In rectangle ABCD,AC = 29 and AB + BC = 41. F | vedclass

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

Question

(A) Let $AB = x$ and $BC = y$. In rectangle $ABCD$,$	riangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$. By the Pythagorean theorem,$

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) Let $AB = x$ and $BC = y$. In rectangle $ABCD$,$	riangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$. By the Pythagorean theorem,$AB^2 + BC^2 = AC^2$. Given $x^2 + y^2 = 29^2 = 841$. Also given $x + y = 41$. We know that $(x + y)^2 = x^2 + y^2 + 2xy$. Substituting the values: $41^2 = 841 + 2xy$. $1681 = 841 + 2xy$. $2xy = 1681 - 841 = 840$. $xy = 420$. Now,we have $x + y = 41$ and $xy = 420$. These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$. $t^2 - 41t + 420 = 0$. Factoring the equation: $(t - 21)(t - 20) = 0$. So,$t = 21$ or $t = 20$. Since $AB > BC$,we have $AB = 21$ and $BC = 20$. Thus,$AB = 21$.

In rectangle $ABCD$,$AC = 29$ and $AB + BC = 41$. Find $AB$ (given $AB > BC$).

Similar Questions

In $\Delta XYZ$,$m\angle X = 90^{\circ}$,$XY = 8$ and $YZ = 17$. Then,the area of $\Delta XYZ$ is.............

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. If $BM = \sqrt{30}$ and $CM = 3$,find $AC$.

In $\Delta PQR$,$PQ = 25$ and $QR = 24$. If $PR = \dots$,then $\angle R$ is a right angle.

In $\Delta ABC$,$m \angle A + m \angle C = m \angle B$ and $\overline{BM}$ is an altitude to $\overline{AC}$. If $AM = 16$ and $CM = 9$,find $BM$,$AB$,and $BC$.

If in two right triangles,one of the acute angles of one triangle is equal to an acute angle of the other triangle,can you say that the two triangles will be similar? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module