In Delta ABC,mangle B = 90^{circ},AB = 2x + 3 | vedclass

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

Question

(A) In $\Delta ABC$,since $m\angle B = 90^{\circ}$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$(3x - 1)^2 = (2x + 3)^2 + (x + 2)^2$$9x^2 - 6x + 1

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In $\Delta ABC$,since $m\angle B = 90^{\circ}$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$(3x - 1)^2 = (2x + 3)^2 + (x + 2)^2$$9x^2 - 6x + 1 = (4x^2 + 12x + 9) + (x^2 + 4x + 4)$$9x^2 - 6x + 1 = 5x^2 + 16x + 13$$4x^2 - 22x - 12 = 0$Dividing by $2$:$2x^2 - 11x - 6 = 0$$2x^2 - 12x + x - 6 = 0$$2x(x - 6) + 1(x - 6) = 0$$(x - 6)(2x + 1) = 0$Thus,$x = 6$ or $x = -\frac{1}{2}$.Since side lengths must be positive,$x = -\frac{1}{2}$ is rejected because it results in negative side lengths.Therefore,$x = 6$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$,$AB = 2x + 3$,$BC = x + 2$,and $AC = 3x - 1$. Find the value of $x$.

Similar Questions

In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$SY = 4.5$ and $XT = 8$,find $XZ$.

In rectangle $ABCD$,$AB + BC = 23$ and its area is $120$. If $AB > BC$,then $AB = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$,and $AM > CM$. If $BM = 12$ and $AC = 25$,find $AB$ and $BC$.

In $\square XYZW$,$\overline{XY} \parallel \overline{ZW}$ and $\overline{XZ} \cap \overline{YW} = \{P\}$. If $PX = 3x - 1$,$PY = 2x + 1$,$PZ = 5x - 3$,and $PW = 6x - 5$,find the value of $x$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2$,$MB = 5$ and $MN = 4$,find $BC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module