure,$AB = 12 \, cm$,$CD = 8 \, cm$,$BD = 20 \, cm$,and $\angle ABD = \angle AEC = \angle EDC = 90^{\circ}$. If $BE = x$,then:

Similar Questions

Straight lines $x-y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,-7)$ is

If the equations of three sides of a triangle are $x = 2$,$y + 1 = 0$,and $x + 2y = 4$,then the coordinates of the circumcentre of this triangle are:

$A$ straight line $4x+y-1=0$ passing through the point $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y+1=0$ at the point $B$. Then the equation of the line $AC$ such that $AB=AC$,is

In a triangle $ABC,$ side $AB$ has the equation $2x + 3y = 29$ and the side $AC$ has the equation $x + 2y = 16.$ If the mid-point of $BC$ is $(5, 6),$ then the equation of $BC$ is:

The area of a triangle is $5$ and two of its vertices are $A(2, 1)$ and $B(3, -2)$. The third vertex,which lies on the line $y = x + 3$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module